2 Vectors in Space

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-vectors-in-space

（该页为章节总览/导航页，无独立正文；本章内容请见其下各小节。）

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Introduction

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-introduction

Chapter Outline

• 2.1 Vectors in the Plane
• 2.2 Vectors in Three Dimensions
• 2.3 The Dot Product
• 2.4 The Cross Product
• 2.5 Equations of Lines and Planes in Space
• 2.6 Quadric Surfaces
• 2.7 Cylindrical and Spherical Coordinates

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2.1 Vectors in the Plane

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-1-vectors-in-the-plane

2.1 Vectors in the Plane

• 2.1.1 Describe a plane vector, using correct notation.
• 2.1.2 Perform basic vector operations (scalar multiplication, addition, subtraction).
• 2.1.3 Express a vector in component form.
• 2.1.4 Explain the formula for the magnitude of a vector.
• 2.1.5 Express a vector in terms of unit vectors.
• 2.1.6 Give two examples of vector quantities.

When describing the movement of an airplane in flight, it is important to communicate two pieces of information: the direction in which the plane is traveling and the plane’s speed. When measuring a force, such as the thrust of the plane’s engines, it is important to describe not only the strength of that force, but also the direction in which it is applied. Some quantities, such as velocity or force, are defined in terms of both size (also called *magnitude*) and direction. A quantity that has magnitude and direction is called a vector. In this text, we denote vectors by boldface letters, such as v.

A vector is a quantity that has both magnitude and direction.

Vector Representation

A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initial point and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction of the vector. The length of the line segment represents its magnitude. We use the notation $\left\| \mathbf{\text{v}} \right\|$ to denote the magnitude of the vector $\mathbf{\text{v}}.$ A vector with an initial point and terminal point that are the same is called the zero vector, denoted $\mathbf{0}.$ The zero vector is the only vector without a direction, and by convention can be considered to have any direction convenient to the problem at hand.

Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even if they have different initial points. Thus, if $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ are equivalent, we write

$$\mathbf{\text{v}} = \mathbf{\text{w}}.$$

Vectors are said to be equivalent vectors if they have the same magnitude and direction.

The arrows in Figure 2.2(b) are equivalent. Each arrow has the same length and direction. A closely related concept is the idea of parallel vectors. Two vectors are said to be parallel if they have the same or opposite directions. We explore this idea in more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point is located.

The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternative notations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to simulate boldface type: $\overset{\rightarrow}{v}.$ When a vector has initial point $P$ and terminal point $Q,$ the notation $\overset{\rightarrow}{PQ}$ is useful because it indicates the direction and location of the vector.

Sketching Vectors

Sketch a vector in the plane from initial point $P(1,1)$ to terminal point $Q(8,5).$

Solution

See Figure 2.3. Because the vector goes from point $P$ to point $Q,$ we name it $\overset{\rightarrow}{PQ}.$

Sketch the vector $\overset{\rightarrow}{ST}$ where $S$ is point $\left( {3,-1} \right)$ and $T$ is point $\left( {-2,3} \right).$

Combining Vectors

Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates a force in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into account if we want to know where the boat will go.

A second example that involves vectors is a quarterback throwing a football. The quarterback does not throw the ball parallel to the ground; instead, he aims up into the air. The velocity of his throw can be represented by a vector. If we know how hard he throws the ball (magnitude—in this case, speed), and the angle (direction), we can tell how far the ball will travel down the field.

A real number is often called a scalar in mathematics and physics. Unlike vectors, scalars are generally considered to have a magnitude only, but no direction. Multiplying a vector by a scalar changes the vector’s magnitude. This is called scalar multiplication. Note that changing the magnitude of a vector does not indicate a change in its direction. For example, wind blowing from north to south might increase or decrease in speed while maintaining its direction from north to south.

The product $k\mathbf{\text{v}}$ of a vector v and a scalar *k* is a vector with a magnitude that is $|k|$ times the magnitude of $\mathbf{\text{v}},$ and with a direction that is the same as the direction of $\mathbf{\text{v}}$ if $k > 0,$ and opposite the direction of $\mathbf{\text{v}}$ if $k < 0.$ This is called scalar multiplication. If $k = 0$ or $\mathbf{v = 0},$ then $k\mathbf{\text{v}} = \mathbf{0}.$

As you might expect, if $k = -1,$ we denote the product $k\mathbf{\text{v}}$ as

$$k\mathbf{\text{v}} = (-1)\mathbf{\text{v}} = \text{−}\mathbf{\text{v}}.$$

Note that $\text{−}\mathbf{\text{v}}$ has the same magnitude as $\mathbf{\text{v}},$ but has the opposite direction (Figure 2.4).

Another operation we can perform on vectors is to add them together in vector addition, but because each vector may have its own direction, the process is different from adding two numbers. The most common graphical method for adding two vectors is to place the initial point of the second vector at the terminal point of the first, as in Figure 2.5(a). To see why this makes sense, suppose, for example, that both vectors represent displacement. If an object moves first from the initial point to the terminal point of vector $\mathbf{\text{v}},$ then from the initial point to the terminal point of vector $\mathbf{\text{w}},$ the overall displacement is the same as if the object had made just one movement from the initial point to the terminal point of the vector $\mathbf{v + w}.$ For obvious reasons, this approach is called the triangle method. Notice that if we had switched the order, so that $\mathbf{\text{w}}$ was our first vector and v was our second vector, we would have ended up in the same place. (Again, see Figure 2.5(a).) Thus, $\mathbf{v + w = w + v}.$

A second method for adding vectors is called the parallelogram method. With this method, we place the two vectors so they have the same initial point, and then we draw a parallelogram with the vectors as two adjacent sides, as in Figure 2.5(b). The length of the diagonal of the parallelogram is the sum. Comparing Figure 2.5(b) and Figure 2.5(a), we can see that we get the same answer using either method. The vector $\mathbf{v + w}$ is called the vector sum.

The sum of two vectors $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ can be constructed graphically by placing the initial point of $\mathbf{\text{w}}$ at the terminal point of $\mathbf{\text{v}}.$ Then, the vector sum, $\mathbf{v + w},$ is the vector with an initial point that coincides with the initial point of $\mathbf{\text{v}}$ and has a terminal point that coincides with the terminal point of $\mathbf{\text{w}}.$ This operation is known as vector addition.

It is also appropriate here to discuss vector subtraction. We define $\mathbf{\text{v}} - \mathbf{\text{w}}$ as $\mathbf{v +}\left( {\text{−}\mathbf{\text{w}}} \right) = \mathbf{v +}(-1)\mathbf{\text{w}}.$ The vector $\mathbf{\text{v}} - \mathbf{\text{w}}$ is called the vector difference. Graphically, the vector $\mathbf{\text{v}} - \mathbf{\text{w}}$ is depicted by drawing a vector from the terminal point of $\mathbf{\text{w}}$ to the terminal point of $\mathbf{\text{v}}$ (Figure 2.6).

In Figure 2.5(a), the initial point of $\mathbf{\text{v}} + \mathbf{\text{w}}$ is the initial point of $\mathbf{\text{v}}.$ The terminal point of $\mathbf{\text{v}} + \mathbf{\text{w}}$ is the terminal point of $\mathbf{\text{w}}.$ These three vectors form the sides of a triangle. It follows that the length of any one side is less than the sum of the lengths of the remaining sides. So we have

$$\left\| {\mathbf{\text{v}} + \mathbf{\text{w}}} \right\| \leq \left\| \mathbf{\text{v}} \right\| + \left\| \mathbf{\text{w}} \right\|.$$

This is known more generally as the triangle inequality. There is one case, however, when the resultant vector $\mathbf{\text{u}} + \mathbf{\text{v}}$ has the same magnitude as the sum of the magnitudes of $\mathbf{\text{u}}$ and $\mathbf{\text{v}}.$ This happens only when $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have the same direction.

Combining Vectors

Given the vectors $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ shown in Figure 2.7, sketch the vectors

1. $3\textbf{w}$

2. $\mathbf{\text{v}} + \mathbf{\text{w}}$

3. $2\mathbf{\text{v}} - \mathbf{\text{w}}$

Solution

1. The vector $3\mathbf{\text{w}}$ has the same direction as $\mathbf{\text{w}};$ it is three times as long as $\mathbf{\text{w}}.$

Vector $3\mathbf{\text{w}}$ has the same direction as $\mathbf{\text{w}}$ and is three times as long.

2. Use either addition method to find $\mathbf{\text{v}} + \mathbf{\text{w}}.$

3. To find $2\mathbf{\text{v}} - \mathbf{\text{w}},$ we can first rewrite the expression as $2\mathbf{\text{v}} + \left( {\text{−}\mathbf{\text{w}}} \right).$ Then we can draw the vector $\text{−}\mathbf{\text{w}},$ then add it to the vector $2\mathbf{\text{v}}.$

Using vectors $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ from Example 2.2, sketch the vector $2\mathbf{\text{w}} - \mathbf{\text{v}}.$

Vector Components

Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.

Comparing Vectors

Are $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ equivalent vectors?

1. $\mathbf{\text{v}}$ has initial point $\left( {3,2} \right)$ and terminal point $\left( {7,2} \right)$

$\mathbf{\text{w}}$ has initial point $\left( {1,-4} \right)$ and terminal point $\left( {1,0} \right)$

2. $\mathbf{\text{v}}$ has initial point $\left( {0,0} \right)$ and terminal point $\left( {1,1} \right)$

$\mathbf{\text{w}}$ has initial point $\left( {-2,2} \right)$ and terminal point $\left( {-1,3} \right)$

Solution

1. The vectors are each $4$ units long, but they are oriented in different directions. So $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ are not equivalent (Figure 2.10).

2. Based on Figure 2.11, and using a bit of geometry, it is clear these vectors have the same length and the same direction, so $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$ are equivalent.

Which of the following vectors are equivalent?

We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of any vector in standard position is known to be $\left( {0,0} \right),$ we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector v has its initial point at the origin and its terminal point at $\left( {x,y} \right),$ we write the vector in component form as

$$\mathbf{\text{v}} = \left\langle {x,y} \right\rangle.$$

When a vector is written in component form like this, the scalars *x* and *y* are called the components of $\mathbf{\text{v}}.$

The vector with initial point $\left( {0,0} \right)$ and terminal point $\left( {x,y} \right)$ can be written in component form as

$$\mathbf{\text{v}} = \left\langle {x,y} \right\rangle.$$

The scalars $x$ and $y$ are called the components of $\mathbf{\text{v}}.$

Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between $\left\langle {x,y} \right\rangle$ and $\left( {x,y} \right).$ The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of $\left\langle {x,y} \right\rangle$ is $\left( {0,0} \right);$ the terminal point of $\left\langle {x,y} \right\rangle$ is $\left( {x,y} \right).$

When we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the *x*-coordinate of the initial point from the *x*-coordinate of the terminal point to get the *x* component, and we subtract the *y*-coordinate of the initial point from the *y*-coordinate of the terminal point to get the *y* component.

Let v be a vector with initial point $\left( {x_{i},y_{i}} \right)$ and terminal point $\left( {x_{t},y_{t}} \right).$ Then we can express v in component form as $\mathbf{\text{v}} = \left\langle {x_{t} - x_{i},y_{t} - y_{i}} \right\rangle.$

Expressing Vectors in Component Form

Express vector $\mathbf{\text{v}}$ with initial point $\left( {-3,4} \right)$ and terminal point $\left( {1,2} \right)$ in component form.

Solution

1. Geometric

1. Sketch the vector in the coordinate plane (Figure 2.12).

2. The terminal point is 4 units to the right and 2 units down from the initial point.

3. Find the point that is 4 units to the right and 2 units down from the origin.

4. In standard position, this vector has initial point $\left( {0,0} \right)$ and terminal point $\left( {4,-2} \right)\text{:}$

$$\mathbf{\text{v}} = \left\langle {4,-2} \right\rangle.$$

2. Algebraic

In the first solution, we used a sketch of the vector to see that the terminal point lies 4 units to the right. We can accomplish this algebraically by finding the difference of the *x*-coordinates:

$$x_{t} - x_{i} = 1 - (-3) = 4.$$

Similarly, the difference of the *y*-coordinates shows the vertical length of the vector.

$$y_{t} - y_{i} = 2 - 4 = -2.$$

So, in component form,

$$\begin{array}{cl}

\mathbf{\text{v}} & {= \left\langle {x_{t} - x_{i},y_{t} - y_{i}} \right\rangle} \\

& {= \left\langle {1 - (-3),2 - 4} \right\rangle} \\

& {= \left\langle {4,-2} \right\rangle.}

\end{array}$$

Vector $\mathbf{\text{w}}$ has initial point $\left( {-4,-5} \right)$ and terminal point $\left( {-1,2} \right).$ Express $\mathbf{\text{w}}$ in component form.

To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector $\mathbf{\text{v}} = \left\langle {x,y} \right\rangle$ is denoted $\left\| \mathbf{\text{v}} \right\|,$ or $\left| \mathbf{\text{v}} \right|,$ and can be computed using the formula

$$\left\| \mathbf{\text{v}} \right\| = \sqrt{x^{2} + y^{2}}.$$

Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point $\left( {x,y} \right).$ Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point $\left( {0,0} \right)$ and terminal point $\left( {x,y} \right),$ we have

$$\begin{array}{cl}

\left\| \mathbf{\text{v}} \right\| & {= \sqrt{\left( {x - 0} \right)^{2} + \left( {y - 0} \right)^{2}}} \\

& {= \sqrt{x^{2} + y^{2}}.}

\end{array}$$

Based on this formula, it is clear that for any vector $\mathbf{\text{v}},$ $\left\| \mathbf{\text{v}} \right\| \geq 0,$ and $\left\| \mathbf{\text{v}} \right\| = 0$ if and only if $\mathbf{v = 0}.$

The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.

We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.

Let $\mathbf{\text{v}} = \left\langle {x_{1},y_{1}} \right\rangle$ and $\mathbf{\text{w}} = \left\langle {x_{2},y_{2}} \right\rangle$ be vectors, and let $k$ be a scalar.

Scalar multiplication:$k\mathbf{\text{v}} = \left\langle {kx_{1},ky_{1}} \right\rangle$

Vector addition:$\mathbf{\text{v}} + \mathbf{\text{w}} = \left\langle {x_{1},y_{1}} \right\rangle + \left\langle {x_{2},y_{2}} \right\rangle = \left\langle {x_{1} + x_{2},y_{1} + y_{2}} \right\rangle$

Performing Operations in Component Form

Let $\mathbf{\text{v}}$ be the vector with initial point $\left( {2,5} \right)$ and terminal point $\left( {8,13} \right),$ and let $\mathbf{\text{w}} = \left\langle {-2,4} \right\rangle.$

1. Express $\mathbf{\text{v}}$ in component form and find $\left\| \mathbf{\text{v}} \right\|.$ Then, using algebra, find

2. $\mathbf{\text{v}} + \mathbf{\text{w}},$

3. $3\mathbf{\text{v}},$ and

4. $\mathbf{\text{v}} - 2\mathbf{\text{w}}.$

Solution

1. To place the initial point of $\mathbf{\text{v}}$ at the origin, we must translate the vector $2$ units to the left and $5$ units down (Figure 2.15). Using the algebraic method, we can express $\mathbf{\text{v}}$ as $\mathbf{\text{v}} = \left\langle {8 - 2,13 - 5} \right\rangle = \left\langle {6,8} \right\rangle\text{:}$

$$\left\| \mathbf{\text{v}} \right\| = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10.$$

2. To find $\mathbf{\text{v}} + \mathbf{\text{w}},$ add the *x*-components and the *y*-components separately:

$$\mathbf{\text{v}} + \mathbf{\text{w}} = \left\langle {6,8} \right\rangle + \left\langle {-2,4} \right\rangle = \left\langle {4,12} \right\rangle.$$

3. To find $3\mathbf{\text{v}},$ multiply $\mathbf{\text{v}}$ by the scalar $k = 3\text{:}$

$$3\mathbf{\text{v}} = 3 \cdot \left\langle {6,8} \right\rangle = \left\langle {3 \cdot 6,3 \cdot 8} \right\rangle = \left\langle {18,24} \right\rangle.$$

4. To find $\mathbf{\text{v}} - 2\mathbf{\text{w}},$ find $-2\mathbf{\text{w}}$ and add it to $\mathbf{\text{v}}\text{:}$

$$\mathbf{\text{v}} - 2\mathbf{\text{w}} = \left\langle {6,8} \right\rangle - 2 \cdot \left\langle {-2,4} \right\rangle = \left\langle {6,8} \right\rangle + \left\langle {4,-8} \right\rangle = \left\langle {10,0} \right\rangle.$$

Let $\mathbf{\text{a}} = \left\langle {7,1} \right\rangle$ and let $\mathbf{\text{b}}$ be the vector with initial point $\left( {3,2} \right)$ and terminal point $\left( {-1,-1} \right).$

1. Find $\left\| \mathbf{\text{a}} \right\|.$

2. Express $\mathbf{\text{b}}$ in component form.

3. Find $3\mathbf{\text{a}} - 4\mathbf{\text{b}}.$

Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.

Properties of Vector Operations

Let $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ be vectors in a plane. Let $\text{r and s}$ be scalars.

$$\begin{matrix}

\text{i.} & & & {\textbf{u} + \textbf{v}} & = & {\textbf{v} + \textbf{u}} & & & \text{Commutative property} \\

\text{ii.} & & & {\left( \textbf{u} + \textbf{v} \right) + \textbf{w}} & = & {\textbf{u} + \left( \textbf{v} + \textbf{w} \right)} & & & \text{Associative property} \\

\text{iii.} & & & {\textbf{u} + 0} & = & \mathbf{\text{u}} & & & \text{Additive identity property} \\

\text{iv.} & & & {\textbf{u} + \left( \text{−}\textbf{u} \right)} & = & 0 & & & \text{Additive inverse property} \\

\text{v.} & & & {r\left( s\textbf{u} \right)} & = & {(rs)\textbf{u}} & & & \text{Associativity of scalar multiplication} \\

\text{vi.} & & & {(r + s)\textbf{u}} & = & {r\textbf{u} + s\textbf{u}} & & & \text{Distributive property} \\

\text{vii.} & & & {r\left( \textbf{u} + \textbf{v} \right)} & = & {r\textbf{u} + r\textbf{v}} & & & \text{Distributive property} \\

\text{viii.} & & & {1\textbf{u}} & = & {\textbf{u},0\textbf{u} = 0} & & & \text{Identity and zero properties}

\end{matrix}$$

Proof of Commutative Property

Let $\mathbf{\text{u}} = \left\langle {x_{1},y_{1}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {x_{2},y_{2}} \right\rangle.$ Apply the commutative property for real numbers:

$$\mathbf{\text{u}} + \mathbf{\text{v}} = \left\langle {x_{1} + x_{2},y_{1} + y_{2}} \right\rangle = \left\langle {x_{2} + x_{1},y_{2} + y_{1}} \right\rangle = \mathbf{\text{v}} + \mathbf{\text{u}}.$$

□

Proof of Distributive Property

Apply the distributive property for real numbers:

$$\begin{array}{cl}

{r\left( {\mathbf{\text{u}} + \mathbf{\text{v}}} \right)} & {= r \cdot \left\langle {x_{1} + x_{2},y_{1} + y_{2}} \right\rangle} \\

& {= \left\langle {r\left( {x_{1} + x_{2}} \right),r\left( {y_{1} + y_{2}} \right)} \right\rangle} \\

& {= \left\langle {rx_{1} + rx_{2},ry_{1} + ry_{2}} \right\rangle} \\

& {= \left\langle {rx_{1},ry_{1}} \right\rangle + \left\langle {rx_{2},ry_{2}} \right\rangle} \\

& {= r\mathbf{\text{u}} + r\mathbf{\text{v}}.}

\end{array}$$

□

Prove the additive inverse property.

We have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry (Figure 2.15).

Consider the angle $\theta$ formed by the vector v and the positive *x*-axis. We can see from the triangle that the components of vector $\mathbf{\text{v}}$ are $\left\langle {\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta,\left\| \mathbf{\text{v}} \right\|\text{sin}\ \theta} \right\rangle.$ Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.

Finding the Component Form of a Vector Using Trigonometry

Find the component form of a vector with magnitude 4 that forms an angle of $-45\text{°}$ with the *x*-axis.

Solution

Let $x$ and $y$ represent the components of the vector (Figure 2.16). Then $x = 4\ \text{cos}\left( {-45\text{°}} \right) = 2\sqrt{2}$ and $y = 4\ \text{sin}\left( {-45\text{°}} \right) = -2\sqrt{2}.$ The component form of the vector is $\left\langle {2\sqrt{2},-2\sqrt{2}} \right\rangle.$

Find the component form of vector $\mathbf{\text{v}}$ with magnitude $10$ that forms an angle of $120\text{°}$ with the positive *x*-axis.

Unit Vectors

A unit vector is a vector with magnitude $1.$ For any nonzero vector $\mathbf{\text{v}},$ we can use scalar multiplication to find a unit vector $\mathbf{\text{u}}$ that has the same direction as $\mathbf{\text{v}}.$ To do this, we multiply the vector by the reciprocal of its magnitude:

$$\mathbf{\text{u}} = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\mathbf{\text{v}}.$$

Recall that when we defined scalar multiplication, we noted that $\left\| {k\mathbf{\text{v}}} \right\| = |k| \cdot \left\| \mathbf{\text{v}} \right\|.$ For $\mathbf{\text{u}} = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\mathbf{\text{v}},$ it follows that $\left\| \mathbf{\text{u}} \right\| = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\left( \left\| \mathbf{\text{v}} \right\| \right) = 1.$ We say that $\mathbf{\text{u}}$ is the *unit vector in the direction of*$\mathbf{\text{v}}$ (Figure 2.17). The process of using scalar multiplication to find a unit vector with a given direction is called normalization.

Finding a Unit Vector

Let $\mathbf{\text{v}} = \left\langle {1,2} \right\rangle.$

1. Find a unit vector with the same direction as $\mathbf{\text{v}}.$

2. Find a vector $\mathbf{\text{w}}$ with the same direction as $\mathbf{\text{v}}$ such that $\left\| \mathbf{\text{w}} \right\| = 7.$

Solution

1. First, find the magnitude of $\mathbf{\text{v}},$ then divide the components of $\mathbf{\text{v}}$ by the magnitude:

$$\left\| \mathbf{\text{v}} \right\| = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5}$$

$$\mathbf{\text{u}} = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\mathbf{\text{v}} = \frac{1}{\sqrt{5}}\left\langle {1,2} \right\rangle = \left\langle {\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}} \right\rangle.$$

2. The vector $\mathbf{\text{u}}$ is in the same direction as $\mathbf{\text{v}}$ and $\left\| \mathbf{\text{u}} \right\| = 1.$ Use scalar multiplication to increase the length of $\mathbf{\text{u}}$ without changing direction:

$$\mathbf{\text{w}} = 7\mathbf{\text{u}} = 7\left\langle {\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}} \right\rangle = \left\langle {\frac{7}{\sqrt{5}},\frac{14}{\sqrt{5}}} \right\rangle.$$

Let $\mathbf{\text{v}} = \left\langle {9,2} \right\rangle.$ Find a vector with magnitude $5$ in the opposite direction as $\mathbf{\text{v}}.$

We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let’s look at standard unit vectors. The standard unit vectors are the vectors $\mathbf{\text{i}} = \left\langle {1,0} \right\rangle$ and $\mathbf{\text{j}} = \left\langle {0,1} \right\rangle$ (Figure 2.18).

By applying the properties of vectors, it is possible to express any vector in terms of $\mathbf{\text{i}}$ and $\mathbf{\text{j}}$ in what we call a *linear combination*:

$$\mathbf{\text{v}} = \left\langle {x,y} \right\rangle = \left\langle {x,0} \right\rangle + \left\langle {0,y} \right\rangle = x\left\langle {1,0} \right\rangle + y\left\langle {0,1} \right\rangle = x\mathbf{\text{i}} + y\mathbf{\text{j}}.$$

Thus, $\mathbf{\text{v}}$ is the sum of a horizontal vector with magnitude $x,$ and a vertical vector with magnitude $y,$ as in the following figure.

Using Standard Unit Vectors

1. Express the vector $\mathbf{\text{w}} = \left\langle {3,-4} \right\rangle$ in terms of standard unit vectors.

2. Vector $\mathbf{\text{u}}$ is a unit vector that forms an angle of $60\text{°}$ with the positive *x*-axis. Use standard unit vectors to describe $\mathbf{\text{u}}.$

Solution

1. Resolve vector $\mathbf{\text{w}}$ into a vector with a zero *y*-component and a vector with a zero *x*-component:

$$\mathbf{\text{w}} = \left\langle {3,-4} \right\rangle = 3\mathbf{\text{i}} - 4\mathbf{\text{j}}.$$

2. Because $\mathbf{\text{u}}$ is a unit vector, the terminal point lies on the unit circle when the vector is placed in standard position (Figure 2.20).

$$\begin{array}{cl}

u & {= \left\langle {\text{cos}\ 60\text{°},\text{sin}\ 60\text{°}} \right\rangle} \\

& {= \left\langle {\frac{1}{2},\frac{\sqrt{3}}{2}} \right\rangle} \\

& {= \frac{1}{2}\mathbf{\text{i}} + \frac{\sqrt{3}}{2}\mathbf{\text{j}}.}

\end{array}$$

Let $\mathbf{\text{a}} = \left\langle {16,-11} \right\rangle$ and let $\mathbf{\text{b}}$ be a unit vector that forms an angle of $225\text{°}$ with the positive *x*-axis. Express $\mathbf{\text{a}}$ and $\mathbf{\text{b}}$ in terms of the standard unit vectors.

Applications of Vectors

Because vectors have both direction and magnitude, they are valuable tools for solving problems involving such applications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we look at two other examples in detail.

Finding Resultant Force

Jane’s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck’s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of $300$ lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of $150$ lb on the car. These forces can be represented by vectors, as shown in Figure 2.21. The angle between these vectors is $15\text{°}.$ Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive *x*-axis.

Solution

To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive *x*-axis. The horizontal vector, then, has initial point $\left( {0,0} \right)$ and terminal point $\left( {300,0} \right).$ It can be expressed as $\left\langle {300,0} \right\rangle$ or $300\mathbf{\text{i}}.$

The second vector has magnitude $150$ and makes an angle of $15\text{°}$ with the first, so we can express it as $\left\langle {150\ \text{cos}\left( {15\text{°}} \right),150\ \text{sin}\left( {15\text{°}} \right)} \right\rangle,$ or $150\ \text{cos}\left( {15\text{°}} \right)\mathbf{\text{i}} + 150\ \text{sin}\left( {15\text{°}} \right)\mathbf{\text{j}}.$ Then, the sum of the vectors, or resultant vector, is $\mathbf{\text{r}} = \left\langle {300,0} \right\rangle + \left\langle {150\ \text{cos}\left( {15\text{°}} \right),150\ \text{sin}\left( {15\text{°}} \right)} \right\rangle,$ and we have

$$\begin{array}{cl}

\left\| \mathbf{\text{r}} \right\| & {= \sqrt{\left( {300 + 150\ \text{cos}\left( {15\text{°}} \right)} \right)^{2} + \left( {150\ \text{sin}\left( {15\text{°}} \right)} \right)^{2}}} \\

& {\approx 446.6.}

\end{array}$$

The angle $\theta$ made by $\mathbf{\text{r}}$ and the positive *x*-axis has $\text{tan}\ \theta = \frac{150\ \text{sin}\ 15\text{°}}{\left( {300 + 150\ \text{cos}\ 15\text{°}} \right)} \approx 0.09,$ so $\theta \approx tan^{-1}(0.09) \approx 5\text{°},$ which means the resultant force $\mathbf{\text{r}}$ has an angle of $5\text{°}$ above the horizontal axis.

Finding Resultant Velocity

An airplane flies due west at an airspeed of $425$ mph. The wind is blowing from the northeast at $40$ mph. What is the ground speed of the airplane? What is the bearing of the airplane?

Solution

Let’s start by sketching the situation described (Figure 2.22).

Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane’s velocity vector is $\mathbf{\text{p}} = -425\mathbf{\text{i}}.$ The vector describing the wind makes an angle of $225\text{°}$ with the positive *x*-axis:

$$\mathbf{\text{w}} = \left\langle {40\ \text{cos}\left( {225\text{°}} \right),40\ \text{sin}\left( {225\text{°}} \right)} \right\rangle = \left\langle {- \frac{40}{\sqrt{2}}, - \frac{40}{\sqrt{2}}} \right\rangle = - \frac{40}{\sqrt{2}}\mathbf{\text{i}} - \frac{40}{\sqrt{2}}\mathbf{\text{j}}.$$

When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:

$$\mathbf{p + w} = -425\mathbf{\text{i}} + \left( {- \frac{40}{\sqrt{2}}\mathbf{\text{i}} - \frac{40}{\sqrt{2}}\mathbf{\text{j}}} \right) = \left( {-425 - \frac{40}{\sqrt{2}}} \right)\mathbf{\text{i}} - \frac{40}{\sqrt{2}}\mathbf{\text{j}}.$$

The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:

$$\left\| \mathbf{p + w} \right\| = \sqrt{\left( {-425 - \frac{40}{\sqrt{2}}} \right)^{2} + \left( {- \frac{40}{\sqrt{2}}} \right)^{2}} \approx 454.17\ \text{mph}$$

As a result of the wind, the plane is traveling at approximately $454$ mph relative to the ground.

To determine the bearing of the airplane, we want to find the direction of the vector $\mathbf{p + w}\text{:}$

$$\begin{array}{rll}

{\text{tan}\ \theta} & = & {\frac{- \frac{40}{\sqrt{2}}}{\left( {-425 - \frac{40}{\sqrt{2}}} \right)} \approx 0.06} \\

\theta & \approx & {3.57\text{°}.}

\end{array}$$

The overall direction of the plane is $3.57\text{°}$ south of west.

An airplane flies due north at an airspeed of $550$ mph. The wind is blowing from the northwest at $50$ mph. What is the ground speed of the airplane?

Section 2.1 Exercises

For the following exercises, consider points $P\left( {-1,3} \right),$ $Q\left( {1,5} \right),$ and $R\left( {-3,7} \right).$ Determine the requested vectors and express each of them a. in component form and b. by using the standard unit vectors.

1.

$\overset{\rightarrow}{PQ}$

2\.

$\overset{\rightarrow}{PR}$

3.

$\overset{\rightarrow}{QP}$

4\.

$\overset{\rightarrow}{RP}$

5.

$\overset{\rightarrow}{PQ} + \overset{\rightarrow}{PR}$

6\.

$\overset{\rightarrow}{PQ} - \overset{\rightarrow}{PR}$

7.

$2\overset{\rightarrow}{PQ} - 2\overset{\rightarrow}{PR}$

8\.

$2\overset{\rightarrow}{PQ} + \frac{1}{2}\overset{\rightarrow}{PR}$

9.

The unit vector in the direction of $\overset{\rightarrow}{PQ}$

10\.

The unit vector in the direction of $\overset{\rightarrow}{PR}$

11.

A vector $\mathbf{\text{v}}$ has initial point $\left( {-1,-3} \right)$ and terminal point $\left( {2,1} \right).$ Find the unit vector in the direction of $\mathbf{\text{v}}.$ Express the answer in component form.

12\.

A vector $\mathbf{\text{v}}$ has initial point $\left( {-2,5} \right)$ and terminal point $\left( {3,-1} \right).$ Find the unit vector in the direction of $\mathbf{\text{v}}.$ Express the answer in component form.

13.

The vector $\mathbf{\text{v}}$ has initial point $P(1,0)$ and terminal point $Q$ that is on the *y*-axis and above the initial point. Find the coordinates of terminal point $Q$ such that the magnitude of the vector $\mathbf{\text{v}}$ is $\sqrt{5}.$

14\.

The vector $\mathbf{\text{v}}$ has initial point $P(1,1)$ and terminal point $Q$ that is on the *x*-axis and left of the initial point. Find the coordinates of terminal point $Q$ such that the magnitude of the vector $\mathbf{\text{v}}$ is $\sqrt{10}.$

For the following exercises, use the given vectors $\textbf{a}$ and $\textbf{b}.$

1. Determine the vector sum $\textbf{a} + \textbf{b}$ and express it in both the component form and by using the standard unit vectors.

2. Find the vector difference $\textbf{a} - \textbf{b}$ and express it in both the component form and by using the standard unit vectors.

3. Verify that the vectors $\textbf{a},$ $\textbf{b},$ and $\textbf{a} + \textbf{b},$ and, respectively, $\textbf{a},$ $\textbf{b},$ and $\textbf{a} - \textbf{b}$ satisfy the triangle inequality.

4. Determine the vectors $2\textbf{a},$ $\text{−}\textbf{b},$ and $2\textbf{a} - \textbf{b}.$ Express the vectors in both the component form and by using standard unit vectors.

15.

$\textbf{a} = 2\textbf{i} + \textbf{j},$ $\textbf{b} = \textbf{i} + 3\textbf{j}$

16\.

$\textbf{a} = 2\textbf{i},$ $\textbf{b} = -2\textbf{i} + 2\textbf{j}$

17.

Let $\textbf{a}$ be a standard-position vector with terminal point $\left( {-2,-4} \right).$ Let $\textbf{b}$ be a vector with initial point $\left( {1,2} \right)$ and terminal point $\left( {-1,4} \right).$ Find the magnitude of vector $-3\textbf{a} + \textbf{b} - 4\textbf{i} + \textbf{j}.$

18\.

Let $\textbf{a}$ be a standard-position vector with terminal point at $\left( {2,5} \right).$ Let $\textbf{b}$ be a vector with initial point $\left( {-1,3} \right)$ and terminal point $\left( {1,0} \right).$ Find the magnitude of vector $\textbf{a} - 3\textbf{b} + 14\textbf{i} - 14\textbf{j}.$

19.

Let $\textbf{u}$ and $\textbf{v}$ be two nonzero vectors that are nonequivalent. Consider the vectors $\textbf{a} = 4\textbf{u} + 5\textbf{v}$ and $\textbf{b} = \textbf{u} + 2\textbf{v}$ defined in terms of $\textbf{u}$ and $\textbf{v}.$ Find the scalar $\lambda$ such that vectors $\textbf{a} + \lambda\textbf{b}$ and $\textbf{u} - \textbf{v}$ are equivalent.

20\.

Let $\textbf{u}$ and $\textbf{v}$ be two nonzero vectors that are nonequivalent. Consider the vectors $\textbf{a} = 2\textbf{u} - 4\textbf{v}$ and $\textbf{b} = 3\textbf{u} - 7\textbf{v}$ defined in terms of $\textbf{u}$ and $\textbf{v}.$ Find the scalars $\alpha$ and $\beta$ such that vectors $\alpha\textbf{a} + \beta\textbf{b}$ and $\textbf{u} - \textbf{v}$ are equivalent.

21.

Consider the vector $\textbf{a}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle$ with components that depend on a real number $t.$ As the number $t$ varies, the components of $\textbf{a}(t)$ change as well, depending on the functions that define them.

1. Write the vectors $\textbf{a}(0)$ and $\textbf{a}(\pi)$ in component form.

2. Show that the magnitude $\left\| {\textbf{a}(t)} \right\|$ of vector $\textbf{a}(t)$ remains constant for any real number $t.$

3. As $t$ varies, show that the terminal point of vector $\textbf{a}(t)$ describes a circle centered at the origin of radius $1.$

22\.

Consider vector $\textbf{a}(x) = \left\langle {x,\sqrt{1 - x^{2}}} \right\rangle$ with components that depend on a real number $x \in \lbrack-1,1\rbrack.$ As the number $x$ varies*,* the components of $\textbf{a}(x)$ change as well, depending on the functions that define them.

1. Write the vectors $\textbf{a}(0)$ and $\textbf{a}(1)$ in component form.

2. Show that the magnitude $\left\| {\textbf{a}(x)} \right\|$ of vector $\textbf{a}(x)$ remains constant for any real number $x$

3. As $x$ varies, show that if $\textbf{a}(x)$ is in standard position, then its terminal point describes a semicircle. Why is it only a semicircle?

23.

Show that vectors $\textbf{a}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle$ and $\textbf{a}(x) = \left\langle {x,\sqrt{1 - x^{2}}} \right\rangle$ are equivalent for $x = 1$ and $t = 2k\pi,$ where $k$ is an integer.

24\.

Show that vectors $\textbf{a}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle$ and $\textbf{a}(x) = \left\langle {x,\sqrt{1 - x^{2}}} \right\rangle$ are opposite for $x = 1$ and $t = \pi + 2k\pi,$ where $k$ is an integer.

For the following exercises, find vector $\mathbf{\text{v}}$ with the given magnitude and in the same direction as vector $\mathbf{\text{u}}.$

25.

$\left\| \mathbf{\text{v}} \right\| = 7,\mathbf{\text{u}} = \left\langle {3,4} \right\rangle$

26\.

$\left\| \mathbf{\text{v}} \right\| = 3,\mathbf{\text{u}} = \left\langle {-2,5} \right\rangle$

27.

$\left\| \mathbf{\text{v}} \right\| = 7,\mathbf{\text{u}} = \left\langle {3,-5} \right\rangle$

28\.

$\left\| \mathbf{\text{v}} \right\| = 10,\mathbf{\text{u}} = \left\langle {2,-1} \right\rangle$

For the following exercises, find the component form of vector $\mathbf{\text{u}},$ given its magnitude and the angle the vector makes with the positive *x*-axis. Give exact answers when possible.

29.

$\left\| \textbf{u} \right\| = 2,$ $\theta = 30\text{°}$

30\.

$\left\| \textbf{u} \right\| = 6,$ $\theta = 60\text{°}$

31.

$\left\| \textbf{u} \right\| = 5,$ $\theta = \frac{\pi}{2}$

32\.

$\left\| \textbf{u} \right\| = 8,$ $\theta = \pi$

33.

$\left\| \mathbf{\text{u}} \right\| = 10,$ $\theta = \frac{5\pi}{6}$

34\.

$\left\| \mathbf{\text{u}} \right\| = 50,$ $\theta = \frac{3\pi}{4}$

For the following exercises, vector $\textbf{u}$ is given. Find the angle $\theta \in \lbrack 0,2\pi)$ that vector $\textbf{u}$ makes with the positive direction of the *x*-axis, in a counter-clockwise direction.

35.

$\textbf{u} = 5\sqrt{2}\textbf{i} - 5\sqrt{2}\textbf{j}$

36\.

$\textbf{u} = \text{−}\sqrt{3}\textbf{i} - \textbf{j}$

37.

Let $\textbf{a} = \left\langle {a_{1},a_{2}} \right\rangle,$ $\textbf{b} = \left\langle {b_{1},b_{2}} \right\rangle,$ and $\textbf{c} = \left\langle {c_{1},c_{2}} \right\rangle$ be three nonzero vectors. If $a_{1}b_{2} - a_{2}b_{1} \neq 0,$ then show there are two scalars, $\alpha$ and $\beta,$ such that $\textbf{c} = \alpha\textbf{a} + \beta\textbf{b}.$

38\.

Consider vectors $\textbf{a} = \left\langle {2,-4} \right\rangle,$ $\mathbf{\text{b}} = \left\langle {-1,2} \right\rangle,$ and c = 0 Determine the scalars $\alpha$ and $\beta$ such that $\textbf{c} = \alpha\textbf{a} + \beta\textbf{b}.$

39.

Let $P\left( {x_{0},f\left( x_{0} \right)} \right)$ be a fixed point on the graph of the differentiable function $f$ with a domain that is the set of real numbers.

1. Determine the real number $z_{0}$ such that point $Q\left( {x_{0} + 1,z_{0}} \right)$ is situated on the line tangent to the graph of $f$ at point $P.$

2. Determine the unit vector $\mathbf{\text{u}}$ with initial point $P$ in the direction of vector $PQ.$

40\.

Consider the function $f(x) = x^{4},$ where $x \in \mathbb{R}.$

1. Determine the real number $z_{0}$ such that point $Q\left( {2,z_{0}} \right)$ s situated on the line tangent to the graph of $f$ at point $P\left( {1,1} \right).$

2. Determine the unit vector $\mathbf{\text{u}}$ with initial point $P$ and terminal point $Q.$

41\.

Consider $f$ and $g$ two functions defined on the same set of real numbers $D.$ Let $\textbf{a} = \left\langle {x,f(x)} \right\rangle$ and $\textbf{b} = \left\langle {x,g(x)} \right\rangle$ be two vectors that describe the graphs of the functions, where $x \in D.$ Show that if the graphs of the functions $f$ and $g$ do not intersect, then the vectors $\mathbf{\text{a}}$ and $\mathbf{\text{b}}$ are not equivalent.

42\.

Find $x \in \mathbb{R}$ such that vectors $\textbf{a} = \left\langle {x,\text{sin}\ x} \right\rangle$ and $\textbf{b} = \left\langle {x,\text{cos}\ x} \right\rangle$ are equivalent.

43.

Calculate the coordinates of point $D$ such that $ABCD$ is a parallelogram, with $A(1,1),$ $B(2,4),$ and $C(7,4).$

44\.

Consider the points $A(2,1),$ $B(10,6),$ $C(13,4),$ and $D(16,-2).$ Determine the component form of vector $\overset{\rightarrow}{AD}.$

45.

The speed of an object is the magnitude of its related velocity vector. A football thrown by a quarterback has an initial speed of $70$ mph and an angle of elevation of $30\text{°}.$ Determine the velocity vector in mph and express it in component form. (Round to two decimal places.)

46\.

A baseball player throws a baseball at an angle of $30\text{°}$ with the horizontal. If the initial speed of the ball is $100$ mph, find the horizontal and vertical components of the initial velocity vector of the baseball. (Round to two decimal places.)

47.

A bullet is fired with an initial velocity of $1500$ ft/sec at an angle of $60\text{°}$ with the horizontal. Find the horizontal and vertical components of the velocity vector of the bullet. (Round to two decimal places.)

48\.

\[T\] A 65-kg sprinter exerts a force of $798$ N at a $19\text{°}$ angle with respect to the ground on the starting block at the instant a race begins. Find the horizontal component of the force. (Round to two decimal places.)

49.

\[T\] Two forces, a horizontal force of $45$ lb and another of $52$ lb, act on the same object. The angle between these forces is $25\text{°}.$ Find the magnitude and direction angle from the positive *x*-axis of the resultant force that acts on the object. (Round to two decimal places.)

50\.

\[T\] Two forces, a vertical force of $26$ lb and another of $45$ lb, act on the same object. The angle between these forces is $55\text{°}.$ Find the magnitude and direction angle from the positive *x*-axis of the resultant force that acts on the object. (Round to two decimal places.)

51.

\[T\] Three forces act on object. Two of the forces have the magnitudes $58$ N and $27$ N, and make angles $53\text{°}$ and $152\text{°},$ respectively, with the positive *x*-axis. Find the magnitude and the direction angle from the positive *x*-axis of the third force such that the resultant force acting on the object is zero. (Round to two decimal places.)

52\.

Three forces with magnitudes $80$ lb, $120$ lb, and $60$ lb act on an object at angles of $45\text{°},$ $60\text{°}$ and $30\text{°},$ respectively, with the positive *x*-axis. Find the magnitude and direction angle from the positive *x*-axis of the resultant force. (Round to two decimal places.)

53.

\[T\] An airplane is flying in the direction of $43\text{°}$ east of north (also abbreviated as $\text{N}43\text{E})$ at a speed of $550$ mph. A wind with speed $25$ mph comes from the southwest at a bearing of $\text{N}15\text{E}.$ What are the ground speed and new direction of the airplane?

54\.

\[T\] A boat is traveling in the water at $30$ mph in a direction of $\text{N}20\text{E}$ (that is, $20\text{°}$ east of north). A strong current is moving at $15$ mph in a direction of $\text{N}45\text{E}.$ What are the new speed and direction of the boat?

55.

\[T\] A 50-lb weight is hung by a cable so that the two portions of the cable make angles of $40\text{°}$ and $53\text{°},$ respectively, with the horizontal. Find the magnitudes of the forces of tension $\text{T}_{1}$ and $\text{T}_{2}$ in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)

56\.

\[T\] A 62-lb weight hangs from a rope that makes the angles of $29\text{°}$ and $61\text{°},$ respectively, with the horizontal. Find the magnitudes of the forces of tension $\text{T}_{1}$ and $\text{T}_{2}$ in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)

57.

\[T\] A 1500-lb boat is parked on a ramp that makes an angle of $30\text{°}$ with the horizontal. The boat’s weight vector points downward and is a sum of two vectors: a horizontal vector $\textbf{v}_{1}$ that is parallel to the ramp and a vertical vector $\textbf{v}_{2}$ that is perpendicular to the inclined surface. The magnitudes of vectors $\textbf{v}_{1}$ and $\textbf{v}_{2}$ are the horizontal and vertical component, respectively, of the boat’s weight vector. Find the magnitudes of $\textbf{v}_{1}$ and $\textbf{v}_{2}.$ (Round to the nearest integer.)

58\.

\[T\] An 85-lb box is at rest on a $26\text{°}$ incline. Determine the magnitude of the force parallel to the incline necessary to keep the box from sliding. (Round to the nearest integer.)

59.

A guy-wire supports a pole that is

$75$ ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground $50$ ft from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is $50$ lb. (Round to the nearest integer.)

60\.

A telephone pole guy-wire has an angle of elevation of $35\text{°}$ with respect to the ground. The force of tension in the guy-wire is $120$ lb. Find the horizontal and vertical components of the force of tension. (Round to the nearest integer.)

---

2.2 Vectors in Three Dimensions

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-2-vectors-in-three-dimensions

2.2 Vectors in Three Dimensions

• 2.2.1 Describe three-dimensional space mathematically.
• 2.2.2 Locate points in space using coordinates.
• 2.2.3 Write the distance formula in three dimensions.
• 2.2.4 Write the equations for simple planes and spheres.
• 2.2.5 Perform vector operations in $\mathbb{R}^{3}.$

Vectors are useful tools for solving two-dimensional problems. Life, however, happens in three dimensions. To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases the topography of the land is important. Does your planned route go through the mountains? Do you have to cross a river? To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a natural extension of the two-dimensional Cartesian coordinate plane into three dimensions.

Three-Dimensional Coordinate Systems

As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal *x*-axis and the vertical *y*-axis. We can add a third dimension, the *z*-axis, which is perpendicular to both the *x*-axis and the *y*-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions we encounter in real life.

The three-dimensional rectangular coordinate system consists of three perpendicular axes: the *x*-axis, the *y*-axis, the *z*-axis, and an origin at the point of intersection (0) of the axes. Because each axis is a number line representing all real numbers in $\mathbb{R},$ the three-dimensional system is often denoted by $\mathbb{R}^{3}.$

In Figure 2.23(a), the positive *z*-axis is shown above the plane containing the *x*- and *y*-axes. The positive *x*-axis appears to the left and the positive *y*-axis is to the right. A natural question to ask is: How was arrangement determined? The system displayed follows the right-hand rule. If we take our right hand and align the fingers with the positive *x*-axis, then curl the fingers so they point in the direction of the positive *y*-axis, our thumb points in the direction of the positive *z*-axis. In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the standard representation.

In two dimensions, we describe a point in the plane with the coordinates $\left( {x,y} \right).$ Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate, $z,$ is appended to indicate alignment with the *z*-axis: $\left( {x,y,z} \right).$ A point in space is identified by all three coordinates (Figure 2.24). To plot the point $\left( {x,y,z} \right),$ go *x* units along the *x*-axis, then $y$ units in the direction of the *y*-axis, then $z$ units in the direction of the *z*-axis.

Locating Points in Space

Sketch the point $\left( {1,-2,3} \right)$ in three-dimensional space.

Solution

To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive $x$ direction, $2$ units in the negative $y$ direction, and $3$ units in the positive $z$ direction. Complete the prism to plot the point (Figure 2.25).

Sketch the point $\left( {-2,3,-1} \right)$ in three-dimensional space.

In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a coordinate plane: the *xy*-plane, the *xz*-plane, and the *yz*-plane (Figure 2.26). We define the *xy*-plane formally as the following set: $\left\{ {\left( {x,y,0} \right):x,y \in \mathbb{R}} \right\}.$ Similarly, the *xz*-plane and the *yz*-plane are defined as $\left\{ {\left( {x,0,z} \right):x,z \in \mathbb{R}} \right\}$ and $\left\{ {\left( {0,y,z} \right):y,z \in \mathbb{R}} \right\},$ respectively.

To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the *xy*-plane, the wall to your right is the *xz*-plane, and the wall to your left is the *yz*-plane.

In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants. The octants fill $\mathbb{R}^{3}$ in the same way that quadrants fill $\mathbb{R}^{2},$ as shown in Figure 2.27.

Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space.

If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We know that the distance $d$ between two points $\left( {x_{1},y_{1}} \right)$ and $\left( {x_{2},y_{2}} \right)$ in the *xy*-coordinate plane is given by the formula

$$d = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2}}.$$

The formula for the distance between two points in space is a natural extension of this formula.

The Distance between Two Points in Space

The distance $d$ between points $\left( {x_{1},y_{1},z_{1}} \right)$ and $\left( {x_{2},y_{2},z_{2}} \right)$ is given by the formula

$$d = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2} + \left( {z_{2} - z_{1}} \right)^{2}}.$$ (2.1)

The proof of this theorem is left as an exercise. (*Hint:* First find the distance $d_{1}$ between the points $\left( {x_{1},y_{1},z_{1}} \right)$ and $\left( {x_{2},y_{2},z_{1}} \right)$ as shown in Figure 2.28.)

Distance in Space

Find the distance between points $P_{1} = \left( {3,\text{−}1,5} \right)$ and $P_{2} = \left( {2,1,\text{−}1} \right).$

Solution

Substitute values directly into the distance formula:

$$\begin{array}{cl}

{d\left( {P_{1},P_{2}} \right)} & {= \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2} + \left( {z_{2} - z_{1}} \right)^{2}}} \\

& {= \sqrt{\left( {2 - 3} \right)^{2} + \left( {1 - (-1)} \right)^{2} + \left( {-1 - 5} \right)^{2}}} \\

& {= \sqrt{(–1)^{2} + 2^{2} + (-6)^{2}}} \\

& {= \sqrt{41}.}

\end{array}$$

Find the distance between points $P_{1} = \left( {1,-5,4} \right)$ and $P_{2} = \left( {4,-1,-1} \right).$

Before moving on to the next section, let’s get a feel for how $\mathbb{R}^{3}$ differs from $\mathbb{R}^{2}.$ For example, in $\mathbb{R}^{2},$ lines that are not parallel must always intersect. This is not the case in $\mathbb{R}^{3}.$ For example, consider the line shown in Figure 2.30. These two lines are not parallel, nor do they intersect.

You can also have circles that are interconnected but have no points in common, as in Figure 2.31.

We have a lot more flexibility working in three dimensions than we do if we stick with only two dimensions.

Writing Equations in ℝ³

Now that we can represent points in space and find the distance between them, we can learn how to write equations of geometric objects such as lines, planes, and curved surfaces in $\mathbb{R}^{3}.$ First, we start with a simple equation. Compare the graphs of the equation $x = 0$ in $\mathbb{R},\mathbb{R}^{2},\ \text{and}\ \mathbb{R}^{3}$ (Figure 2.32). From these graphs, we can see the same equation can describe a point, a line, or a plane.

In space, the equation $x = 0$ describes all points $\left( {0,y,z} \right).$ This equation defines the *yz*-plane. Similarly, the *xy*-plane contains all points of the form $\left( {x,y,0} \right).$ The equation $z = 0$ defines the *xy*-plane and the equation $y = 0$ describes the *xz*-plane (Figure 2.33).

Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one of the coordinate planes. When a plane is parallel to the *xy*-plane, for example, the *z*-coordinate of each point in the plane has the same constant value. Only the *x*- and *y*-coordinates of points in that plane vary from point to point.

1. The plane in space that is parallel to the *xy*-plane and contains point $\left( {a,b,c} \right)$ can be represented by the equation $z = c.$

2. The plane in space that is parallel to the *xz*-plane and contains point $\left( {a,b,c} \right)$ can be represented by the equation $y = b.$

3. The plane in space that is parallel to the *yz*-plane and contains point $\left( {a,b,c} \right)$ can be represented by the equation $x = a.$

Writing Equations of Planes Parallel to Coordinate Planes

1. Write an equation of the plane passing through point $\left( {3,11,7} \right)$ that is parallel to the *yz*-plane.

2. Find an equation of the plane passing through points $\left( {6,-2,9} \right),$ $\left( {0,-2,4} \right),$ and $\left( {1,-2,-3} \right).$

Solution

1. When a plane is parallel to the *yz*-plane, only the *y*- and *z*-coordinates may vary. The *x*-coordinate has the same constant value for all points in this plane, so this plane can be represented by the equation $x = 3.$

2. Each of the points $\left( {6,-2,9} \right),$ $\left( {0,-2,4} \right),$ and $\left( {1,-2,-3} \right)$ has the same *y*-coordinate. This plane can be represented by the equation $y = -2.$

Write an equation of the plane passing through point $\left( {1,-6,-4} \right)$ that is parallel to the *xy*-plane.

As we have seen, in $\mathbb{R}^{2}$ the equation $x = 5$ describes the vertical line passing through point $\left( {5,0} \right).$ This line is parallel to the *y*-axis. In a natural extension, the equation $x = 5$ in $\mathbb{R}^{3}$ describes the plane passing through point $\left( {5,0,0} \right),$ which is parallel to the *yz*-plane. Another natural extension of a familiar equation is found in the equation of a sphere.

A sphere is the set of all points in space equidistant from a fixed point, the center of the sphere (Figure 2.34), just as the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, the distance from the center to a point on the sphere is called the *radius*.

The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance.

The sphere with center $\left( {a,b,c} \right)$ and radius $r$ can be represented by the equation

$$\left( {x - a} \right)^{2} + \left( {y - b} \right)^{2} + \left( {z - c} \right)^{2} = r^{2}.$$ (2.2)

This equation is known as the standard equation of a sphere.

Finding an Equation of a Sphere

Find the standard equation of the sphere with center $\left( {10,7,4} \right)$ and point $\left( {-1,3,-2} \right),$ as shown in Figure 2.35.

Solution

Use the distance formula to find the radius $r$ of the sphere:

$$\begin{array}{cl}

r & {= \sqrt{\left( {-1 - 10} \right)^{2} + \left( {3 - 7} \right)^{2} + \left( {-2 - 4} \right)^{2}}} \\

& {= \sqrt{{(-11)}^{2} + {(-4)}^{2} + {(-6)}^{2}}} \\

& {= \sqrt{173}.}

\end{array}$$

The standard equation of the sphere is

$$\left( {x - 10} \right)^{2} + \left( {y - 7} \right)^{2} + \left( {z - 4} \right)^{2} = 173.$$

Find the standard equation of the sphere with center $\left( {-2,4,-5} \right)$ containing point $\left( {4,4,-1} \right).$

Finding the Equation of a Sphere

Let $P = \left( {-5,2,3} \right)$ and $Q = \left( {3,4,-1} \right),$ and suppose line segment $PQ$ forms the diameter of a sphere (Figure 2.36). Find an equation of the sphere.

Solution

Since $PQ$ is a diameter of the sphere, we know the center of the sphere is the midpoint of $PQ.$ Then,

$$\begin{array}{cl}

C & {= \left( {\frac{-5 + 3}{2},\frac{2 + 4}{2},\frac{3 + (-1)}{2}} \right)} \\

& {= \left( {-1,3,1} \right).}

\end{array}$$

Furthermore, we know the radius of the sphere is half the length of the diameter. This gives

$$\begin{array}{cl}

r & {= \frac{1}{2}\sqrt{\left( {-5 - 3} \right)^{2} + \left( {2 - 4} \right)^{2} + \left( {3 - (-1)} \right)^{2}}} \\

& {= \frac{1}{2}\sqrt{64 + 4 + 16}} \\

& {= \sqrt{21}.}

\end{array}$$

Then, the equation of the sphere is $\left( {x + 1} \right)^{2} + \left( {y - 3} \right)^{2} + \left( {z - 1} \right)^{2} = 21.$

Find an equation of the sphere with diameter $PQ,$ where $P = \left( {2,-1,-3} \right)$ and $Q = \left( {-2,5,-1} \right).$

Graphing Other Equations in Three Dimensions

Describe the set of points that satisfies $\left( {x - 4} \right)\left( {z - 2} \right) = 0,$ and graph the set.

Solution

We must have either $x - 4 = 0$ or $z - 2 = 0,$ so the set of points forms the two planes $x = 4$ and $z = 2$ (Figure 2.37).

Describe the set of points that satisfies $\left( {y + 2} \right)\left( {z - 3} \right) = 0,$ and graph the set.

Graphing Other Equations in Three Dimensions

Describe the set of points in three-dimensional space that satisfies $\left( {x - 2} \right)^{2} + \left( {y - 1} \right)^{2} = 4,$ and graph the set.

Solution

The *x*- and *y*-coordinates form a circle in the *xy*-plane of radius $2,$ centered at $\left( {2,1} \right).$ Since there is no restriction on the *z*-coordinate, the three-dimensional result is a circular cylinder of radius $2$ centered on the line with $x = 2\ \text{and}\ y = 1.$ The cylinder extends indefinitely in the *z*-direction (Figure 2.38).

Describe the set of points in three dimensional space that satisfies $x^{2} + {(z - 2)}^{2} = 16,$ and graph the surface.

Working with Vectors in ℝ³

Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.

Three-dimensional vectors can also be represented in component form. The notation $\mathbf{\text{v}} = \left\langle {x,y,z} \right\rangle$ is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, $\left( {0,0,0} \right),$ and terminal point $\left( {x,y,z} \right).$ The zero vector is $\mathbf{0} = \left\langle {0,0,0} \right\rangle.$ So, for example, the three dimensional vector $\mathbf{\text{v}} = \left\langle {2,4,1} \right\rangle$ is represented by a directed line segment from point $\left( {0,0,0} \right)$ to point $\left( {2,4,1} \right)$ (Figure 2.39).

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If $\mathbf{\text{v}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle$ and $\mathbf{\text{w}} = \left\langle {x_{2},y_{2},z_{2}} \right\rangle$ are vectors, and $k$ is a scalar, then

$$\mathbf{\text{v}} + \mathbf{\text{w}} = \left\langle {x_{1} + x_{2},y_{1} + y_{2},z_{1} + z_{2}} \right\rangle\ \text{and}\ k\mathbf{\text{v}} = \left\langle {kx_{1},ky_{1},kz_{1}} \right\rangle.$$

If $k = -1,$ then $k\mathbf{\text{v}} = (-1)\mathbf{\text{v}}$ is written as $\text{−}\mathbf{\text{v}},$ and vector subtraction is defined by $\mathbf{\text{v}} - \mathbf{w = v +}\left( {\text{−}\mathbf{\text{w}}} \right)\mathbf{= v +}(-1)\mathbf{\text{w}}.$

The standard unit vectors extend easily into three dimensions as well—$\mathbf{\text{i}} = \left\langle {1,0,0} \right\rangle,$ $\mathbf{\text{j}} = \left\langle {0,1,0} \right\rangle,$ and $\mathbf{\text{k}} = \left\langle {0,0,1} \right\rangle$—and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in $\mathbb{R}^{3}$ in the following ways:

$$\mathbf{\text{v}} = \left\langle {x,y,z} \right\rangle = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}.$$

Vector Representations

Let $\overset{\rightarrow}{PQ}$ be the vector with initial point $P = (3,12,6)$ and terminal point $Q = \left( {-4,-3,2} \right)$ as shown in Figure 2.40. Express $\overset{\rightarrow}{PQ}$ in both component form and using standard unit vectors.

Solution

In component form,

$$\begin{array}{cl}

\overset{\rightarrow}{PQ} & {= \left\langle {x_{2} - x_{1},y_{2} - y_{1},z_{2} - z_{1}} \right\rangle} \\

& {= \left\langle {-4 - 3,-3 - 12,2 - 6} \right\rangle = \left\langle {-7,-15,-4} \right\rangle.}

\end{array}$$

In standard unit form,

$$\overset{\rightarrow}{PQ} = -7\mathbf{\text{i}} - 15\mathbf{\text{j}} - 4\mathbf{\text{k}}.$$

Let $S = \left( {3,8,2} \right)$ and $T = \left( {2,-1,3} \right).$ Express $\overset{\rightarrow}{ST}$ in component form and in standard unit form.

As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space (Figure 2.41).

We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.

Let $\mathbf{\text{v}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle$ and $\mathbf{\text{w}} = \left\langle {x_{2},y_{2},z_{2}} \right\rangle$ be vectors, and let $k$ be a scalar.

Scalar multiplication: $k\mathbf{\text{v}} = \left\langle {kx_{1},ky_{1},kz_{1}} \right\rangle$

Vector addition: $\mathbf{\text{v}} + \mathbf{\text{w}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle + \left\langle {x_{2},y_{2},z_{2}} \right\rangle = \left\langle {x_{1} + x_{2},y_{1} + y_{2},z_{1} + z_{2}} \right\rangle$

Vector subtraction: $\mathbf{\text{v}} - \mathbf{\text{w}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle - \left\langle {x_{2},y_{2},z_{2}} \right\rangle = \left\langle {x_{1} - x_{2},y_{1} - y_{2},z_{1} - z_{2}} \right\rangle$

Vector magnitude: $\left\| \mathbf{\text{v}} \right\| = \sqrt{x_{1}{}^{2} + y_{1}{}^{2} + z_{1}{}^{2}}$

Unit vector in the direction of v: $\frac{1}{\left\| \mathbf{\text{v}} \right\|}\mathbf{\text{v}} = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\left\langle {x_{1},y_{1},z_{1}} \right\rangle = \left\langle {\frac{x_{1}}{\left\| \mathbf{\text{v}} \right\|},\frac{y_{1}}{\left\| \mathbf{\text{v}} \right\|},\frac{z_{1}}{\left\| \mathbf{\text{v}} \right\|}} \right\rangle,$ if $\mathbf{\text{v}} \neq \mathbf{0}$

We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

Vector Operations in Three Dimensions

Let $\mathbf{\text{v}} = \left\langle {-2,9,5} \right\rangle$ and $\mathbf{\text{w}} = \left\langle {1,-1,0} \right\rangle$ (Figure 2.42). Find the following vectors.

1. $3\mathbf{\text{v}} - 2\mathbf{\text{w}}$

2. $5\left\| \mathbf{\text{w}} \right\|$

3. $\left\| {5\mathbf{\text{w}}} \right\|$

4. A unit vector in the direction of $\mathbf{\text{v}}$

Solution

1. First, use scalar multiplication of each vector, then subtract:

$$\begin{array}{cl}

{3\mathbf{\text{v}} - 2\mathbf{\text{w}}} & {= 3\left\langle {-2,9,5} \right\rangle - 2\left\langle {1,-1,0} \right\rangle} \\

& {= \left\langle {-6,27,15} \right\rangle - \left\langle {2,-2,0} \right\rangle} \\

& {= \left\langle {-6 - 2,27 - (-2),15 - 0} \right\rangle} \\

& {= \left\langle {-8,29,15} \right\rangle.}

\end{array}$$

2. Write the equation for the magnitude of the vector, then use scalar multiplication:

$$5\left\| \mathbf{\text{w}} \right\| = 5\sqrt{1^{2} + (-1)^{2} + 0^{2}} = 5\sqrt{2}.$$

3. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:

$$\left\| {5\mathbf{\text{w}}} \right\| = \left\| \left\langle {5,-5,0} \right\rangle \right\| = \sqrt{5^{2} + (-5)^{2} + 0^{2}} = \sqrt{50} = 5\sqrt{2}.$$

4. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:

$$\begin{array}{cl}

\frac{\mathbf{\text{v}}}{\left\| \mathbf{\text{v}} \right\|} & {= \frac{1}{\left\| \mathbf{\text{v}} \right\|}\left\langle {-2,9,5} \right\rangle} \\

& {= \frac{1}{\sqrt{(-2)^{2} + 9^{2} + 5^{2}}}\left\langle {-2,9,5} \right\rangle} \\

& {= \frac{1}{\sqrt{110}}\left\langle {-2,9,5} \right\rangle} \\

& {= \left\langle {\frac{-2}{\sqrt{110}},\frac{9}{\sqrt{110}},\frac{5}{\sqrt{110}}} \right\rangle.}

\end{array}$$

Let $\mathbf{\text{v}} = \left\langle {-1,-1,1} \right\rangle$ and $\mathbf{\text{w}} = \left\langle {2,0,1} \right\rangle.$ Find a unit vector in the direction of $5\mathbf{\text{v}} + 3\mathbf{\text{w}}.$

Throwing a Forward Pass

A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of $30\text{°}$ (see the following figure). Write the initial velocity vector of the ball, $\mathbf{\text{v}},$ in component form.

Solution

The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector $\mathbf{\text{w}}$ extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of $30\text{°}$ (see the following figure). This vector would have the same direction as $\mathbf{\text{v}},$ but it may not have the right magnitude.

The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is

$$\text{Dist from QB to receiver} = \sqrt{15^{2} + 20^{2}} = \sqrt{225 + 400} = \sqrt{625} = 25\ \text{yd}.$$

We have $\frac{25}{\left\| \mathbf{\text{w}} \right\|} = \text{cos}\ 30\text{°}.$ Then the magnitude of $\mathbf{\text{w}}$ is given by

$$\left\| \mathbf{\text{w}} \right\| = \frac{25}{\text{cos}\ 30\text{°}} = \frac{25 \cdot 2}{\sqrt{3}} = \frac{50}{\sqrt{3}}\ \text{yd}$$

and the vertical distance from the receiver to the terminal point of $\mathbf{\text{w}}$ is

$$\text{Vert dist from receiver to terminal point of}\ \mathbf{\text{w}} = \left\| \mathbf{\text{w}} \right\|\text{sin}\ 30\text{°} = \frac{50}{\sqrt{3}} \cdot \frac{1}{2} = \frac{25}{\sqrt{3}}\ \text{yd}.$$

Then $\mathbf{\text{w}} = \left\langle {20,15,\frac{25}{\sqrt{3}}} \right\rangle,$ and has the same direction as $\mathbf{\text{v}}.$

Recall, though, that we calculated the magnitude of $\mathbf{\text{w}}$ to be $\left\| \mathbf{\text{w}} \right\| = \frac{50}{\sqrt{3}},$ and $\mathbf{\text{v}}$ has magnitude $60$ mph. So, we need to multiply vector $\mathbf{\text{w}}$ by an appropriate constant, $k.$ We want to find a value of $k$ so that $\left\| {k\mathbf{\text{w}}} \right\| = 60$ mph. We have

$$\left\| {k\mathbf{\text{w}}} \right\| = k\left\| \mathbf{\text{w}} \right\| = k\frac{50}{\sqrt{3}}\ \text{mph,}$$

so we want

$$\begin{array}{rll}

& & \\

& & \\

{k\frac{50}{\sqrt{3}}} & = & 60 \\

k & = & \frac{60\sqrt{3}}{50} \\

k & = & {\frac{6\sqrt{3}}{5}.}

\end{array}$$

Then

$$\mathbf{\text{v}} = k\mathbf{\text{w}} = k\left\langle {20,15,\frac{25}{\sqrt{3}}} \right\rangle = \frac{6\sqrt{3}}{5}\left\langle {20,15,\frac{25}{\sqrt{3}}} \right\rangle = \left\langle {24\sqrt{3},18\sqrt{3},30} \right\rangle.$$

Let’s double-check that $\left\| \mathbf{\text{v}} \right\| = 60.$ We have

$$\left\| \mathbf{\text{v}} \right\| = \sqrt{\left( {24\sqrt{3}} \right)^{2} + \left( {18\sqrt{3}} \right)^{2} + (30)^{2}} = \sqrt{1728 + 972 + 900} = \sqrt{3600} = 60\ \text{mph}.$$

So, we have found the correct components for $\mathbf{\text{v}}.$

Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of $40$ mph and an angle of $45\text{°}.$ Write the initial velocity vector of the ball, $\mathbf{\text{v}},$ in component form.

Section 2.2 Exercises

61.

Consider a rectangular box with one of the vertices at the origin, as shown in the following figure. If point $A(2,3,5)$ is the opposite vertex to the origin, then find

1. the coordinates of the other six vertices of the box and

2. the length of the diagonal of the box determined by the vertices $O$ and $A.$

62\.

Find the coordinates of point $P$ and determine its distance to the origin.

For the following exercises, describe and graph the set of points that satisfies the given equation.

63.

$\left( {y - 5} \right)\left( {z - 6} \right) = 0$

64\.

$\left( {z - 2} \right)\left( {z - 5} \right) = 0$

65.

$\left( {y - 1} \right)^{2} + {(z - 1)}^{2} = 1$

66\.

$\left( {x - 2} \right)^{2} + {(z - 5)}^{2} = 4$

67.

Write the equation of the plane passing through point $(1,1,1)$ that is parallel to the *xy*-plane.

68\.

Write the equation of the plane passing through point $(1,-3,2)$ that is parallel to the *xz*-plane.

69.

Find an equation of the plane passing through points $(1,-3,-2),$ $(0,3,-2),$ and $(1,0,-2).$

70\.

Find an equation of the plane passing through points $(1,9,2),$ $(1,3,6),$ and $(1,-7,8).$

For the following exercises, find an equation of the sphere in standard form that satisfies the given conditions.

71.

Center $C\left( {-1,7,4} \right)$ and radius $4$

72\.

Center $C\left( {-4,7,2} \right)$ and radius $6$

73.

Diameter $PQ,$ where $P\left( {-1,5,7} \right)$ and $Q\left( {-5,2,9} \right)$

74\.

Diameter $PQ,$ where $P\left( {-16,-3,9} \right)$ and $Q\left( {-2,3,5} \right)$

For the following exercises, find the center and radius of the sphere with an equation in general form that is given.

75.

$x^{2} + y^{2} + z^{2} - 4z + 3 = 0$

76\.

$x^{2} + y^{2} + z^{2} - 6x + 8y - 10z + 25 = 0$

For the following exercises, express vector $\overset{\rightarrow}{PQ}$ with the initial point at $P$ and the terminal point at $Q$

1. in component form and

2. by using standard unit vectors.

77.

$P\left( {3,0,2} \right)$ and $Q\left( {-1,-1,4} \right)$

78\.

$P\left( {0,10,5} \right)$ and $Q\left( {1,1,-3} \right)$

79.

$P\left( {-2,5,-8} \right)$ and $M\left( {1,-7,4} \right),$ where $M$ is the midpoint of the line segment $PQ$

80\.

$Q\left( {0,7,-6} \right)$ and $M\left( {-1,3,2} \right),$ where $M$ is the midpoint of the line segment $PQ$

81.

Find terminal point $Q$ of vector $\overset{\rightarrow}{PQ} = \left\langle {7,-1,3} \right\rangle$ with the initial point at $P\left( {-2,3,5} \right).$

82\.

Find initial point $P$ of vector $\overset{\rightarrow}{PQ} = \left\langle {-9,1,2} \right\rangle$ with the terminal point at $Q\left( {10,0,-1} \right).$

For the following exercises, use the given vectors $\mathbf{\text{a}}$ and $\mathbf{\text{b}}$ to find and express the vectors $\mathbf{\text{a}} + \textbf{b},$ $4\mathbf{\text{a}},$ and $-5\mathbf{\text{a}} + 3\mathbf{\text{b}}$ in component form.

83.

$\mathbf{\text{a}} = \left\langle {-1,-2,4} \right\rangle,$ $\mathbf{\text{b}} = \left\langle {-5,6,-7} \right\rangle$

84\.

$\mathbf{\text{a}} = \left\langle {3,-2,4} \right\rangle,$ $\mathbf{\text{b}} = \left\langle {-5,6,-9} \right\rangle$

85.

$\mathbf{\text{a}} = \text{−}\mathbf{\text{k}},$ $\mathbf{\text{b}} = \text{−}\textbf{i}$

86\.

$\mathbf{\text{a}} = \mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}},$ $\mathbf{\text{b}} = 2\mathbf{\text{i}} - 3\textbf{j} + 2\mathbf{\text{k}}$

For the following exercises, vectors u and v are given. Find the magnitudes of vectors $\mathbf{\text{u}} - \mathbf{\text{v}}$ and $-2\mathbf{\text{u}}.$

87.

$\mathbf{\text{u}} = 2\mathbf{\text{i}} + 3\mathbf{\text{j}} + 4\mathbf{\text{k}},$ $\mathbf{\text{v}} = \text{−}\mathbf{\text{i}} + 5\mathbf{\text{j}} - \mathbf{\text{k}}$

88\.

$\mathbf{\text{u}} = \mathbf{\text{i}} + \mathbf{\text{j}},$ $\mathbf{\text{v}} = \mathbf{\text{j}} - \mathbf{\text{k}}$

89.

$\mathbf{\text{u}} = \left\langle {2\ \text{cos}\ t,-2\ \text{sin}\ t,3} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {0,0,3} \right\rangle,$ where $t$ is a real number.

90\.

$\mathbf{\text{u}} = \left\langle {0,1,\ \text{sinh}\ t} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {1,1,0} \right\rangle,$ where $t$ is a real number.

For the following exercises, find the unit vector in the direction of the given vector $\mathbf{\text{a}}$ and express it using standard unit vectors.

91.

$\mathbf{\text{a}} = 3\mathbf{\text{i}} - 4\mathbf{\text{j}}$

92\.

$\mathbf{\text{a}} = \left\langle {4,-3,6} \right\rangle$

93.

$\mathbf{\text{a}} = \overset{\rightarrow}{PQ},$ where $P\left( {-2,3,1} \right)$ and $Q\left( {0,-4,4} \right)$

94\.

$\mathbf{\text{a}} = \overset{\rightarrow}{OP},$ where $P\left( {-1,-1,1} \right)$

95.

$\mathbf{\text{a}} = \mathbf{\text{u}} - \mathbf{\text{v}} + \textbf{w},$ where $\mathbf{\text{u}} = \mathbf{\text{i}} - \mathbf{\text{j}} - \mathbf{\text{k}},$ $\mathbf{\text{v}} = 2\mathbf{\text{i}} - \mathbf{\text{j}} + \mathbf{\text{k}},$ and $\mathbf{\text{w}} = \text{−}\mathbf{\text{i}} + \mathbf{\text{j}} + 3\mathbf{\text{k}}$

96\.

$\mathbf{\text{a}} = 2\mathbf{\text{u}} + \mathbf{\text{v}} - \textbf{w},$ where $\mathbf{\text{u}} = \mathbf{\text{i}} - \mathbf{\text{k}},$ $\mathbf{\text{v}} = 2\mathbf{\text{j}},$ and $\mathbf{\text{w}} = \mathbf{\text{i}} - \mathbf{\text{j}}$

97.

Determine whether $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{PQ}$ are equivalent vectors, where $A\left( {1,1,1} \right),B\left( {3,3,3} \right),P\left( {1,4,5} \right),$ and $Q\left( {3,6,7} \right).$

98\.

Determine whether the vectors $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{PQ}$ are equivalent, where $A\left( {1,4,1} \right),$ $B\left( {-2,2,0} \right),$ $P\left( {2,5,7} \right),$ and $Q\left( {-3,2,1} \right).$

For the following exercises, find vector $\mathbf{\text{u}}$ with a magnitude that is given and satisfies the given conditions.

99.

$\mathbf{\text{v}} = \left\langle {7,-1,3} \right\rangle,$ $\left\| \mathbf{\text{u}} \right\| = 10,$ $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have the same direction

100\.

$\mathbf{\text{v}} = \left\langle {2,4,1} \right\rangle,$ $\left\| \mathbf{\text{u}} \right\| = 15,$ $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have the same direction

101.

$\mathbf{\text{v}} = \left\langle {2\ \text{sin}\ t,2\ \text{cos}\ t,1} \right\rangle,$ $\left\| \mathbf{\text{u}} \right\| = 2,$ $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have opposite directions for any $t,$ where $t$ is a real number

102\.

$\mathbf{\text{v}} = \left\langle {3\ \text{sinh}\ t,0,3} \right\rangle,$ $\left\| \mathbf{\text{u}} \right\| = 5,$ $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have opposite directions for any $t,$ where $t$ is a real number

103.

Determine a vector of magnitude $5$ in the direction of vector $\overset{\rightarrow}{AB},$ where $A(2,1,5)$ and $B(3,4,-7).$

104\.

Find a vector of magnitude $2$ that points in the opposite direction than vector $\overset{\rightarrow}{AB},$ where $A(-1,-1,1)$ and $B(0,1,1).$ Express the answer in component form.

105.

Consider the points $A\left( {2,\alpha,0} \right),B\left( {0,1,\beta} \right),$ and $C\left( {1,1,\beta} \right),$ where $\alpha$ and $\beta$ are negative real numbers. Find $\alpha$ and $\beta$ such that $\left\| {\overset{\rightarrow}{OA} - \overset{\rightarrow}{OB} + \overset{\rightarrow}{OC}} \right\| = \left\| \overset{\rightarrow}{OB} \right\| = 4.$

106\.

Consider points $A\left( {\alpha,0,0} \right),B\left( {0,\beta,0} \right),$ and $C\left( {\alpha,\beta,\beta} \right),$ where $\alpha$ and $\beta$ are positive real numbers. Find $\alpha$ and $\beta$ such that $\left\| {\overset{—}{OA} + \overset{—}{OB}} \right\| = \sqrt{2}\ \text{and}\ \left\| \overset{—}{OC} \right\| = \sqrt{3}.$

107\.

Let $P\left( {x,y,z} \right)$ be a point situated at an equal distance from points $A\left( {1,-1,0} \right)$ and $B\left( {-1,2,1} \right).$ Show that point $P$ lies on the plane of equation $-2x + 3y + z = 2.$

108\.

Let $P\left( {x,y,z} \right)$ be a point situated at an equal distance from the origin and point $A\left( {4,1,2} \right).$ Show that the coordinates of point $P$ satisfy the equation $8x + 2y + 4z = 21.$

109\.

The points $A,B,$ and $C$ are collinear (in this order) if the relation $\left\| \overset{\rightarrow}{AB} \right\| + \left\| \overset{\rightarrow}{BC} \right\| = \left\| \overset{\rightarrow}{AC} \right\|$ is satisfied. Show that $A(5,3,-1),$ $B(-5,-3,1),$ and $C(-15,-9,3)$ are collinear points.

110\.

Show that points $A(1,0,1),$ $B(0,1,1),$ and $C(1,1,1)$ are not collinear.

111.

\[T\] A force $\mathbf{\text{F}}$ of $50\ \text{N}$ acts on a particle in the direction of the vector $\overset{\rightarrow}{OP},$ where $P(3,4,0).$

1. Express the force as a vector in component form.

2. Find the angle between force $\mathbf{\text{F}}$ and the positive direction of the *x*-axis. Express the answer in degrees rounded to the nearest integer.

112\.

\[T\] A force $\mathbf{\text{F}}$ of $40\ \text{N}$ acts on a box in the direction of the vector $\overset{\rightarrow}{OP},$ where $P(1,0,2).$

1. Express the force as a vector by using standard unit vectors.

2. Find the angle between force $\mathbf{\text{F}}$ and the positive direction of the *x*-axis.

113.

If $\mathbf{\text{F}}$ is a force that moves an object from point $P_{1}\left( {x_{1},y_{1},z_{1}} \right)$ to another point $P_{2}\left( {x_{2},y_{2},z_{2}} \right),$ then the displacement vector is defined as $\textbf{D} = \left( {x_{2} - x_{1}} \right)\mathbf{\text{i}} + \left( {y_{2} - y_{1}} \right)\textbf{j} + \left( {z_{2} - z_{1}} \right)\textbf{k}.$ A metal container is lifted $10$ m vertically by a constant force $\mathbf{\text{F}}.$ Express the displacement vector $\mathbf{\text{D}}$ by using standard unit vectors.

114\.

A box is pulled $4$ yd horizontally in the *x*-direction by a constant force $\mathbf{\text{F}}.$ Find the displacement vector in component form.

115.

The sum of the forces acting on an object is called the *resultant* or *net force*. An object is said to be in static equilibrium if the resultant force of the forces that act on it is zero. Let $\mathbf{\text{F}}_{1} = \left\langle {10,6,3} \right\rangle,$ $\mathbf{\text{F}}_{2} = \left\langle {0,4,9} \right\rangle,$ and $\mathbf{\text{F}}_{3} = \left\langle {10,-3,-9} \right\rangle$ be three forces acting on a box. Find the force $\mathbf{\text{F}}_{4}$ acting on the box such that the box is in static equilibrium. Express the answer in component form.

116\.

\[T\] Let $\mathbf{\text{F}}_{k} = \left\langle {1,k,k^{2}} \right\rangle,$ $k = 1\text{,...},n$ be $n$ forces acting on a particle, with $n \geq 2.$

1. Find the net force $\mathbf{\text{F}} = {\sum\limits_{k = 1}^{n}F_{k}}.$ Express the answer using standard unit vectors.

2. Use a computer algebra system (CAS) to find *n* such that $\left\| \textbf{F} \right\| < 100.$

117.

The force of gravity $\mathbf{\text{F}}$ acting on an object is given by $\mathbf{\text{F}} = m\mathbf{\text{g}},$ where *m* is the mass of the object (expressed in kilograms) and $\mathbf{\text{g}}$ is acceleration resulting from gravity, with $\left\| \textbf{g} \right\| = 9.8$ $\text{N/kg}.$ A 2-kg disco ball hangs by a chain from the ceiling of a room.

1. Find the force of gravity $\mathbf{\text{F}}$ acting on the disco ball and find its magnitude.

2. Find the force of tension $\mathbf{\text{T}}$ in the chain and its magnitude.

Express the answers using standard unit vectors.

118\.

A 5-kg pendant chandelier is designed such that the alabaster bowl is held by four chains of equal length, as shown in the following figure.

1. Find the magnitude of the force of gravity acting on the chandelier.

2. Find the magnitudes of the forces of tension for each of the four chains (assume chains are essentially vertical).

119.

\[T\] A 30-kg block of cement is suspended by three cables of equal length that are anchored at points $P(-2,0,0),$ $Q\left( {1,\sqrt{3},0} \right),$ and $R\left( {1,\text{−}\sqrt{3},0} \right).$ The load is located at $S\left( {0,0,-2\sqrt{3}} \right),$ as shown in the following figure. Let $\mathbf{\text{F}}_{1},$ $\mathbf{\text{F}}_{2},$ and $\mathbf{\text{F}}_{3}$ be the forces of tension resulting from the load in cables $RS,QS,$ and $PS,$ respectively.

1. Find the gravitational force $\mathbf{\text{F}}$ acting on the block of cement that counterbalances the sum $\mathbf{\text{F}}_{1} + \textbf{F}_{2} + \textbf{F}_{3}$ of the forces of tension in the cables.

2. Find forces $\mathbf{\text{F}}_{1},$ $\mathbf{\text{F}}_{2},$ and $\mathbf{\text{F}}_{3}.$ Express the answer in component form.

120\.

Two soccer players are practicing for an upcoming game. One of them runs 10 m from point *A* to point *B*. She then turns left at $90\text{°}$ and runs 10 m until she reaches point *C*. Then she kicks the ball with a speed of 10 m/sec at an upward angle of $45\text{°}$ to her teammate, who is located at point *A*. Write the velocity of the ball in component form.

121.

Let $\textbf{r}(t) = \left\langle {x(t),y(t),z(t)} \right\rangle$ be the position vector of a particle at the time $t \in \lbrack 0,T\rbrack,$ where $x,y,$ and $z$ are smooth functions on $\lbrack 0,T\rbrack.$ The instantaneous velocity of the particle at time $t$ is defined by vector $\mathbf{\text{v}}(t) = \left\langle {x\text{'}(t),y\text{'}(t),z\text{'}(t)} \right\rangle,$ with components that are the derivatives with respect to $t,$ of the functions *x*, *y*, and *z*, respectively. The magnitude $\left\| {\mathbf{\text{v}}(t)} \right\|$ of the instantaneous velocity vector is called the *speed of the particle at time* t. Vector $\mathbf{\text{a}}(t) = \left\langle {x^{''}(t),y^{''}(t),z^{''}(t)} \right\rangle,$ with components that are the second derivatives with respect to $t,$ of the functions $x,y,$ and $z,$ respectively, gives the acceleration of the particle at time $t.$ Consider $\textbf{r}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t,2t} \right\rangle$ the position vector of a particle at time $t \in \lbrack 0,30\rbrack,$ where the components of $\mathbf{\text{r}}$ are expressed in centimeters and time is expressed in seconds.

1. Find the instantaneous velocity, speed, and acceleration of the particle after the first second. Round your answer to two decimal places.

2. Use a CAS to visualize the path of the particle—that is, the set of all points of coordinates $\left( {\text{cos}\ t,\text{sin}\ t,2t} \right),$ where $t \in \lbrack 0,30\rbrack.$

122\.

\[T\] Let $\textbf{r}(t) = \left\langle {t,2t^{2},4t^{2}} \right\rangle$ be the position vector of a particle at time $t$ (in seconds), where $t \in \lbrack 0,10\rbrack$ (here the components of $\mathbf{\text{r}}$ are expressed in centimeters).

1. Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Round your answer to two decimal places.

2. Use a CAS to visualize the path of the particle defined by the points $\left( {t,2t^{2},4t^{2}} \right),$ where $t \in \lbrack 0,60\rbrack.$

---

2.3 The Dot Product

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-3-the-dot-product

2.3 The Dot Product

• 2.3.1 Calculate the dot product of two given vectors.
• 2.3.2 Determine whether two given vectors are perpendicular.
• 2.3.3 Find the direction cosines of a given vector.
• 2.3.4 Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it.
• 2.3.5 Calculate the work done by a given force.

If we apply a force to an object so that the object moves, we say that *work* is done by the force. In Introduction to Applications of Integration on integration applications, we looked at a constant force and we assumed the force was applied in the direction of motion of the object. Under those conditions, work can be expressed as the product of the force acting on an object and the distance the object moves. In this chapter, however, we have seen that both force and the motion of an object can be represented by vectors.

In this section, we develop an operation called the *dot product*, which allows us to calculate work in the case when the force vector and the motion vector have different directions. The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector. The dot product can also help us measure the angle formed by a pair of vectors and the position of a vector relative to the coordinate axes. It even provides a simple test to determine whether two vectors meet at a right angle.

The Dot Product and Its Properties

We have already learned how to add and subtract vectors. In this chapter, we investigate two types of vector multiplication. The first type of vector multiplication is called the dot product, based on the notation we use for it, and it is defined as follows:

The dot product of vectors $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ is given by the sum of the products of the components

$$\mathbf{\text{u}} \cdot \mathbf{\text{v}} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}.$$ (2.3)

Note that if $\textbf{u}$ and $\textbf{v}$ are two-dimensional vectors, we calculate the dot product in a similar fashion. Thus, if $\mathbf{\text{u}} = \left\langle {u_{1},u_{2}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2}} \right\rangle,$ then

$$\mathbf{\text{u}} \cdot \mathbf{\text{v}} = u_{1}v_{1} + u_{2}v_{2}.$$

When two vectors are combined under addition or subtraction, the result is a vector. When two vectors are combined using the dot product, the result is a scalar. For this reason, the dot product is often called the *scalar product*. It may also be called the *inner product*.

Calculating Dot Products

1. Find the dot product of $\mathbf{\text{u}} = \left\langle {3,5,2} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {-1,3,0} \right\rangle.$

2. Find the scalar product of $\mathbf{\text{p}} = 10\mathbf{\text{i}} - 4\mathbf{\text{j}} + 7\mathbf{\text{k}}$ and $\mathbf{\text{q}} = -2\mathbf{\text{i}} + \mathbf{\text{j}} + 6\mathbf{\text{k}}.$

Solution

1. Substitute the vector components into the formula for the dot product:

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \mathbf{\text{v}}} & {= u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}} \\

& {= 3(-1) + 5(3) + 2(0) = -3 + 15 + 0 = 12.}

\end{array}$$

2. The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:

$$\begin{array}{cl}

{\mathbf{\text{p}} \cdot \mathbf{\text{q}}} & {= u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}} \\

& {= 10(-2) + (-4)(1) + (7)(6) = -20 - 4 + 42 = 18.}

\end{array}$$

Find $\mathbf{\text{u}} \cdot \mathbf{\text{v}},$ where $\mathbf{\text{u}} = \left\langle {2,9,-1} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {-3,1,-4} \right\rangle.$

Like vector addition and subtraction, the dot product has several algebraic properties. We prove three of these properties and leave the rest as exercises.

Properties of the Dot Product

Let $\mathbf{\text{u}},$ $\mathbf{\text{v}},$ and $\mathbf{\text{w}}$ be vectors, and let *c* be a scalar.

$$\begin{array}{lccrllccl}

\text{i.} & & & {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} & = & {\mathbf{\text{v}} \cdot \mathbf{\text{u}}} & & & \text{Commutative property} \\

\text{ii.} & & & {\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}} + \mathbf{\text{w}}} \right)} & = & {\mathbf{\text{u}} \cdot \mathbf{\text{v}} + \mathbf{\text{u}} \cdot \mathbf{\text{w}}} & & & \text{Distributive property} \\

\text{iii.} & & & {c\left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right)} & = & {\left( {c\mathbf{\text{u}}} \right) \cdot \mathbf{\text{v}} = \mathbf{\text{u}} \cdot \left( {c\mathbf{\text{v}}} \right)} & & & \text{Associative property} \\

\text{iv.} & & & {\mathbf{\text{v}} \cdot \mathbf{\text{v}}} & = & \left\| \mathbf{\text{v}} \right\|^{2} & & & \text{Property of magnitude}

\end{array}$$

Proof

Let $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle.$ Then

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \mathbf{\text{v}}} & {= \left\langle {u_{1},u_{2},u_{3}} \right\rangle \cdot \left\langle {v_{1},v_{2},v_{3}} \right\rangle} \\

& {= u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}} \\

& {= v_{1}u_{1} + v_{2}u_{2} + v_{3}u_{3}} \\

& {= \left\langle {v_{1},v_{2},v_{3}} \right\rangle \cdot \left\langle {u_{1},u_{2},u_{3}} \right\rangle} \\

& {= \mathbf{\text{v}} \cdot \mathbf{\text{u}}.}

\end{array}$$

The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:

$$\begin{array}{cl}

{c\left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right)} & {= c\left( {u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}} \right)} \\

& {= c\left( {u_{1}v_{1}} \right) + c\left( {u_{2}v_{2}} \right) + c\left( {u_{3}v_{3}} \right)} \\

& {= \left( {cu_{1}} \right)v_{1} + \left( {cu_{2}} \right)v_{2} + \left( {cu_{3}} \right)v_{3}} \\

& {= \left\langle {cu_{1},cu_{2},cu_{3}} \right\rangle \cdot \left\langle {v_{1},v_{2},v_{3}} \right\rangle} \\

& {= c\left\langle {u_{1},u_{2},u_{3}} \right\rangle \cdot \left\langle {v_{1},v_{2},v_{3}} \right\rangle} \\

& {= (c\mathbf{\text{u}}) \cdot \mathbf{\text{v}}.}

\end{array}$$

The proof that $c\left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right) = \mathbf{\text{u}} \cdot \left( {c\mathbf{\text{v}}} \right)$ is similar.

The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:

$$\begin{array}{cl}

{\mathbf{\text{v}} \cdot \mathbf{\text{v}}} & {= \left\langle {v_{1},v_{2},v_{3}} \right\rangle \cdot \left\langle {v_{1},v_{2},v_{3}} \right\rangle} \\

& {= \left( v_{1} \right)^{2} + \left( v_{2} \right)^{2} + \left( v_{3} \right)^{2}} \\

& {= \left\lbrack \sqrt{\left( v_{1} \right)^{2} + \left( v_{2} \right)^{2} + \left( v_{3} \right)^{2}} \right\rbrack^{2}} \\

& {= \left\| \mathbf{\text{v}} \right\|^{2}.}

\end{array}$$

□

Note that the definition of the dot product yields $\mathbf{0} \cdot \mathbf{\text{v}} = 0.$ By property iv., if $\mathbf{\text{v}} \cdot \mathbf{\text{v}} = 0,$ then $\mathbf{\text{v}} = \mathbf{0}.$

Using Properties of the Dot Product

Let $\mathbf{\text{a}} = \left\langle {1,2,-3} \right\rangle,$ $\mathbf{\text{b}} = \left\langle {0,2,4} \right\rangle,$ and $\mathbf{\text{c}} = \left\langle {5,-1,3} \right\rangle.$ Find each of the following products.

1. $\left( {\mathbf{\text{a}} \cdot \mathbf{\text{b}}} \right)\mathbf{\text{c}}$

2. $\mathbf{\text{a}} \cdot \left( {2\mathbf{\text{c}}} \right)$

3. $\left\| \mathbf{\text{b}} \right\|^{2}$

Solution

1. Note that this expression asks for the scalar multiple of c by $\mathbf{\text{a}} \cdot \mathbf{\text{b}}\text{:}$

$$\begin{array}{cl}

{\left( {\mathbf{\text{a}} \cdot \mathbf{\text{b}}} \right)\mathbf{\text{c}}} & {= \left( {\left\langle {1,2,-3} \right\rangle \cdot \left\langle {0,2,4} \right\rangle} \right)\left\langle {5,-1,3} \right\rangle} \\

& {= \left( {1(0) + 2(2) + (-3)(4)} \right)\left\langle {5,-1,3} \right\rangle} \\

& {= -8\left\langle {5,-1,3} \right\rangle} \\

& {= \left\langle {-40,8,-24} \right\rangle.}

\end{array}$$

2. This expression is a dot product of vector a and scalar multiple 2c:

$$\begin{array}{cl}

{\mathbf{\text{a}} \cdot \left( {2\mathbf{\text{c}}} \right)} & {= 2\left( {\mathbf{\text{a}} \cdot \mathbf{\text{c}}} \right)} \\

& {= 2\left( {\left\langle {1,2,-3} \right\rangle \cdot \left\langle {5,-1,3} \right\rangle} \right)} \\

& {= 2\left( {1(5) + 2(-1) + (-3)(3)} \right)} \\

& {= 2(-6) = -12.}

\end{array}$$

3. Simplifying this expression is a straightforward application of the dot product:

$$\left\| \mathbf{\text{b}} \right\|^{2} = \mathbf{\text{b}} \cdot \mathbf{\text{b}} = \left\langle {0,2,4} \right\rangle \cdot \left\langle {0,2,4} \right\rangle = 0^{2} + 2^{2} + 4^{2} = 0 + 4 + 16 = 20.$$

Find the following products for $\mathbf{\text{p}} = \left\langle {7,0,2} \right\rangle,$ $\mathbf{\text{q}} = \left\langle {-2,2,-2} \right\rangle,$ and $\mathbf{\text{r}} = \left\langle {0,2,-3} \right\rangle.$

1. $\left( {\mathbf{\text{r}} \cdot \mathbf{\text{p}}} \right)\mathbf{\text{q}}$

2. $\left\| \mathbf{\text{p}} \right\|^{2}$

Using the Dot Product to Find the Angle between Two Vectors

When two nonzero vectors are placed in standard position, whether in two dimensions or three dimensions, they form an angle between them (Figure 2.44). The dot product provides a way to find the measure of this angle. This property is a result of the fact that we can express the dot product in terms of the cosine of the angle formed by two vectors.

Evaluating a Dot Product

The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle between them:

$$\mathbf{\text{u}} \cdot \mathbf{\text{v}} = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta.$$ (2.4)

Proof

Place vectors $\textbf{u}$ and $\textbf{v}$ in standard position and consider the vector $\mathbf{\text{v}} - \mathbf{\text{u}}$ (Figure 2.45). These three vectors form a triangle with side lengths $\left\| \mathbf{\text{u}} \right\|,\left\| \mathbf{\text{v}} \right\|,\ \text{and}\ \left\| {\mathbf{\text{v}} - \mathbf{\text{u}}} \right\|.$

Recall from trigonometry that the law of cosines describes the relationship among the side lengths of the triangle and the angle *θ*. Applying the law of cosines here gives

$$\left\| {\mathbf{\text{v}} - \mathbf{\text{u}}} \right\|^{2} = \left\| \mathbf{\text{u}} \right\|^{2} + \left\| \mathbf{\text{v}} \right\|^{2} - 2\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta.$$

The dot product provides a way to rewrite the left side of this equation:

$$\begin{array}{cl}

\left\| {\mathbf{\text{v}} - \mathbf{\text{u}}} \right\|^{2} & {= \left( {\mathbf{\text{v}} - \mathbf{\text{u}}} \right) \cdot \left( {\mathbf{\text{v}} - \mathbf{\text{u}}} \right)} \\

& {= \left( {\mathbf{\text{v}} - \mathbf{\text{u}}} \right) \cdot \mathbf{\text{v}} - \left( {\mathbf{\text{v}} - \mathbf{\text{u}}} \right) \cdot \mathbf{\text{u}}} \\

& {= \mathbf{\text{v}} \cdot \mathbf{\text{v}} - \mathbf{\text{u}} \cdot \mathbf{\text{v}} - \mathbf{\text{v}} \cdot \mathbf{\text{u}} + \mathbf{\text{u}} \cdot \mathbf{\text{u}}} \\

& {= \mathbf{\text{v}} \cdot \mathbf{\text{v}} - \mathbf{\text{u}} \cdot \mathbf{\text{v}} - \mathbf{\text{u}} \cdot \mathbf{\text{v}} + \mathbf{\text{u}} \cdot \mathbf{\text{u}}} \\

& {= \left\| \mathbf{\text{v}} \right\|^{2} - 2\mathbf{\text{u}} \cdot \mathbf{\text{v}} + \left\| \mathbf{\text{u}} \right\|^{2}.}

\end{array}$$

Substituting into the law of cosines yields

$$\begin{array}{rll}

\left\| {\mathbf{\text{v}} - \mathbf{\text{u}}} \right\|^{2} & = & {\left\| \mathbf{\text{u}} \right\|^{2} + \left\| \mathbf{\text{v}} \right\|^{2} - 2\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta} \\

{\left\| \mathbf{\text{v}} \right\|^{2} - 2\mathbf{\text{u}} \cdot \mathbf{\text{v}} + \left\| \mathbf{\text{u}} \right\|^{2}} & = & {\left\| \mathbf{\text{u}} \right\|^{2} + \left\| \mathbf{\text{v}} \right\|^{2} - 2\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta} \\

{- 2\mathbf{\text{u}} \cdot \mathbf{\text{v}}} & = & {-2\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta} \\

{\mathbf{\text{u}} \cdot \mathbf{\text{v}}} & = & {\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta.}

\end{array}$$

□

We can use this form of the dot product to find the measure of the angle between two nonzero vectors. The following equation rearranges Equation 2.3 to solve for the cosine of the angle:

$$\text{cos}\ \theta = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|}.$$ (2.5)

Using this equation, we can find the cosine of the angle between two nonzero vectors. Since we are considering the smallest angle between the vectors, we assume $0\text{°} \leq \theta \leq 180\text{°}$ (or $0 \leq \theta \leq \pi$ if we are working in radians). The inverse cosine is unique over this range, so we are then able to determine the measure of the angle $\theta.$

Finding the Angle between Two Vectors

Find the measure of the angle between each pair of vectors.

1. i + j + k and 2i – j – 3k

2. $\left\langle {2,5,6} \right\rangle$ and $\left\langle {-2,-4,4} \right\rangle$

Solution

1. To find the cosine of the angle formed by the two vectors, substitute the components of the vectors into Equation 2.5:

$$\begin{array}{cl}

{\text{cos}\ \theta} & {= \frac{(\mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}) \cdot \left( {2\mathbf{\text{i}} - \mathbf{\text{j}} - 3\mathbf{\text{k}}} \right)}{\left\| {\mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}} \right\| \cdot \left\| {2\mathbf{\text{i}} - \mathbf{\text{j}} - 3\mathbf{\text{k}}} \right\|}} \\

& {= \frac{1(2) + (1)(-1) + (1)(-3)}{\sqrt{1^{2} + 1^{2} + 1^{2}}\ \sqrt{2^{2} + {(-1)}^{2} + {(-3)}^{2}}}} \\

& {= \frac{-2}{\sqrt{3}\ \sqrt{14}} = \frac{-2}{\sqrt{42}}.}

\end{array}$$

Therefore, $\theta = \text{arccos}\ \frac{-2}{\sqrt{42}}$.

2. Start by finding the value of the cosine of the angle between the vectors:

$$\begin{array}{cl}

{\text{cos}\ \theta} & {= \frac{\left\langle {2,5,6} \right\rangle \cdot \left\langle {-2,-4,4} \right\rangle}{\left\| \left\langle {2,5,6} \right\rangle \right\| \cdot \left\| \left\langle {-2,-4,4} \right\rangle \right\|}} \\

& {= \frac{2(-2) + (5)(-4) + (6)(4)}{\sqrt{2^{2} + 5^{2} + 6^{2}}\ \sqrt{{(-2)}^{2} + {(-4)}^{2} + 4^{2}}}} \\

& {= \frac{0}{\sqrt{65}\ \sqrt{36}} = 0.}

\end{array}$$

Now, $\text{cos}\ \theta = 0$ and $0 \leq \theta \leq \pi,$ so $\theta = {\pi\text{/}2.}$

Find the measure of the angle, in radians, formed by vectors $\mathbf{\text{a}} = \left\langle {1,2,0} \right\rangle$ and $\mathbf{\text{b}} = \left\langle {2,4,1} \right\rangle.$ Round to the nearest hundredth.

The angle between two vectors can be acute $\left( {0 < \text{cos}\ \theta < 1} \right),$ obtuse $\left( {-1 < \text{cos}\ \theta < 0} \right),$ or straight $\left( {\text{cos}\ \theta = -1} \right).$ If $\text{cos}\ \theta = 1,$ then both vectors have the same direction. If $\text{cos}\ \theta = 0,$ then the vectors, when placed in standard position, form a right angle (Figure 2.46). We can formalize this result into a theorem regarding orthogonal (perpendicular) vectors.

Orthogonal Vectors

The nonzero vectors $\textbf{u}$ and $\textbf{v}$ are orthogonal vectors if and only if $\mathbf{\text{u}} \cdot \mathbf{\text{v}} = 0.$

Proof

Let $\textbf{u}$ and $\textbf{v}$ be nonzero vectors, and let $\theta$ denote the angle between them. First, assume $\mathbf{\text{u}} \cdot \mathbf{\text{v}} = 0.$ Then

$$\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta = 0.$$

However, $\left\| \mathbf{\text{u}} \right\| \neq 0$ and $\left\| \mathbf{\text{v}} \right\| \neq 0,$ so we must have $\text{cos}\ \theta = 0.$ Hence, $\theta = 90\text{°},$ and the vectors are orthogonal.

Now assume $\textbf{u}$ and $\textbf{v}$ are orthogonal. Then $\theta = 90\text{°}$ and we have

$$\mathbf{\text{u}} \cdot \mathbf{\text{v}} = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ 90\text{°} = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|(0) = 0.$$

□

The terms *orthogonal*, *perpendicular*, and *normal* each indicate that mathematical objects are intersecting at right angles. The use of each term is determined mainly by its context. We say that vectors are orthogonal and lines are perpendicular. The term *normal* is used most often when measuring the angle made with a plane or other surface.

Identifying Orthogonal Vectors

Determine whether $\mathbf{\text{p}} = \left\langle {1,0,5} \right\rangle$ and $\mathbf{\text{q}} = \left\langle {10,3,-2} \right\rangle$ are orthogonal vectors.

Solution

Using the definition, we need only check the dot product of the vectors:

$$\mathbf{\text{p}} \cdot \mathbf{\text{q}} = 1(10) + (0)(3) + (5)(-2) = 10 + 0 - 10 = 0.$$

Because $\mathbf{\text{p}} \cdot \mathbf{\text{q}} = 0,$ the vectors are orthogonal (Figure 2.47).

For which value of *x* is $\mathbf{\text{p}} = \left\langle {2,8,-1} \right\rangle$ orthogonal to $\mathbf{\text{q}} = \left\langle {x,-1,2} \right\rangle?$

Measuring the Angle Formed by Two Vectors

Let $\mathbf{\text{v}} = \left\langle {2,3,3} \right\rangle.$ Find the measures of the angles formed by the following vectors.

1. $\textbf{v}$ and i

2. $\textbf{v}$ and j

3. $\textbf{v}$ and k

Solution

1. Let *α* be the angle formed by $\textbf{v}$ and i:

$$\begin{array}{cl}

{\text{cos}\ \alpha} & {= \frac{\mathbf{\text{v}} \cdot \mathbf{\text{i}}}{\left\| \mathbf{\text{v}} \right\| \cdot \left\| \mathbf{\text{i}} \right\|}} \\

& {= \frac{\left\langle {2,3,3} \right\rangle \cdot \left\langle {1,0,0} \right\rangle}{\sqrt{2^{2} + 3^{2} + 3^{2}}\ \sqrt{1}}} \\

& {= \frac{2}{\sqrt{22}}.}

\end{array}$$

$$\alpha = \text{arccos}\ \frac{2}{\sqrt{22}} \approx 1.130\ \text{rad}.$$

2. Let *β* represent the angle formed by $\textbf{v}$ and j:

$$\begin{array}{cl}

{\text{cos}\ \beta} & {= \frac{\mathbf{\text{v}} \cdot \mathbf{\text{j}}}{\left\| \mathbf{\text{v}} \right\| \cdot \left\| \mathbf{\text{j}} \right\|}} \\

& {= \frac{\left\langle {2,3,3} \right\rangle \cdot \left\langle {0,1,0} \right\rangle}{\sqrt{2^{2} + 3^{2} + 3^{2}}\ \sqrt{1}}} \\

& {= \frac{3}{\sqrt{22}}.}

\end{array}$$

$$\beta = \text{arccos}\ \frac{3}{\sqrt{22}} \approx 0.877\ \text{rad.}$$

3. Let *γ* represent the angle formed by $\textbf{v}$ and k:

$$\begin{array}{cl}

{\text{cos}\ \gamma} & {= \frac{\mathbf{\text{v}} \cdot \mathbf{\text{k}}}{\left\| \mathbf{\text{v}} \right\| \cdot \left\| \mathbf{\text{k}} \right\|}} \\

& {= \frac{\left\langle {2,3,3} \right\rangle \cdot \left\langle {0,0,1} \right\rangle}{\sqrt{2^{2} + 3^{2} + 3^{2}}\ \sqrt{1}}} \\

& {= \frac{3}{\sqrt{22}}.}

\end{array}$$

$$\gamma = \text{arccos}\ \frac{3}{\sqrt{22}} \approx 0.877\ \text{rad.}$$

Let $\mathbf{\text{v}} = \left\langle {3,-5,1} \right\rangle.$ Find the measure of the angles formed by each pair of vectors.

1. $\textbf{v}$ and i

2. $\textbf{v}$ and j

3. $\textbf{v}$ and k

The angle a vector makes with each of the coordinate axes, called a direction angle, is very important in practical computations, especially in a field such as engineering. For example, in astronautical engineering, the angle at which a rocket is launched must be determined very precisely. A very small error in the angle can lead to the rocket going hundreds of miles off course. Direction angles are often calculated by using the dot product and the cosines of the angles, called the direction cosines. Therefore, we define both these angles and their cosines.

The angles formed by a nonzero vector and the coordinate axes are called the direction angles for the vector (Figure 2.48). The cosines for these angles are called the direction cosines.

In Example 2.25, the direction cosines of $\mathbf{\text{v}} = \left\langle {2,3,3} \right\rangle$ are $\text{cos}\ \alpha = \frac{2}{\sqrt{22}},$ $\text{cos}\ \beta = \frac{3}{\sqrt{22}},$ and $\text{cos}\ \gamma = \frac{3}{\sqrt{22}}.$ The direction angles of $\textbf{v}$ are $\alpha = 1.130\ \text{rad},$ $\beta = 0.877\ \text{rad},$ and $\gamma = 0.877\ \text{rad}.$

So far, we have focused mainly on vectors related to force, movement, and position in three-dimensional physical space. However, vectors are often used in more abstract ways. For example, suppose a fruit vendor sells apples, bananas, and oranges. On a given day, he sells 30 apples, 12 bananas, and 18 oranges. He might use a quantity vector, $\mathbf{\text{q}} = \left\langle {30,12,18} \right\rangle,$ to represent the quantity of fruit he sold that day. Similarly, he might want to use a price vector, $\mathbf{\text{p}} = \left\langle {0.50,0.25,1} \right\rangle,$ to indicate that he sells his apples for 50¢ each, bananas for 25¢ each, and oranges for \$1 apiece. In this example, although we could still graph these vectors, we do not interpret them as literal representations of position in the physical world. We are simply using vectors to keep track of particular pieces of information about apples, bananas, and oranges.

This idea might seem a little strange, but if we simply regard vectors as a way to order and store data, we find they can be quite a powerful tool. Going back to the fruit vendor, let’s think about the dot product, $\mathbf{\text{q}} \cdot \mathbf{\text{p}}.$ We compute it by multiplying the number of apples sold (30) by the price per apple (50¢), the number of bananas sold by the price per banana, and the number of oranges sold by the price per orange. We then add all these values together. So, in this example, the dot product tells us how much money the fruit vendor had in sales on that particular day.

When we use vectors in this more general way, there is no reason to limit the number of components to three. What if the fruit vendor decides to start selling grapefruit? In that case, he would want to use four-dimensional quantity and price vectors to represent the number of apples, bananas, oranges, and grapefruit sold, and their unit prices. As you might expect, to calculate the dot product of four-dimensional vectors, we simply add the products of the components as before, but the sum has four terms instead of three.

Using Vectors in an Economic Context

AAA Party Supply Store sells invitations, party favors, decorations, and food service items such as paper plates and napkins. When AAA buys its inventory, it pays 25¢ per package for invitations and party favors. Decorations cost AAA 50¢ each, and food service items cost 20¢ per package. AAA sells invitations for \$2.50 per package and party favors for \$1.50 per package. Decorations sell for \$4.50 each and food service items for \$1.25 per package.

During the month of May, AAA Party Supply Store sells 1258 invitations, 342 party favors, 2426 decorations, and 1354 food service items. Use vectors and dot products to calculate how much money AAA made in sales during the month of May. How much did the store make in profit?

Solution

The cost, price, and quantity vectors are

$$\begin{array}{l}

{\mathbf{\text{c}} = \left\langle {0.25,0.25,0.50,0.20} \right\rangle} \\

{\mathbf{\text{p}} = \left\langle {2.50,1.50,4.50,1.25} \right\rangle} \\

{\mathbf{\text{q}} = \left\langle {1258,342,2426,1354} \right\rangle.}

\end{array}$$

AAA sales for the month of May can be calculated using the dot product $\mathbf{\text{p}} \cdot \mathbf{\text{q}}.$ We have

$$\begin{array}{cl}

{\mathbf{\text{p}} \cdot \mathbf{\text{q}}} & {= \left\langle {2.50,1.50,4.50,1.25} \right\rangle \cdot \left\langle {1258,342,2426,1354} \right\rangle} \\

& {= 3145 + 513 + 10917 + 1692.5} \\

& {= 16267.5.}

\end{array}$$

So, AAA took in \$16,267.50 during the month of May.

To calculate the profit, we must first calculate how much AAA paid for the items sold. We use the dot product $\mathbf{\text{c}} \cdot \mathbf{\text{q}}$ to get

$$\begin{array}{cl}

{\mathbf{\text{c}} \cdot \mathbf{\text{q}}} & {= \left\langle {0.25,0.25,0.50,0.20} \right\rangle \cdot \left\langle {1258,342,2426,1354} \right\rangle} \\

& {= 314.5 + 85.5 + 1213 + 270.8} \\

& {= 1883.8.}

\end{array}$$

So, AAA paid \$1,883.80 for the items they sold. Their profit, then, is given by

$$\begin{array}{cl}

{\mathbf{\text{p}} \cdot \mathbf{\text{q}} - \mathbf{\text{c}} \cdot \mathbf{\text{q}}} & {= 16267.5 - 1883.8} \\

& {= 14383.7.}

\end{array}$$

Therefore, AAA Party Supply Store made \$14,383.70 in May.

On June 1, AAA Party Supply Store decided to increase the price they charge for party favors to \$2 per package. They also changed suppliers for their invitations, and are now able to purchase invitations for only 10¢ per package. All their other costs and prices remain the same. If AAA sells 1408 invitations, 147 party favors, 2112 decorations, and 1894 food service items in the month of June, use vectors and dot products to calculate their total sales and profit for June.

Projections

As we have seen, addition combines two vectors to create a resultant vector. But what if we are given a vector and we need to find its component parts? We use vector projections to perform the opposite process; they can break down a vector into its components. The magnitude of a vector projection is a scalar projection. For example, if a child is pulling the handle of a wagon at a 55° angle, we can use projections to determine how much of the force on the handle is actually moving the wagon forward (Figure 2.49). We return to this example and learn how to solve it after we see how to calculate projections.

The vector projection of $\textbf{v}$ onto $\textbf{u}$ is the vector labeled proj_uv in Figure 2.50. It has the same initial point as $\textbf{u}$ and $\textbf{v}$ and the same direction as $\textbf{u}$, and represents the component of $\textbf{v}$ that acts in the direction of $\textbf{u}$. If $\theta$ represents the angle between $\textbf{u}$ and $\textbf{v}$, then, by properties of triangles, we know the length of $\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}$ is $\left\| \text{proj}_{\mathbf{\text{u}}}\textbf{v} \right\| = {\left\| \textbf{v} \right\| \cdot}\left| \text{cos} \right|\ \theta.$ Note that when the angle $\theta$ between $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ is an obtuse angle, the projection will be in the opposite direction of $\mathbf{\text{u}}$. When expressing $\text{cos}\ \theta$ in terms of the dot product, this becomes

$$\begin{matrix}

\left\| \text{proj}_{\mathbf{\text{u}}}\textbf{v} \right\| & {= {\left\| \textbf{v} \right\| \cdot}\left| \text{cos} \right|\ \theta} \\

& {= \left\| \textbf{v} \right\|\left( \frac{\left| \textbf{u} \cdot \textbf{v} \right|}{\left\| \textbf{u} \right\|\left\| \textbf{v} \right\|} \right)} \\

& {= \frac{\left| \textbf{u} \cdot \textbf{v} \right|}{\left\| \textbf{u} \right\|}.}

\end{matrix}$$

We now multiply by a unit vector in the direction of $\textbf{u}$ to get $\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}\text{:}$

$$\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}} = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|}\left( {\frac{1}{\left\| \mathbf{\text{u}} \right\|}\mathbf{\text{u}}} \right) = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}.$$ (2.6)

The length of this vector is also known as the scalar projection of $\textbf{v}$ onto $\textbf{u}$ and is denoted by

$$\text{comp}_{\mathbf{\text{u}}}\textbf{v} = \frac{\textbf{u} \cdot \textbf{v}}{\left\| \textbf{u} \right\|}.$$ (2.7)

Finding Projections

Find the projection of $\textbf{v}$ onto u.

1. $\mathbf{\text{v}} = \left\langle {3,5,1} \right\rangle$ and $\mathbf{\text{u}} = \left\langle {-1,4,3} \right\rangle$

2. $\mathbf{\text{v}} = 3\mathbf{\text{i}} - 2\mathbf{\text{j}}$ and $\mathbf{\text{u}} = \mathbf{\text{i}} + 6\mathbf{\text{j}}$

Solution

1. Substitute the components of $\textbf{v}$ and $\textbf{u}$ into the formula for the projection:

$$\begin{array}{cl}

{\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}} & {= \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}} \\

& {= \frac{\left\langle {-1,4,3} \right\rangle \cdot \left\langle {3,5,1} \right\rangle}{\left\| \left\langle {-1,4,3} \right\rangle \right\|^{2}}\left\langle {-1,4,3} \right\rangle} \\

& {= \frac{-3 + 20 + 3}{(-1)^{2} + 4^{2} + 3^{2}}\left\langle {-1,4,3} \right\rangle} \\

& {= \frac{20}{26}\left\langle {-1,4,3} \right\rangle} \\

& {= \left\langle {- \frac{10}{13},\frac{40}{13},\frac{30}{13}} \right\rangle.}

\end{array}$$

2. To find the two-dimensional projection, simply adapt the formula to the two-dimensional case:

$$\begin{array}{cl}

{\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}} & {= \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}} \\

& {= \frac{\left( {\mathbf{\text{i}} + 6\mathbf{\text{j}}} \right) \cdot \left( {3\mathbf{\text{i}} - 2\mathbf{\text{j}}} \right)}{\left\| {\mathbf{\text{i}} + 6\mathbf{\text{j}}} \right\|^{2}}\left( {\mathbf{\text{i}} + 6\mathbf{\text{j}}} \right)} \\

& {= \frac{1(3) + 6(-2)}{1^{2} + 6^{2}}\left( {\mathbf{\text{i}} + 6\mathbf{\text{j}}} \right)} \\

& {= - \frac{9}{37}\left( {\mathbf{\text{i}} + 6\mathbf{\text{j}}} \right)} \\

& {= - \frac{9}{37}\mathbf{\text{i}} - \frac{54}{37}\mathbf{\text{j}}.}

\end{array}$$

Sometimes it is useful to decompose vectors—that is, to break a vector apart into a sum. This process is called the *resolution of a vector into components*. Projections allow us to identify two orthogonal vectors having a desired sum. For example, let $\mathbf{\text{v}} = \left\langle {6,-4} \right\rangle$ and let $\mathbf{\text{u}} = \left\langle {3,1} \right\rangle.$ We want to decompose the vector $\textbf{v}$ into orthogonal components such that one of the component vectors has the same direction as $\textbf{u}$.

We first find the component that has the same direction as $\textbf{u}$ by projecting $\textbf{v}$ onto $\textbf{u}$. Let $\mathbf{\text{p}} = \text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}.$ Then, we have

$$\begin{array}{cl}

\mathbf{\text{p}} & {= \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}} \\

& {= \frac{18 - 4}{9 + 1}\mathbf{\text{u}}} \\

& {= \frac{7}{5}\mathbf{\text{u}} = \frac{7}{5}\left\langle {3,1} \right\rangle = \left\langle {\frac{21}{5},\frac{7}{5}} \right\rangle.}

\end{array}$$

Now consider the vector $\mathbf{\text{q}} = \mathbf{\text{v}} - \mathbf{\text{p}}.$ We have

$$\begin{array}{cl}

\mathbf{\text{q}} & {= \mathbf{\text{v}} - \mathbf{\text{p}}} \\

& {= \left\langle {6,-4} \right\rangle - \left\langle {\frac{21}{5},\frac{7}{5}} \right\rangle} \\

& {= \left\langle {\frac{9}{5}, - \frac{27}{5}} \right\rangle.}

\end{array}$$

Clearly, by the way we defined $\textbf{q}$, we have $\mathbf{\text{v}} = \mathbf{\text{q}} + \mathbf{\text{p}},$ and

$$\begin{array}{cl}

{\mathbf{\text{q}} \cdot \mathbf{\text{p}}} & {= \left\langle {\frac{9}{5}, - \frac{27}{5}} \right\rangle \cdot \left\langle {\frac{21}{5},\frac{7}{5}} \right\rangle} \\

& {= \frac{9(21)}{25} + \frac{-27(7)}{25}} \\

& {= \frac{189}{25} - \frac{189}{25} = 0.}

\end{array}$$

Therefore, $\textbf{q}$ and p are orthogonal.

Resolving Vectors into Components

Express $\mathbf{\text{v}} = \left\langle {8,-3,-3} \right\rangle$ as a sum of orthogonal vectors such that one of the vectors has the same direction as $\mathbf{\text{u}} = \left\langle {2,3,2} \right\rangle.$

Solution

Let p represent the projection of $\textbf{v}$ onto $\textbf{u}$:

$$\begin{array}{cl}

\mathbf{\text{p}} & {= \text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}} \\

& {= \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}} \\

& {= \frac{\left\langle {2,3,2} \right\rangle \cdot \left\langle {8,-3,-3} \right\rangle}{\left\| \left\langle {2,3,2} \right\rangle \right\|^{2}}\left\langle {2,3,2} \right\rangle} \\

& {= \frac{16 - 9 - 6}{2^{2} + 3^{2} + 2^{2}}\left\langle {2,3,2} \right\rangle} \\

& {= \frac{1}{17}\left\langle {2,3,2} \right\rangle} \\

& {= \left\langle {\frac{2}{17},\frac{3}{17},\frac{2}{17}} \right\rangle.}

\end{array}$$

Then,

$$\mathbf{\text{q}} = \mathbf{\text{v}} - \mathbf{\text{p}} = \left\langle {8,-3,-3} \right\rangle - \left\langle {\frac{2}{17},\frac{3}{17},\frac{2}{17}} \right\rangle = \left\langle {\frac{134}{17}, - \frac{54}{17}, - \frac{53}{17}} \right\rangle.$$

To check our work, we can use the dot product to verify that p and $\textbf{q}$ are orthogonal vectors:

$$\mathbf{\text{p}} \cdot \mathbf{\text{q}} = \left\langle {\frac{2}{17},\frac{3}{17},\frac{2}{17}} \right\rangle \cdot \left\langle {\frac{134}{17}, - \frac{54}{17}, - \frac{53}{17}} \right\rangle = \frac{268}{289} - \frac{162}{289} - \frac{106}{289} = 0.$$

Then,

$$\mathbf{\text{v}} = \mathbf{\text{p}} + \mathbf{\text{q}} = \left\langle {\frac{2}{17},\frac{3}{17},\frac{2}{17}} \right\rangle + \left\langle {\frac{134}{17}, - \frac{54}{17}, - \frac{53}{17}} \right\rangle.$$

Express $\mathbf{\text{v}} = 5\mathbf{\text{i}} - \mathbf{\text{j}}$ as a sum of orthogonal vectors such that one of the vectors has the same direction as $\mathbf{\text{u}} = 4\mathbf{\text{i}} + 2\mathbf{\text{j}}.$

Scalar Projection of Velocity

A container ship leaves port traveling $15\text{°}$ north of east. Its engine generates a speed of 20 knots along that path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of 2 knots. Considering both the engine and the current, how fast is the ship moving in the direction $15\text{°}$ north of east? Round the answer to two decimal places.

Solution

Let $\textbf{v}$ be the velocity vector generated by the engine, and let $\textbf{w}$ be the velocity vector of the current. We already know $\left\| \mathbf{\text{v}} \right\| = 20$ along the desired route. We just need to add in the scalar projection of $\textbf{w}$ onto $\textbf{v}$. We get

$$\begin{array}{cl}

{\text{comp}_{\mathbf{\text{v}}}\mathbf{\text{w}}} & {= \frac{\mathbf{\text{v}} \cdot \mathbf{\text{w}}}{\left\| \mathbf{\text{v}} \right\|}} \\

& {= \frac{\left\| \mathbf{\text{v}} \right\|\left\| \mathbf{\text{w}} \right\|\text{cos}\left( {30\text{°}} \right)}{\left\| \mathbf{\text{v}} \right\|}} \\

& {= \left\| \mathbf{\text{w}} \right\|\text{cos}\left( {30\text{°}} \right)} \\

& {= 2\frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.73\ \text{knots}.}

\end{array}$$

The ship is moving at 21.73 knots in the direction $15\text{°}$ north of east.

Repeat the previous example, but assume the ocean current is moving southeast instead of northeast, as shown in the following figure.

Work

Now that we understand dot products, we can see how to apply them to real-life situations. The most common application of the dot product of two vectors is in the calculation of work.

From physics, we know that work is done when an object is moved by a force. When the force is constant and applied in the same direction the object moves, then we define the work done as the product of the force and the distance the object travels: $W = Fd.$ We saw several examples of this type in earlier chapters. Now imagine the direction of the force is different from the direction of motion, as with the example of a child pulling a wagon. To find the work done, we need to multiply the component of the force that acts in the direction of the motion by the magnitude of the displacement. The dot product allows us to do just that. If we represent an applied force by a vector F and the displacement of an object by a vector s, then the work done by the force is the dot product of F and s.

When a constant force is applied to an object so the object moves in a straight line from point *P* to point *Q*, the work *W* done by the force F, acting at an angle *θ* from the line of motion, is given by

$$W = \mathbf{\text{F}} \cdot \overset{\rightarrow}{PQ} = \left\| \mathbf{\text{F}} \right\|\left\| \overset{\rightarrow}{PQ} \right\|\text{cos}\ \theta.$$ (2.8)

Let’s revisit the problem of the child’s wagon introduced earlier. Suppose a child is pulling a wagon with a force having a magnitude of 8 lb on the handle at an angle of 55°. If the child pulls the wagon 50 ft, find the work done by the force (Figure 2.51).

We have

$$W = \left\| \mathbf{\text{F}} \right\|\left\| \overset{\rightarrow}{PQ} \right\|\text{cos}\ \theta = 8(50)\left( {\text{cos}\left( {55\text{°}} \right)} \right) \approx 229\ \text{ft} \cdot \text{lb}.$$

In U.S. standard units, we measure the magnitude of force $\left\| \mathbf{\text{F}} \right\|$ in pounds. The magnitude of the displacement vector $\left\| \overset{\rightarrow}{PQ} \right\|$ tells us how far the object moved, and it is measured in feet. The customary unit of measure for work, then, is the foot-pound. One foot-pound is the amount of work required to move an object weighing 1 lb a distance of 1 ft straight up. In the metric system, the unit of measure for force is the newton (N), and the unit of measure of magnitude for work is a newton-meter (N·m), or a joule (J).

Calculating Work

A conveyor belt generates a force $\mathbf{\text{F}} = 5\mathbf{\text{i}} - 3\mathbf{\text{j}} + \mathbf{\text{k}}$ that moves a suitcase from point $\left( {1,1,1} \right)$ to point $\left( {9,4,7} \right)$ along a straight line. Find the work done by the conveyor belt. The distance is measured in meters and the force is measured in newtons.

Solution

The displacement vector $\overset{\rightarrow}{PQ}$ has initial point $\left( {1,1,1} \right)$ and terminal point $\left( {9,4,7} \right)\text{:}$

$$\overset{\rightarrow}{PQ} = \left\langle {9 - 1,4 - 1,7 - 1} \right\rangle = \left\langle {8,3,6} \right\rangle = 8\mathbf{\text{i}} + 3\mathbf{\text{j}} + 6\mathbf{\text{k}}.$$

Work is the dot product of force and displacement:

$$\begin{array}{cl}

W & {= \mathbf{\text{F}} \cdot \overset{\rightarrow}{PQ}} \\

& {= \left( {5\mathbf{\text{i}} - 3\mathbf{\text{j}} + \mathbf{\text{k}}} \right) \cdot \left( {8\mathbf{\text{i}} + 3\mathbf{\text{j}} + 6\mathbf{\text{k}}} \right)} \\

& {= 5(8) + (-3)(3) + 1(6)} \\

& {= 37\text{N} \cdot \text{m}} \\

& {= 37\ \text{J}.}

\end{array}$$

A constant force of 30 lb is applied at an angle of 60° to pull a handcart 10 ft across the ground (Figure 2.52). What is the work done by this force?

Section 2.3 Exercises

For the following exercises, the vectors $\textbf{u}$ and $\textbf{v}$ are given. Calculate the dot product $\mathbf{\text{u}} \cdot \mathbf{\text{v}}.$

123.

$\mathbf{\text{u}} = \left\langle {3,0} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {2,2} \right\rangle$

124\.

$\mathbf{\text{u}} = \left\langle {3,-4} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {4,3} \right\rangle$

125.

$\mathbf{\text{u}} = \left\langle {2,2,-1} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {-1,2,2} \right\rangle$

126\.

$\mathbf{\text{u}} = \left\langle {4,5,-6} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {0,-2,-3} \right\rangle$

For the following exercises, the vectors a, b, and c are given. Determine the vectors $\left( {\mathbf{\text{a}} \cdot \textbf{b}} \right)\textbf{c}$ and $\left( {\mathbf{\text{a}} \cdot \textbf{c}} \right)\textbf{b}.$ Express the vectors in component form.

127.

$\mathbf{\text{a}} = \left\langle {2,0,-3} \right\rangle,$ $\textbf{b} = \left\langle {-4,-7,1} \right\rangle,$ $\textbf{c} = \left\langle {1,1,-1} \right\rangle$

128\.

$\mathbf{\text{a}} = \left\langle {0,1,2} \right\rangle,$ $\textbf{b} = \left\langle {-1,0,1} \right\rangle,$ $\textbf{c} = \left\langle {1,0,-1} \right\rangle$

129.

$\mathbf{\text{a}} = \mathbf{\text{i}} + \mathbf{\text{j}},$ $\textbf{b} = \mathbf{\text{i}} - \mathbf{\text{k}},$ $\textbf{c} = \mathbf{\text{i}} - 2\mathbf{\text{k}}$

130\.

$\mathbf{\text{a}} = \mathbf{\text{i}} - \mathbf{\text{j}} + \mathbf{\text{k}},$ $\textbf{b} = \mathbf{\text{j}} + 3\mathbf{\text{k}},$ $\textbf{c} = \text{−}\mathbf{\text{i}} + 2\textbf{j} - 4\mathbf{\text{k}}$

For the following exercises, the two-dimensional vectors a and b are given.

1. Find the measure of the angle $\theta$ between a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.

2. Is $\theta$ an acute angle?

131.

\[T\] $\mathbf{\text{a}} = \left\langle {3,-1} \right\rangle,$ $\textbf{b} = \left\langle {-4,0} \right\rangle$

132\.

\[T\] $\mathbf{\text{a}} = \left\langle {2,1} \right\rangle,$ $\textbf{b} = \left\langle {-1,3} \right\rangle$

133.

$\mathbf{\text{u}} = 3\mathbf{\text{i}},$ $\mathbf{\text{v}} = 4\mathbf{\text{i}} + 4\mathbf{\text{j}}$

134\.

$\mathbf{\text{u}} = 5\textbf{i},$ $\mathbf{\text{v}} = -6\mathbf{\text{i}} + 6\mathbf{\text{j}}$

For the following exercises, find the measure of the angle between the three-dimensional vectors a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.

135.

$\mathbf{\text{a}} = \left\langle {3,-1,2} \right\rangle,$ $\textbf{b} = \left\langle {1,-1,-2} \right\rangle$

136\.

$\mathbf{\text{a}} = \left\langle {0,-1,-3} \right\rangle,$ $\textbf{b} = \left\langle {2,3,-1} \right\rangle$

137.

$\mathbf{\text{a}} = \mathbf{\text{i}} + \mathbf{\text{j}},$ $\textbf{b} = \mathbf{\text{j}} - \textbf{k}$

138\.

$\mathbf{\text{a}} = \mathbf{\text{i}} - 2\textbf{j} + \mathbf{\text{k}},$ $\textbf{b} = \mathbf{\text{i}} + \mathbf{\text{j}} - 2\mathbf{\text{k}}$

139.

\[T\] $\mathbf{\text{a}} = 3\mathbf{\text{i}} - \mathbf{\text{j}} - 2\mathbf{\text{k}},$ $\textbf{b} = \mathbf{\text{v}} + \textbf{w},$ where $\mathbf{\text{v}} = -2\mathbf{\text{i}} - 3\mathbf{\text{j}} + 2\mathbf{\text{k}}$ and $\mathbf{\text{w}} = \mathbf{\text{i}} + 2\mathbf{\text{k}}$

140\.

\[T\] $\mathbf{\text{a}} = 3\mathbf{\text{i}} - \mathbf{\text{j}} + 2\mathbf{\text{k}},$ $\textbf{b} = \mathbf{\text{v}} - \textbf{w},$ where $\mathbf{\text{v}} = 2\mathbf{\text{i}} + \mathbf{\text{j}} + 4\mathbf{\text{k}}$ and $\mathbf{\text{w}} = 6\mathbf{\text{i}} + \mathbf{\text{j}} + 2\mathbf{\text{k}}$

For the following exercises determine whether the given vectors are orthogonal.

141.

$\mathbf{\text{a}} = \left\langle {x,y} \right\rangle,$ $\textbf{b} = \left\langle {\text{−}y,x} \right\rangle,$ where *x* and *y* are nonzero real numbers

142\.

$\mathbf{\text{a}} = \left\langle {x,x} \right\rangle,$ $\textbf{b} = \left\langle {\text{−}y,y} \right\rangle,$ where *x* and *y* are nonzero real numbers

143.

$\mathbf{\text{a}} = 3\mathbf{\text{i}} - \mathbf{\text{j}} - 2\mathbf{\text{k}},$ $\textbf{b} = -2\mathbf{\text{i}} - 3\textbf{j} + \textbf{k}$

144\.

$\mathbf{\text{a}} = \mathbf{\text{i}} - \mathbf{\text{j}},$ $\textbf{b} = 7\mathbf{\text{i}} + 2\textbf{j} - \textbf{k}$

145.

Find all two-dimensional vectors a orthogonal to vector $\textbf{b} = \left\langle {3,4} \right\rangle.$ Express the answer in component form.

146\.

Find all two-dimensional vectors a orthogonal to vector $\textbf{b} = \left\langle {5,-6} \right\rangle.$ Express the answer by using standard unit vectors.

147.

Determine all three-dimensional vectors $\textbf{u}$ orthogonal to vector $\mathbf{\text{v}} = \left\langle {1,1,0} \right\rangle.$ Express the answer by using standard unit vectors.

148\.

Determine all three-dimensional vectors $\textbf{u}$ orthogonal to vector $\mathbf{\text{v}} = \mathbf{\text{i}} - \mathbf{\text{j}} - \textbf{k}.$ Express the answer in component form.

149.

Determine the real number $\alpha$ such that vectors $\mathbf{\text{a}} = 2\mathbf{\text{i}} + 3\textbf{j}$ and $\textbf{b} = 9\mathbf{\text{i}} + \alpha\textbf{j}$ are orthogonal.

150\.

Determine the real number $\alpha$ such that vectors $\mathbf{\text{a}} = -3\mathbf{\text{i}} + 2\textbf{j}$ and $\textbf{b} = 2\mathbf{\text{i}} + \alpha\textbf{j}$ are orthogonal.

151.

\[T\] Consider the points $P(4,5)$ and $Q(5,-7).$

1. Determine vectors $\overset{\rightarrow}{OP}$ and $\overset{\rightarrow}{OQ}.$ Express the answer by using standard unit vectors.

2. Determine the measure of angle *O* in triangle *OPQ*. Express the answer in degrees rounded to two decimal places.

152\.

\[T\] Consider points $A(1,1),$ $B(2,-7),$ and $C(6,3).$

1. Determine vectors $\overset{\rightarrow}{BA}$ and $\overset{\rightarrow}{BC}.$ Express the answer in component form.

2. Determine the measure of angle *B* in triangle *ABC*. Express the answer in degrees rounded to two decimal places.

153.

Determine the measure of angle *A* in triangle *ABC*, where $A(1,1,8),$ $B(4,-3,-4),$ and $C(-3,1,5).$ Express your answer in degrees rounded to two decimal places.

154\.

Consider points $P(3,7,-2)$ and $Q(1,1,-3).$ Determine the angle between vectors $\overset{\rightarrow}{OP}$ and $\overset{\rightarrow}{OQ}.$ Express the answer in degrees rounded to two decimal places.

For the following exercises, determine which (if any) pairs of the following vectors are orthogonal.

155.

$\mathbf{\text{u}} = \left\langle {3,7,-2} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {5,-3,-3} \right\rangle,$ $\mathbf{\text{w}} = \left\langle {0,1,-1} \right\rangle$

156\.

$\mathbf{\text{u}} = \mathbf{\text{i}} - \mathbf{\text{k}},$ $\mathbf{\text{v}} = 5\textbf{j} - 5\mathbf{\text{k}},$ $\mathbf{\text{w}} = 10\textbf{j}$

157\.

Use vectors to show that a parallelogram with equal diagonals is a rectangle.

158\.

Use vectors to show that the diagonals of a rhombus are perpendicular.

159\.

Show that $\mathbf{\text{u}} \cdot (\mathbf{\text{v}} + \mathbf{\text{w}}) = \mathbf{\text{u}} \cdot \mathbf{\text{v}} + \mathbf{\text{u}} \cdot \mathbf{\text{w}}$ is true for any vectors $\textbf{u}$, $\textbf{v}$, and $\textbf{w}$.

160\.

Verify the identity $\mathbf{\text{u}} \cdot (\mathbf{\text{v}} + \mathbf{\text{w}}) = \mathbf{\text{u}} \cdot \mathbf{\text{v}} + \mathbf{\text{u}} \cdot \mathbf{\text{w}}$ for vectors $\mathbf{\text{u}} = \left\langle {1,0,4} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {-2,3,5} \right\rangle,$ and $\mathbf{\text{w}} = \left\langle {4,-2,6} \right\rangle.$

For the following problems, the vector $\textbf{u}$ is given.

1. Find the direction cosines for the vector $\textbf{u}$.

2. Find the direction angles for the vector $\textbf{u}$ expressed in degrees. (Round the answer to the nearest integer.)

161.

$\mathbf{\text{u}} = \left\langle {2,2,1} \right\rangle$

162\.

$\mathbf{\text{u}} = \mathbf{\text{i}} - 2\mathbf{\text{j}} + 2\mathbf{\text{k}}$

163.

$\mathbf{\text{u}} = \left\langle {-1,5,2} \right\rangle$

164\.

$\mathbf{\text{u}} = \left\langle {2,3,4} \right\rangle$

165\.

Consider $\mathbf{\text{u}} = \left\langle {a,b,c} \right\rangle$ a nonzero three-dimensional vector. Let $\text{cos}\ \alpha,$ $\text{cos}\ \beta,$ and $\text{cos}\ \gamma$ be the direction cosines of $\textbf{u}$. Show that $\text{cos}^{2}\alpha + \text{cos}^{2}\beta + \text{cos}^{2}\gamma = 1.$

166\.

Determine the direction cosines of vector $\mathbf{\text{u}} = \mathbf{\text{i}} + 2\mathbf{\text{j}} + 2\mathbf{\text{k}}$ and show they satisfy $\text{cos}^{2}\alpha + \text{cos}^{2}\beta + \text{cos}^{2}\gamma = 1.$

For the following exercises, the vectors $\textbf{u}$ and $\textbf{v}$ are given.

1. Find the vector projection $\textbf{w} = \text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}}$ of vector $\textbf{v}$ onto vector $\textbf{u}$. Express your answer in component form.

2. Find the scalar projection $\text{comp}_{\textbf{u}}\textbf{v}$ of vector $\textbf{v}$ onto vector u.

167.

$\mathbf{\text{u}} = 5\mathbf{\text{i}} + 2\mathbf{\text{j}},$ $\mathbf{\text{v}} = 2\mathbf{\text{i}} + 3\mathbf{\text{j}}$

168\.

$\mathbf{\text{u}} = \left\langle {-4,7} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {3,5} \right\rangle$

169.

$\mathbf{\text{u}} = 3\mathbf{\text{i}} + 2\mathbf{\text{k}},$ $\mathbf{\text{v}} = 2\textbf{j} + 4\mathbf{\text{k}}$

170\.

$\mathbf{\text{u}} = \left\langle {4,4,0} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {0,4,1} \right\rangle$

171.

Consider the vectors $\mathbf{\text{u}} = 4\mathbf{\text{i}} - 3\mathbf{\text{j}}$ and $\mathbf{\text{v}} = 3\mathbf{\text{i}} + 2\mathbf{\text{j}}.$

1. Find the component form of vector $\mathbf{\text{w}} = \text{proj}_{\textbf{u}}\textbf{v}$ that represents the projection of $\textbf{v}$ onto $\textbf{u}$.

2. Write the decomposition $\mathbf{\text{v}} = \textbf{w} + \textbf{q}$ of vector $\textbf{v}$ into the orthogonal components $\textbf{w}$ and $\textbf{q}$, where $\textbf{w}$ is the projection of $\textbf{v}$ onto $\textbf{u}$ and $\textbf{q}$ is a vector orthogonal to the direction of $\textbf{u}$.

172\.

Consider vectors $\mathbf{\text{u}} = 2\mathbf{\text{i}} + 4\textbf{j}$ and $\mathbf{\text{v}} = 4\textbf{j} + 2\mathbf{\text{k}}.$

1. Find the component form of vector $\mathbf{\text{w}} = \text{proj}_{\textbf{u}}\textbf{v}$ that represents the projection of $\textbf{v}$ onto $\textbf{u}$.

2. Write the decomposition $\mathbf{\text{v}} = \textbf{w} + \textbf{q}$ of vector $\textbf{v}$ into the orthogonal components $\textbf{w}$ and $\textbf{q}$, where $\textbf{w}$ is the projection of $\textbf{v}$ onto $\textbf{u}$ and $\textbf{q}$ is a vector orthogonal to the direction of $\textbf{u}$.

173.

A methane molecule has a carbon atom situated at the origin and four hydrogen atoms located at points $P\left( {1,1,-1} \right),Q\left( {1,-1,1} \right),R\left( {-1,1,1} \right),\ \text{and}\ S\left( {-1,-1,-1} \right)$ (see figure).

1. Find the distance between the hydrogen atoms located at *P* and *R*.

2. Find the angle between vectors $\overset{\rightarrow}{OS}$ and $\overset{\rightarrow}{OR}$ that connect the carbon atom with the hydrogen atoms located at *S* and *R*, which is also called the *bond angle*. Express the answer in degrees rounded to two decimal places.

174\.

\[T\] Find the vectors that join the center of a clock to the hours 1:00, 2:00, and 3:00. Assume the clock is circular with a radius of 1 unit.

175.

Find the work done by force $\mathbf{\text{F}} = \left\langle {5,6,-2} \right\rangle$ (measured in Newtons) that moves a particle from point $P\left( {3,-1,0} \right)$ to point $Q\left( {2,3,1} \right)$ along a straight line (the distance is measured in meters).

176\.

\[T\] A sled is pulled by exerting a force of 100 N on a rope that makes an angle of $25\text{°}$ with the horizontal. Find the work done in pulling the sled 40 m. (Round the answer to one decimal place.)

177.

\[T\] A father is pulling his son on a sled at an angle of $20\text{°}$ with the horizontal with a force of 25 lb (see the following image). He pulls the sled in a straight path of 50 ft. How much work was done by the man pulling the sled? (Round the answer to the nearest integer.)

178\.

\[T\] A car is towed using a force of 1600 N. The rope used to pull the car makes an angle of 25° with the horizontal. Find the work done in towing the car 2 km. Express the answer in joules $(1\text{J} = 1\text{N} \cdot \text{m})$ rounded to the nearest integer.

179.

\[T\] A boat sails north aided by a wind blowing in a direction of $\text{N3}0\text{°}\text{E}$ with a magnitude of 500 lb. How much work is performed by the wind as the boat moves 100 ft? (Round the answer to two decimal places.)

180\.

Vector $\mathbf{\text{p}} = \left\langle {150,225,375} \right\rangle$ represents the price of certain models of bicycles sold by a bicycle shop. Vector $\mathbf{\text{n}} = \left\langle {10,7,9} \right\rangle$ represents the number of bicycles sold of each model, respectively. Compute the dot product $\mathbf{\text{p}} \cdot \mathbf{\text{n}}$ and state its meaning.

181.

\[T\] Two forces $\mathbf{\text{F}}_{1}$ and $\mathbf{\text{F}}_{2}$ are represented by vectors with initial points that are at the origin. The first force has a magnitude of 20 lb and the terminal point of the vector is point $P(1,1,0).$ The second force has a magnitude of 40 lb and the terminal point of its vector is point $Q(0,1,1).$ Let F be the resultant force of forces $\mathbf{\text{F}}_{1}$ and $\mathbf{\text{F}}_{2}.$

1. Find the magnitude of F. (Round the answer to one decimal place.)

2. Find the direction angles of F. (Express the answer in degrees rounded to one decimal place.)

182\.

\[T\] Consider $\textbf{r}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t,2t} \right\rangle$ the position vector of a particle at time $t \in \lbrack 0,30\rbrack,$ where the components of r are expressed in centimeters and time in seconds. Let $\overset{\rightarrow}{OP}$ be the position vector of the particle after 1 sec.

1. Show that all vectors $\overset{\rightarrow}{PQ},$ where $Q(x,y,z)$ is an arbitrary point, orthogonal to the instantaneous velocity vector $\mathbf{\text{v}}(1)$ of the particle after 1 sec, can be expressed as $\overset{\rightarrow}{PQ} = \left\langle {x - \text{cos}\ 1,y - \text{sin}\ 1,z - 2} \right\rangle,$ where $x\ \text{sin}\ 1 - y\ \text{cos}\ 1 - 2z + 4 = 0.$ The set of point *Q* describes a plane called the *normal plane* to the path of the particle at point *P*.

2. Use a CAS to visualize the instantaneous velocity vector and the normal plane at point *P* along with the path of the particle.

---

2.4 The Cross Product

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-4-the-cross-product

2.4 The Cross Product

• 2.4.1 Calculate the cross product of two given vectors.
• 2.4.2 Use determinants to calculate a cross product.
• 2.4.3 Find a vector orthogonal to two given vectors.
• 2.4.4 Determine areas and volumes by using the cross product.
• 2.4.5 Calculate the torque of a given force and position vector.

Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies a force at the end of the wrench. This creates rotation, or torque, which tightens the bolt. We can use vectors to represent the force applied by the mechanic, and the distance (radius) from the bolt to the end of the wrench. Then, we can represent torque by a vector oriented along the axis of rotation. Note that the torque vector is orthogonal to both the force vector and the radius vector.

In this section, we develop an operation called the *cross product,* which allows us to find a vector orthogonal to two given vectors. Calculating torque is an important application of cross products, and we examine torque in more detail later in the section. This material uses a 3 × 3 determinant of the form $\left| \begin{array}{lll}

a & b & c \\

d & e & f \\

g & h & i

\end{array} \right|$, which expands by minors to $\left. a \middle| \begin{array}{ll}

e & f \\

h & i

\end{array} \middle| - b \middle| \begin{array}{ll}

d & f \\

g & i

\end{array} \middle| + c \middle| \begin{array}{ll}

d & e \\

g & h

\end{array} \middle| = a(ei–fh)–b(di–fg) + c(dh–eg). \right.$

The Cross Product and Its Properties

The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ be nonzero vectors. We want to find a vector $\mathbf{\text{w}} = \left\langle {w_{1},w_{2},w_{3}} \right\rangle$ orthogonal to both $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$—that is, we want to find $\mathbf{\text{w}}$ such that $\mathbf{\text{u}} \cdot \mathbf{\text{w}} = 0$ and $\mathbf{\text{v}} \cdot \mathbf{\text{w}} = 0.$ Therefore, $w_{1},$ $w_{2},$ and $w_{3}$ must satisfy

$$\begin{array}{rll}

{u_{1}w_{1} + u_{2}w_{2} + u_{3}w_{3}} & = & 0 \\

{v_{1}w_{1} + v_{2}w_{2} + v_{3}w_{3}} & = & 0.

\end{array}$$

If we multiply the top equation by $v_{3}$ and the bottom equation by $u_{3}$ and subtract, we can eliminate the variable $w_{3},$ which gives

$$\left( {u_{1}v_{3} - v_{1}u_{3}} \right)w_{1} + \left( {u_{2}v_{3} - v_{2}u_{3}} \right)w_{2} = 0.$$

If we select

$$\begin{array}{rll}

w_{1} & = & {u_{2}v_{3} - u_{3}v_{2}} \\

w_{2} & = & {\text{−}\left( {u_{1}v_{3} - u_{3}v_{1}} \right),}

\end{array}$$

we get a possible solution vector. Substituting these values back into the original equations gives

$$w_{3} = u_{1}v_{2} - u_{2}v_{1}.$$

That is, vector

$$\mathbf{\text{w}} = \left\langle {u_{2}v_{3} - u_{3}v_{2},\text{−}\left( {u_{1}v_{3} - u_{3}v_{1}} \right),u_{1}v_{2} - u_{2}v_{1}} \right\rangle$$

is orthogonal to both $\mathbf{\text{u}}$ and $\mathbf{\text{v}},$ which leads us to define the following operation, called the cross product.

Let $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle\ \text{and}\ \mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle.$ Then, the cross product $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ is vector

$$\begin{array}{cl}

{\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} & {= \left( {u_{2}v_{3} - u_{3}v_{2}} \right)\mathbf{\text{i}} - \left( {u_{1}v_{3} - u_{3}v_{1}} \right)\mathbf{\text{j}} + \left( {u_{1}v_{2} - u_{2}v_{1}} \right)\mathbf{\text{k}}} \\

& {= \left\langle {u_{2}v_{3} - u_{3}v_{2},\text{−}\left( {u_{1}v_{3} - u_{3}v_{1}} \right),u_{1}v_{2} - u_{2}v_{1}} \right\rangle.}

\end{array}$$ (2.9)

From the way we have developed $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}},$ it should be clear that the cross product is orthogonal to both $\mathbf{\text{u}}$ and $\mathbf{\text{v}}.$ However, it never hurts to check. To show that $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ is orthogonal to $\mathbf{\text{u}},$ we calculate the dot product of $\mathbf{\text{u}}$ and $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}.$

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right)} & {= \left\langle {u_{1},u_{2},u_{3}} \right\rangle \cdot \left\langle {u_{2}v_{3} - u_{3}v_{2},\text{−}u_{1}v_{3} + u_{3}v_{1},u_{1}v_{2} - u_{2}v_{1}} \right\rangle} \\

& {= u_{1}\left( {u_{2}v_{3} - u_{3}v_{2}} \right) + u_{2}\left( {\text{−}u_{1}v_{3} + u_{3}v_{1}} \right) + u_{3}\left( {u_{1}v_{2} - u_{2}v_{1}} \right)} \\

& {= u_{1}u_{2}v_{3} - u_{1}u_{3}v_{2} - u_{1}u_{2}v_{3} + u_{2}u_{3}v_{1} + u_{1}u_{3}v_{2} - u_{2}u_{3}v_{1}} \\

& {= \left( {u_{1}u_{2}v_{3} - u_{1}u_{2}v_{3}} \right) + \left( {\text{−}u_{1}u_{3}v_{2} + u_{1}u_{3}v_{2}} \right) + \left( {u_{2}u_{3}v_{1} - u_{2}u_{3}v_{1}} \right)} \\

& {= 0}

\end{array}$$

In a similar manner, we can show that the cross product is also orthogonal to $\mathbf{\text{v}}.$

Finding a Cross Product

Let $\mathbf{\text{p}} = \left\langle {-1,2,5} \right\rangle\ \text{and}\ \mathbf{\text{q}} = \left\langle {4,0,-3} \right\rangle$ (Figure 2.53). Find $\mathbf{\text{p}}\ \times \ \mathbf{\text{q}}.$

Solution

Substitute the components of the vectors into Equation 2.9:

$$\begin{matrix}

{\textbf{p}\ \times \ \textbf{q}} & {= \left\langle -1,2,5 \right\rangle\ \times \ \left\langle 4,0,-3 \right\rangle} \\

& {= \left\langle p_{2}q_{3} - p_{3}q_{2}, - \left( {p_{1}q_{3}~–~p_{3}q_{1}} \right),p_{1}q_{2} - p_{2}q_{1} \right\rangle} \\

& {= \left\langle p_{2}q_{3} - p_{3}q_{2},p_{3}q_{1} - p_{1}q_{3},p_{1}q_{2} - p_{2}q_{1} \right\rangle} \\

& {= \left\langle 2(-3) - 5(0),5(4) - (-1)(-3),(-1)0 - 2(4) \right\rangle} \\

& {= \left\langle -6,17,-8 \right\rangle.}

\end{matrix}$$

Find $\mathbf{\text{p}}\ \times \ \mathbf{\text{q}}$ for $\mathbf{\text{p}} = \left\langle {5,1,2} \right\rangle$ and $\mathbf{\text{q}} = \left\langle {-2,0,1} \right\rangle.$ Express the answer using standard unit vectors.

Although it may not be obvious from Equation 2.9, the direction of $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ is given by the right-hand rule. If we hold the right hand out with the fingers pointing in the direction of $\mathbf{\text{u}},$ then curl the fingers toward vector $\mathbf{\text{v}},$ the thumb points in the direction of the cross product, as shown.

Notice what this means for the direction of $\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}.$ If we apply the right-hand rule to $\mathbf{\text{v}}\ \times \ \mathbf{\text{u}},$ we start with our fingers pointed in the direction of $\mathbf{\text{v}},$ then curl our fingers toward the vector $\mathbf{\text{u}}.$ In this case, the thumb points in the opposite direction of $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}.$ (Try it!)

Anticommutativity of the Cross Product

Let $\mathbf{\text{u}} = \left\langle {0,2,1} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {3,-1,0} \right\rangle.$ Calculate $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ and $\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}$ and graph them.

Solution

We have

$$\begin{array}{rll}

{\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} & = & {\left\langle {\left( {0 + 1} \right),\text{−}\left( {0 - 3} \right),\left( {0 - 6} \right)} \right\rangle = \left\langle {1,3,-6} \right\rangle} \\

{\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}} & = & {\left\langle {\left( {-1 - 0} \right),\text{−}\left( {3 - 0} \right),\left( {6 - 0} \right)} \right\rangle = \left\langle {-1,-3,6} \right\rangle.}

\end{array}$$

We see that, in this case, $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \text{−}\left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}} \right)$ (Figure 2.56). We prove this in general later in this section.

Suppose vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ lie in the *xy*-plane (the *z*-component of each vector is zero). Now suppose the *x*- and *y*-components of $\mathbf{\text{u}}$ and the *y*-component of $\mathbf{\text{v}}$ are all positive, whereas the *x*-component of $\mathbf{\text{v}}$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ point?

The cross products of the standard unit vectors $\mathbf{\text{i}},\mathbf{\text{j}},$ and $\mathbf{\text{k}}$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that

$$\mathbf{\text{i}}\ \times \ \mathbf{\text{i}} = \mathbf{\text{j}}\ \times \ \mathbf{\text{j}} = \mathbf{\text{k}}\ \times \ \mathbf{\text{k}} = \mathbf{0}.$$

(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar $0.)$ It’s up to you to verify the calculations on your own.

Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of $\mathbf{\text{i}}$ and $\mathbf{\text{j}}$ is parallel to $\mathbf{\text{k}}.$ Similarly, the vector product of $\mathbf{\text{i}}$ and $\mathbf{\text{k}}$ is parallel to $\mathbf{\text{j}},$ and the vector product of $\mathbf{\text{j}}$ and $\mathbf{\text{k}}$ is parallel to $\mathbf{\text{i}}.$ We can use the right-hand rule to determine the direction of each product. Then we have

$$\begin{array}{rllccrll}

{\mathbf{\text{i}}\ \times \ \mathbf{\text{j}}} & = & \mathbf{\text{k}} & & & {\mathbf{\text{j}}\ \times \ \mathbf{\text{i}}} & = & {\text{−}\mathbf{\text{k}}} \\

{\mathbf{\text{j}}\ \times \ \mathbf{\text{k}}} & = & \mathbf{\text{i}} & & & {\mathbf{\text{k}}\ \times \ \mathbf{\text{j}}} & = & {\text{−}\mathbf{\text{i}}} \\

{\mathbf{\text{k}}\ \times \ \mathbf{\text{i}}} & = & \mathbf{\text{j}} & & & {\mathbf{\text{i}}\ \times \ \mathbf{\text{k}}} & = & {\text{−}\mathbf{\text{j}}.}

\end{array}$$

These formulas come in handy later.

Cross Product of Standard Unit Vectors

Find $\mathbf{\text{i}}\ \times \ \left( {\mathbf{\text{j}}\ \times \ \mathbf{\text{k}}} \right).$

Solution

We know that $\mathbf{\text{j}}\ \times \ \mathbf{\text{k}} = \mathbf{\text{i}}.$ Therefore, $\mathbf{\text{i}}\ \times \ \left( {\mathbf{\text{j}}\ \times \ \mathbf{\text{k}}} \right) = \mathbf{\text{i}}\ \times \ \mathbf{\text{i}} = \mathbf{0}.$

Find $\left( {\mathbf{\text{i}}\ \times \ \mathbf{\text{j}}} \right)\ \times \ \left( {\mathbf{\text{k}}\ \times \ \mathbf{\text{i}}} \right).$

As we have seen, the dot product is often called the *scalar product* because it results in a scalar. The cross product results in a vector, so it is sometimes called the vector product. These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.

Properties of the Cross Product

Let $\mathbf{\text{u}},\mathbf{\text{v}},$ and $\mathbf{\text{w}}$ be vectors in space, and let $c$ be a scalar.

$$\begin{array}{lccrllcl}

\text{i.} & & & {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} & = & {\text{−}\left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}} \right)} & & \text{Anticommutative property} \\

\text{ii.} & & & {\mathbf{\text{u}}\ \times \ \left( {\mathbf{\text{v}} + \mathbf{\text{w}}} \right)} & = & {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} + \mathbf{\text{u}}\ \times \ \mathbf{\text{w}}} & & \text{Distributive property} \\

\text{iii.} & & & {c\left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right)} & = & {\left( {c\mathbf{\text{u}}} \right)\ \times \ \mathbf{\text{v}} = \mathbf{\text{u}}\ \times \ \left( {c\mathbf{\text{v}}} \right)} & & \text{Multiplication by a constant} \\

\text{iv.} & & & {\mathbf{\text{u}}\ \times \ \mathbf{0}} & = & {\mathbf{0}\ \times \ \mathbf{\text{u}} = \mathbf{0}} & & \text{Cross product of the zero vector} \\

\text{v.} & & & {\mathbf{\text{v}}\ \times \ \mathbf{\text{v}}} & = & \mathbf{0} & & \text{Cross product of a vector with itself} \\

\text{vi.} & & & {\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & = & {\left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) \cdot \mathbf{\text{w}}} & & \text{Scalar triple product}

\end{array}$$

Proof

For property $\text{i}\operatorname{.,}$ we want to show $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \text{−}\left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}} \right).$ We have

$$\begin{array}{cl}

{\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} & {= \left\langle {u_{1},u_{2},u_{3}} \right\rangle\ \times \ \left\langle {v_{1},v_{2},v_{3}} \right\rangle} \\

& {= \left\langle {u_{2}v_{3} - u_{3}v_{2},\text{−}u_{1}v_{3} + u_{3}v_{1},u_{1}v_{2} - u_{2}v_{1}} \right\rangle} \\

& {= \text{−}\left\langle {u_{3}v_{2} - u_{2}v_{3},\text{−}u_{3}v_{1} + u_{1}v_{3},u_{2}v_{1} - u_{1}v_{2}} \right\rangle} \\

& {= \text{−}\left\langle {v_{1},v_{2},v_{3}} \right\rangle\ \times \ \left\langle {u_{1},u_{2},u_{3}} \right\rangle} \\

& {= \text{−}\left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}} \right).}

\end{array}$$

Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.

For property $\text{iv}\operatorname{.,}$ this follows directly from the definition of the cross product. We have

$$\begin{array}{cl}

{\mathbf{\text{u}}\ \times \ \mathbf{0}} & {= \left\langle {u_{2}(0) - u_{3}(0),\text{−}\left( {u_{2}(0) - u_{3}(0)} \right),u_{1}(0) - u_{2}(0)} \right\rangle} \\

& {= \left\langle {0,0,0} \right\rangle = \mathbf{0}.}

\end{array}$$

Then, by property i., $\mathbf{0}\ \times \ \mathbf{\text{u}} = \mathbf{0}$ as well. Remember that the dot product of a vector and the zero vector is the *scalar* $0,$ whereas the cross product of a vector with the zero vector is the *vector* $\mathbf{0}.$

Property $\text{vi}.$ looks like the associative property, but note the change in operations:

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & {= \mathbf{\text{u}} \cdot \left\langle {v_{2}w_{3} - v_{3}w_{2},\text{−}v_{1}w_{3} + v_{3}w_{1},v_{1}w_{2} - v_{2}w_{1}} \right\rangle} \\

& {= u_{1}\left( {v_{2}w_{3} - v_{3}w_{2}} \right) + u_{2}\left( {\text{−}v_{1}w_{3} + v_{3}w_{1}} \right) + u_{3}\left( {v_{1}w_{2} - v_{2}w_{1}} \right)} \\

& {= u_{1}v_{2}w_{3} - u_{1}v_{3}w_{2} - u_{2}v_{1}w_{3} + u_{2}v_{3}w_{1} + u_{3}v_{1}w_{2} - u_{3}v_{2}w_{1}} \\

& {= \left( {u_{2}v_{3} - u_{3}v_{2}} \right)w_{1} + \left( {u_{3}v_{1} - u_{1}v_{3}} \right)w_{2} + \left( {u_{1}v_{2} - u_{2}v_{1}} \right)w_{3}} \\

& {= \left\langle {u_{2}v_{3} - u_{3}v_{2},u_{3}v_{1} - u_{1}v_{3},u_{1}v_{2} - u_{2}v_{1}} \right\rangle \cdot \left\langle {w_{1},w_{2},w_{3}} \right\rangle} \\

& {= \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) \cdot \mathbf{\text{w}}.}

\end{array}$$

□

Using the Properties of the Cross Product

Use the cross product properties to calculate $\left( {2\mathbf{\text{i}}\ \times \ 3\mathbf{\text{j}}} \right)\ \times \ \mathbf{\text{j}}.$

Solution

$$\begin{array}{cl}

{\left( {2\mathbf{\text{i}}\ \times \ 3\mathbf{\text{j}}} \right)\ \times \ \mathbf{\text{j}}} & {= 2\left( {\mathbf{\text{i}}\ \times \ 3\mathbf{\text{j}}} \right)\ \times \ \mathbf{\text{j}}} \\

& {= 2(3)\left( {\mathbf{\text{i}}\ \times \ \mathbf{\text{j}}} \right)\ \times \ \mathbf{\text{j}}} \\

& {= \left( {6\mathbf{\text{k}}} \right)\ \times \ \mathbf{\text{j}}} \\

& {= 6\left( {\mathbf{\text{k}}\ \times \ \mathbf{\text{j}}} \right)} \\

& {= 6\left( {\text{−}\mathbf{\text{i}}} \right) = -6\mathbf{\text{i}}.}

\end{array}$$

Use the properties of the cross product to calculate $\left( {\mathbf{\text{i}}\ \times \ \mathbf{\text{k}}} \right)\ \times \ \left( {\mathbf{\text{k}}\ \times \ \mathbf{\text{j}}} \right).$

So far in this section, we have been concerned with the direction of the vector $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}},$ but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ involving the magnitudes of $\mathbf{\text{u}}$ and $\mathbf{\text{v}},$ and the sine of the angle between them.

Magnitude of the Cross Product

Let $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ be vectors, and let $\theta$ be the angle between them. Then, $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\| = \left\| \mathbf{\text{u}} \right\| \cdot \left\| \mathbf{\text{v}} \right\| \cdot \text{sin}\ \theta.$

Proof

Let $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ be vectors, and let $\theta$ denote the angle between them. Then

$$\begin{array}{cl}

\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|^{2} & {= \left( {u_{2}v_{3} - u_{3}v_{2}} \right)^{2} + \left( {u_{3}v_{1} - u_{1}v_{3}} \right)^{2} + \left( {u_{1}v_{2} - u_{2}v_{1}} \right)^{2}} \\

& {= u_{2}^{2}v_{3}^{2} - 2u_{2}u_{3}v_{2}v_{3} + u_{3}^{2}v_{2}^{2} + u_{3}^{2}v_{1}^{2} - 2u_{1}u_{3}v_{1}v_{3} + u_{1}^{2}v_{3}^{2} + u_{1}^{2}v_{2}^{2} - 2u_{1}u_{2}v_{1}v_{2} + u_{2}^{2}v_{1}^{2}} \\

& {= u_{1}^{2}v_{1}^{2} + u_{1}^{2}v_{2}^{2} + u_{1}^{2}v_{3}^{2} + u_{2}^{2}v_{1}^{2} + u_{2}^{2}v_{2}^{2} + u_{2}^{2}v_{3}^{2} + u_{3}^{2}v_{1}^{2} + u_{3}^{2}v_{2}^{2} + u_{3}^{2}v_{3}^{2}} \\

& {\qquad - \left( {u_{1}^{2}v_{1}^{2} + u_{2}^{2}v_{2}^{2} + u_{3}^{2}v_{3}^{2} + 2u_{1}u_{2}v_{1}v_{2} + 2u_{1}u_{3}v_{1}v_{3} + 2u_{2}u_{3}v_{2}v_{3}} \right)} \\

& {= \left( {u_{1}^{2} + u_{2}^{2} + u_{3}^{2}} \right)\left( {v_{1}^{2} + v_{2}^{2} + v_{3}^{2}} \right) - \left( {u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}} \right)^{2}} \\

& {= \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2} - \left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right)^{2}} \\

& {= \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2} - \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2}\text{cos}^{2}\theta} \\

& {= \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2}\left( {1 - \text{cos}^{2}\theta} \right)} \\

& {= \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2}\left( {\text{sin}^{2}\theta} \right).}

\end{array}$$

Taking square roots and noting that $\sqrt{\text{sin}^{2}\theta} = \text{sin}\ \theta$ for $0 \leq \theta \leq 180\text{°},$ we have the desired result:

$$\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\| = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{sin}\ \theta.$$

□

This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.

Calculating the Cross Product

Use Properties of the Cross Product to find the magnitude of the cross product of $\mathbf{\text{u}} = \left\langle {0,4,0} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {0,0,-3} \right\rangle.$

Solution

We have

$$\begin{array}{cl}

\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\| & {= \left\| \mathbf{\text{u}} \right\| \cdot \left\| \mathbf{\text{v}} \right\| \cdot \text{sin}\ \theta} \\

& {= \sqrt{0^{2} + 4^{2} + 0^{2}} \cdot \sqrt{0^{2} + 0^{2} + (-3)^{2}} \cdot \text{sin}\ \frac{\pi}{2}} \\

& {= 4(3)(1) = 12.}

\end{array}$$

Use Properties of the Cross Product to find the magnitude of $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}},$ where $\mathbf{\text{u}} = \left\langle {-8,0,0} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {0,2,0} \right\rangle.$

Determinants and the Cross Product

Using Equation 2.9 to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using determinant notation.

A $2\ \times \ 2$ determinant is defined by

$$\left| \begin{array}{ll}

a_{1} & a_{2} \\

b_{1} & b_{2}

\end{array} \right| = a_{1}b_{2} - b_{1}a_{2}.$$

For example,

$$\left| \begin{array}{lr}

3 & -2 \\

5 & 1

\end{array} \right| = 3(1) - 5(-2) = 3 + 10 = 13.$$

A $3\ \times \ 3$ determinant is defined in terms of $2\ \times \ 2$ determinants as follows:

$$\left| \begin{array}{lll}

a_{1} & a_{2} & a_{3} \\

b_{1} & b_{2} & b_{3} \\

c_{1} & c_{2} & c_{3}

\end{array} \right| = a_{1}\left| \begin{array}{ll}

b_{2} & b_{3} \\

c_{2} & c_{3}

\end{array} \right| - a_{2}\left| \begin{array}{ll}

b_{1} & b_{3} \\

c_{1} & c_{3}

\end{array} \right| + a_{3}\left| \begin{array}{ll}

b_{1} & b_{2} \\

c_{1} & c_{2}

\end{array} \right|.$$ (2.10)

Equation 2.10 is referred to as the *expansion of the determinant along the first row*. Notice that the multipliers of each of the $2\ \times \ 2$ determinants on the right side of this expression are the entries in the first row of the $3\ \times \ 3$ determinant. Furthermore, each of the $2\ \times \ 2$ determinants contains the entries from the $3\ \times \ 3$ determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, $a_{1}$ is the multiplier, and the $2\ \times \ 2$ determinant contains the entries that remain if you cross out the first row and first column of the $3\ \times \ 3$ determinant. Similarly, for the second term, the multiplier is $a_{2},$ and the $2\ \times \ 2$ determinant contains the entries that remain if you cross out the first row and second column of the $3\ \times \ 3$ determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.

Using Expansion Along the First Row to Compute a $3\ \times \ 3$ Determinant

Evaluate the determinant $\left| \begin{array}{rrr}

2 & 5 & -1 \\

-1 & 1 & 3 \\

-2 & 3 & 4

\end{array} \right|.$

Solution

We have

$$\begin{array}{cl}

\left| \begin{array}{rrr}

2 & 5 & -1 \\

-1 & 1 & 3 \\

-2 & 3 & 4

\end{array} \right| & {= 2\left| \begin{array}{ll}

1 & 3 \\

3 & 4

\end{array} \right| - 5\left| \begin{array}{ll}

-1 & 3 \\

-2 & 4

\end{array} \right| - 1\left| \begin{array}{ll}

-1 & 1 \\

-2 & 3

\end{array} \right|} \\

& {= 2\left( {4 - 9} \right) - 5\left( {-4 + 6} \right) - 1\left( {-3 + 2} \right)} \\

& {= 2(-5) - 5(2) - 1(-1) = -10 - 10 + 1} \\

& {= -19.}

\end{array}$$

Evaluate the determinant $\left| \begin{array}{rrr}

1 & -2 & -1 \\

3 & 2 & -3 \\

1 & 5 & 4

\end{array} \right|.$

Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.

Let $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ be vectors. Then the cross product $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ is given by

$$\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \left| \begin{array}{lll}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

u_{1} & u_{2} & u_{3} \\

v_{1} & v_{2} & v_{3}

\end{array} \right| = \left| \begin{array}{ll}

u_{2} & u_{3} \\

v_{2} & v_{3}

\end{array} \right|\mathbf{\text{i}} - \left| \begin{array}{ll}

u_{1} & u_{3} \\

v_{1} & v_{3}

\end{array} \right|\mathbf{\text{j}} + \left| \begin{array}{ll}

u_{1} & u_{2} \\

v_{1} & v_{2}

\end{array} \right|\mathbf{\text{k}}.$$

Using Determinant Notation to find $\mathbf{\text{p}}\ \times \ \mathbf{\text{q}}$

Let $\mathbf{\text{p}} = \left\langle {-1,2,5} \right\rangle$ and $\mathbf{\text{q}} = \left\langle {4,0,-3} \right\rangle.$ Find $\mathbf{\text{p}}\ \times \ \mathbf{\text{q}}.$

Solution

We set up our determinant by putting the standard unit vectors across the first row, the components of $\mathbf{\text{u}}$ in the second row, and the components of $\mathbf{\text{v}}$ in the third row. Then, we have

$$\begin{array}{cl}

{\mathbf{\text{p}}\ \times \ \mathbf{\text{q}}} & {= \left| \begin{array}{rrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

-1 & 2 & 5 \\

4 & 0 & -3

\end{array} \right| = \left| \begin{array}{rr}

2 & 5 \\

0 & -3

\end{array} \right|\mathbf{\text{i}} - \left| \begin{array}{rr}

-1 & 5 \\

4 & -3

\end{array} \right|\mathbf{\text{j}} + \left| \begin{array}{rr}

-1 & 2 \\

4 & 0

\end{array} \right|\mathbf{\text{k}}} \\

& {= \left( {-6 - 0} \right)\mathbf{\text{i}} - \left( {3 - 20} \right)\mathbf{\text{j}} + \left( {0 - 8} \right)\mathbf{\text{k}}} \\

& {= -6\mathbf{\text{i}} + 17\mathbf{\text{j}} - 8\mathbf{\text{k}}.}

\end{array}$$

Notice that this answer confirms the calculation of the cross product in Example 2.31.

Use determinant notation to find $\mathbf{\text{a}}\ \times \ \mathbf{\text{b}},$ where $\mathbf{\text{a}} = \left\langle {8,2,3} \right\rangle$ and $\mathbf{\text{b}} = \left\langle {-1,0,4} \right\rangle.$

Using the Cross Product

The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a *parallelepiped*. The following examples illustrate these calculations.

Finding a Unit Vector Orthogonal to Two Given Vectors

Let $\mathbf{\text{a}} = \left\langle {5,2,-1} \right\rangle$ and $\mathbf{\text{b}} = \left\langle {0,-1,4} \right\rangle.$ Find a unit vector orthogonal to both $\mathbf{\text{a}}$ and $\mathbf{\text{b}}.$

Solution

The cross product $\mathbf{\text{a}}\ \times \ \mathbf{\text{b}}$ is orthogonal to both vectors $\mathbf{\text{a}}$ and $\mathbf{\text{b}}.$ We can calculate it with a determinant:

$$\begin{array}{cl}

{\mathbf{\text{a}}\ \times \ \mathbf{\text{b}}} & {= \left| \begin{array}{rrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

5 & 2 & -1 \\

0 & -1 & 4

\end{array} \right| = \left| \begin{array}{rr}

2 & -1 \\

-1 & 4

\end{array} \right|\mathbf{\text{i}} - \left| \begin{array}{rr}

5 & -1 \\

0 & 4

\end{array} \right|\mathbf{\text{j}} + \left| \begin{array}{rr}

5 & 2 \\

0 & -1

\end{array} \right|\mathbf{\text{k}}} \\

& {= \left( {8 - 1} \right)\mathbf{\text{i}} - \left( {20 - 0} \right)\mathbf{\text{j}} + \left( {-5 - 0} \right)\mathbf{\text{k}}} \\

& {= 7\mathbf{\text{i}} - 20\mathbf{\text{j}} - 5\mathbf{\text{k}}.}

\end{array}$$

Normalize this vector to find a unit vector in the same direction:

$$\left\| {\mathbf{\text{a}}\ \times \ \mathbf{\text{b}}} \right\| = \sqrt{(7)^{2} + (-20)^{2} + (-5)^{2}} = \sqrt{474}.$$

Thus, $\left\langle {\frac{7}{\sqrt{474}},\frac{-20}{\sqrt{474}},\frac{-5}{\sqrt{474}}} \right\rangle$ is a unit vector orthogonal to $\mathbf{\text{a}}$ and $\mathbf{\text{b}}.$

Find a unit vector orthogonal to both $\mathbf{\text{a}}$ and $\mathbf{\text{b}},$ where $\mathbf{\text{a}} = \left\langle {4,0,3} \right\rangle$ and $\mathbf{\text{b}} = \left\langle {1,1,4} \right\rangle.$

To use the cross product for calculating areas, we state and prove the following theorem.

Area of a Parallelogram

If we locate vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|$ (Figure 2.57).

Proof

We show that the magnitude of the cross product is equal to the base times height of the parallelogram.

$$\begin{array}{cl}

\text{Area of a parallelogram} & {= \text{base}\ \times \ \text{height}} \\

& {= \left\| \mathbf{\text{u}} \right\|\left( {\left\| \mathbf{\text{v}} \right\|\text{sin}\ \theta} \right)} \\

& {= \left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|}

\end{array}$$

□

Finding the Area of a Triangle

Let $P = \left( {1,0,0} \right),Q = \left( {0,1,0} \right),\ \text{and}\ R = \left( {0,0,1} \right)$ be the vertices of a triangle (Figure 2.58). Find its area.

Solution

We have $\overset{\rightarrow}{PQ} = \left\langle {0 - 1,1 - 0,0 - 0} \right\rangle = \left\langle {-1,1,0} \right\rangle$ and $\overset{\rightarrow}{PR} = \left\langle {0 - 1,0 - 0,1 - 0} \right\rangle = \left\langle {-1,0,1} \right\rangle.$ The area of the parallelogram with adjacent sides $\overset{\rightarrow}{PQ}$ and $\overset{\rightarrow}{PR}$ is given by $\left\| {\overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{PR}} \right\|\text{:}$

$$\begin{array}{rll}

{\overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{PR}} & = & {\left| \begin{array}{rrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

-1 & 1 & 0 \\

-1 & 0 & 1

\end{array} \right| = \left( {1 - 0} \right)\mathbf{\text{i}} - \left( {-1 - 0} \right)\mathbf{\text{j}} + \left( {0 - (-1)} \right)\mathbf{\text{k}} = \mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}} \\

\left\| {\overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{PR}} \right\| & = & {\left\| \left\langle {1,1,1} \right\rangle \right\| = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}.}

\end{array}$$

The area of $\text{Δ}PQR$ is half the area of the parallelogram, or $\sqrt{3}\text{/}2.$

Find the area of the parallelogram $PQRS$ with vertices $P\left( {1,1,0} \right),Q\left( {7,1,0} \right),R\left( {9,4,2} \right),$ and $S\left( {3,4,2} \right).$

The Triple Scalar Product

Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.

The triple scalar product of vectors $\mathbf{\text{u}},$ $\mathbf{\text{v}},$ and $\mathbf{\text{w}}$ is $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right).$

Calculating a Triple Scalar Product

The triple scalar product of vectors $\mathbf{\text{u}} = u_{1}\mathbf{\text{i}} + u_{2}\mathbf{\text{j}} + u_{3}\mathbf{\text{k}},$ $\mathbf{\text{v}} = v_{1}\mathbf{\text{i}} + v_{2}\mathbf{\text{j}} + v_{3}\mathbf{\text{k}},$ and $\mathbf{\text{w}} = w_{1}\mathbf{\text{i}} + w_{2}\mathbf{\text{j}} + w_{3}\mathbf{\text{k}}$ is the determinant of the $3\ \times \ 3$ matrix formed by the components of the vectors:

$$\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right) = \left| \begin{array}{lll}

u_{1} & u_{2} & u_{3} \\

v_{1} & v_{2} & v_{3} \\

w_{1} & w_{2} & w_{3}

\end{array} \right|.$$

Proof

The calculation is straightforward.

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & {= \left\langle {u_{1},u_{2},u_{3}} \right\rangle \cdot \left\langle {v_{2}w_{3} - v_{3}w_{2},\text{−}v_{1}w_{3} + v_{3}w_{1},v_{1}w_{2} - v_{2}w_{1}} \right\rangle} \\

& {= u_{1}\left( {v_{2}w_{3} - v_{3}w_{2}} \right) + u_{2}\left( {\text{−}v_{1}w_{3} + v_{3}w_{1}} \right) + u_{3}\left( {v_{1}w_{2} - v_{2}w_{1}} \right)} \\

& {= u_{1}\left( {v_{2}w_{3} - v_{3}w_{2}} \right) - u_{2}\left( {v_{1}w_{3} - v_{3}w_{1}} \right) + u_{3}\left( {v_{1}w_{2} - v_{2}w_{1}} \right)} \\

& {= \left| \begin{array}{lll}

u_{1} & u_{2} & u_{3} \\

v_{1} & v_{2} & v_{3} \\

w_{1} & w_{2} & w_{3}

\end{array} \right|}

\end{array}$$

□

Calculating the Triple Scalar Product

Let $\mathbf{\text{u}} = \left\langle {1,3,5} \right\rangle,\mathbf{\text{v}} = \left\langle {2,-1,0} \right\rangle\ \text{and}\ \mathbf{\text{w}} = \left\langle {-3,0,-1} \right\rangle.$ Calculate the triple scalar product $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right).$

Solution

Apply Calculating a Triple Scalar Product directly:

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & {= \left| \begin{array}{rrr}

1 & 3 & 5 \\

2 & -1 & 0 \\

-3 & 0 & -1

\end{array} \right|} \\

& {= 1\left| \begin{array}{rr}

-1 & 0 \\

0 & -1

\end{array} \right| - 3\left| \begin{array}{rr}

2 & 0 \\

-3 & -1

\end{array} \right| + 5\left| \begin{array}{rr}

2 & -1 \\

-3 & 0

\end{array} \right|} \\

& {= \left( {1 - 0} \right) - 3\left( {-2 - 0} \right) + 5\left( {0 - 3} \right)} \\

& {= 1 + 6 - 15 = -8.}

\end{array}$$

Calculate the triple scalar product $\mathbf{\text{a}} \cdot \left( {\mathbf{\text{b}}\ \times \ \mathbf{\text{c}}} \right),$ where $\mathbf{\text{a}} = \left\langle {2,-4,1} \right\rangle,$ $\mathbf{\text{b}} = \left\langle {0,3,-1} \right\rangle,$ and $\mathbf{\text{c}} = \left\langle {5,-3,3} \right\rangle.$

When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:

$$\left| \begin{array}{lll}

a_{1} & a_{2} & a_{3} \\

b_{1} & b_{2} & b_{3} \\

c_{1} & c_{2} & c_{3}

\end{array} \right| = d\qquad\left| \begin{array}{lll}

b_{1} & b_{2} & b_{3} \\

a_{1} & a_{2} & a_{3} \\

c_{1} & c_{2} & c_{3}

\end{array} \right| = \text{−}d\qquad\left| \begin{array}{lll}

b_{1} & b_{2} & b_{3} \\

c_{1} & c_{2} & c_{3} \\

a_{1} & a_{2} & a_{3}

\end{array} \right| = d\qquad\left| \begin{array}{lll}

c_{1} & c_{2} & c_{3} \\

b_{1} & b_{2} & b_{3} \\

a_{1} & a_{2} & a_{3}

\end{array} \right| = \text{−}d.$$

Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:

$$\begin{array}{cl}

\left| \begin{array}{rlr}

1 & 2 & 1 \\

-2 & 0 & 3 \\

4 & 1 & -1

\end{array} \right| & {= \left| \begin{array}{lr}

0 & 3 \\

1 & -1

\end{array} \right| - 2\left| \begin{array}{rr}

-2 & 3 \\

4 & -1

\end{array} \right| + \left| \begin{aligned}

-2 & 0 \\

4 & 1

\end{aligned} \right|} \\

& {= \left( {0 - 3} \right) - 2\left( {2 - 12} \right) + \left( {-2 - 0} \right) = -3 + 20 - 2 = 15.}

\end{array}$$

Switching the top two rows we have

$$\left| \begin{array}{rlr}

-2 & 0 & 3 \\

1 & 2 & 1 \\

4 & 1 & -1

\end{array} \right| = -2\left| \begin{array}{lr}

2 & 1 \\

1 & -1

\end{array} \right| + 3\left| \begin{array}{ll}

1 & 2 \\

4 & 1

\end{array} \right| = -2\left( {-2 - 1} \right) + 3\left( {1 - 8} \right) = 6 - 21 = -15.$$

Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let $\mathbf{\text{u}} = u_{1}\mathbf{\text{i}} + u_{2}\mathbf{\text{j}} + u_{3}\mathbf{\text{k}},$ $\mathbf{\text{v}} = v_{1}\mathbf{\text{i}} + v_{2}\mathbf{\text{j}} + v_{3}\mathbf{\text{k}},$ and $\mathbf{\text{w}} = w_{1}\mathbf{\text{i}} + w_{2}\mathbf{\text{j}} + w_{3}\mathbf{\text{k}}.$ Applying Calculating a Triple Scalar Product, we have

$$\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right) = \left| \begin{array}{lll}

u_{1} & u_{2} & u_{3} \\

v_{1} & v_{2} & v_{3} \\

w_{1} & w_{2} & w_{3}

\end{array} \right|\quad\text{and}\quad\mathbf{\text{u}} \cdot \left( {\mathbf{\text{w}}\ \times \ \mathbf{\text{v}}} \right) = \left| \begin{array}{lll}

u_{1} & u_{2} & u_{3} \\

w_{1} & w_{2} & w_{3} \\

v_{1} & v_{2} & v_{3}

\end{array} \right|.$$

We can obtain the determinant for calculating $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{w}}\ \times \ \mathbf{\text{v}}} \right)$ by switching the bottom two rows of $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right).$ Therefore, $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right) = \text{−}\mathbf{\text{u}} \cdot \left( {\mathbf{\text{w}}\ \times \ \mathbf{\text{v}}} \right).$

Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:

$$\begin{array}{rll}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & = & {\text{−}\mathbf{\text{u}} \cdot \left( {\mathbf{\text{w}}\ \times \ \mathbf{\text{v}}} \right)} \\

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & = & {\mathbf{\text{v}} \cdot \left( {\mathbf{\text{w}}\ \times \ \mathbf{\text{u}}} \right) = \mathbf{\text{w}} \cdot \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right).}

\end{array}$$

Let $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ be two vectors in standard position. If $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in Area of a Parallelogram that the area of this parallelogram is $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|.$ Now suppose we add a third vector $\mathbf{\text{w}}$ that does not lie in the same plane as $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ but still shares the same initial point. Then these vectors form three edges of a parallelepiped, a three-dimensional prism with six faces that are each parallelograms, as shown in Figure 2.59. The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of $\mathbf{\text{u}},\mathbf{\text{v}},$ and $\mathbf{\text{w}}$ provides a simple method for calculating the volume of the parallelepiped defined by these vectors.

Volume of a Parallelepiped

The volume of a parallelepiped with adjacent edges given by the vectors $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ is the absolute value of the triple scalar product:

$$V = \left| {\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} \right|.$$

See Figure 2.59.

Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.

Proof

The area of the base of the parallelepiped is given by $\left\| {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right\|.$ The height of the figure is given by $\left\| {\text{proj}_{\text{v×w}}\mathbf{\text{u}}} \right\|.$ The volume of the parallelepiped is the product of the height and the area of the base, so we have

$$\begin{array}{cl}

V & {= \left\| {\text{proj}_{\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}}\mathbf{\text{u}}} \right\|\left\| {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right\|} \\

& {= \left| \frac{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)}{\left\| {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right\|} \right|\left\| {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right\|} \\

& {= \left| {\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} \right|.}

\end{array}$$

□

Calculating the Volume of a Parallelepiped

Let $\mathbf{\text{u}} = \left\langle {-1,-2,1} \right\rangle,\mathbf{\text{v}} = \left\langle {4,3,2} \right\rangle,\ \text{and}\ \mathbf{\text{w}} = \left\langle {0,-5,-2} \right\rangle.$ Find the volume of the parallelepiped with adjacent edges $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ (Figure 2.60).

Solution

We have

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & {= \left| \begin{array}{rrr}

-1 & -2 & 1 \\

4 & 3 & 2 \\

0 & -5 & -2

\end{array} \right| = (-1)\left| \begin{array}{rr}

3 & 2 \\

-5 & -2

\end{array} \right| + 2\left| \begin{array}{rr}

4 & 2 \\

0 & -2

\end{array} \right| + \left| \begin{array}{rr}

4 & 3 \\

0 & -5

\end{array} \right|} \\

& {= (-1)\left( {-6 + 10} \right) + 2\left( {-8 - 0} \right) + \left( {-20 - 0} \right)} \\

& {= -4 - 16 - 20} \\

& {= -40.}

\end{array}$$

Thus, the volume of the parallelepiped is $|-40| = 40$ units³.

Find the volume of the parallelepiped formed by the vectors $\mathbf{\text{a}} = 3\mathbf{\text{i}} + 4\mathbf{\text{j}} - \mathbf{\text{k}},$ $\mathbf{\text{b}} = 2\mathbf{\text{i}} - \mathbf{\text{j}} - \mathbf{\text{k}},$ and $\mathbf{\text{c}} = 3\mathbf{\text{j}} + \mathbf{\text{k}}.$

Applications of the Cross Product

The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (Introduction to Vector Calculus).

Using the Triple Scalar Product

Use the triple scalar product to show that vectors $\mathbf{\text{u}} = \left\langle {2,0,5} \right\rangle,\mathbf{\text{v}} = \left\langle {2,2,4} \right\rangle,\ \text{and}\ \mathbf{\text{w}} = \left\langle {1,-1,3} \right\rangle$ are coplanar—that is, show that these vectors lie in the same plane.

Solution

Start by calculating the triple scalar product to find the volume of the parallelepiped defined by $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}\text{:}$

$$\begin{array}{cl}

{\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} & {= \left| \begin{array}{rrr}

2 & 0 & 5 \\

2 & 2 & 4 \\

1 & -1 & 3

\end{array} \right|} \\

& {= \left\lbrack {2(2)(3) + (0)(4)(1) + 5(2)(-1)} \right\rbrack - \left\lbrack {5(2)(1) + (2)(4)(-1) + (0)(2)(3)} \right\rbrack} \\

& {= 2 - 2} \\

& {= 0.}

\end{array}$$

The volume of the parallelepiped is $0$ units³, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.

Are the vectors $\mathbf{\text{a}} = \mathbf{\text{i}} + \mathbf{\text{j}} - \mathbf{\text{k}},$ $\mathbf{\text{b}} = \mathbf{\text{i}} - \mathbf{\text{j}} + \mathbf{\text{k}},$ and $\mathbf{\text{c}} = \mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}$ coplanar?

Finding an Orthogonal Vector

Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points $P = \left( {9,-3,-2} \right),Q = \left( {1,3,0} \right),$ and $R = \left( {-2,5,0} \right).$

Solution

The plane must contain vectors $\overset{\rightarrow}{PQ}$ and $\overset{\rightarrow}{QR}\text{:}$

$$\begin{array}{l}

{\overset{\rightarrow}{PQ} = \left\langle {1 - 9,3 - (-3),0 - (-2)} \right\rangle = \left\langle {-8,6,2} \right\rangle} \\

{\overset{\rightarrow}{QR} = \left\langle {-2 - 1,5 - 3,0 - 0} \right\rangle = \left\langle {-3,2,0} \right\rangle.}

\end{array}$$

The cross product $\overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{QR}$ produces a vector orthogonal to both $\overset{\rightarrow}{PQ}$ and $\overset{\rightarrow}{QR}.$ Therefore, the cross product is orthogonal to the plane that contains these two vectors:

$$\begin{array}{cl}

{\overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{QR}} & {= \left| \left. \begin{array}{rrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

{- 8} & 6 & 2 \\

{- 3} & 2 & 0

\end{array} \right| \right.} \\

& {= 0\mathbf{\text{i}} - 6\mathbf{\text{j}} - 16\mathbf{\text{k}} - \left( {-18\mathbf{\text{k}} + 4\mathbf{\text{i}} + 0\mathbf{\text{j}}} \right)} \\

& {= -4\mathbf{\text{i}} - 6\mathbf{\text{j}} + 2\mathbf{\text{k}}.}

\end{array}$$

We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.

Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.

Torque, $\tau$ (the Greek letter *tau*), measures the tendency of a force to produce rotation about an axis of rotation. Let $\mathbf{\text{r}}$ be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector $\mathbf{\text{F}}$ represent the force. Then torque is equal to the cross product of $\mathbf{\text{r}}$ and $\mathbf{\text{F}}\text{:}$

$$\tau = \mathbf{\text{r}}\ \times \ \mathbf{\text{F}}.$$

See Figure 2.61.

Think about using a wrench to tighten a bolt. The torque $\tau$ applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.

Evaluating Torque

A bolt is tightened by applying a force of $6$ N to a 0.15-m wrench (Figure 2.62). The angle between the wrench and the force vector is $40\text{°}.$ Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.

Solution

Substitute the given information into the equation defining torque:

$$\left\| \tau \right\| = \left\| {\mathbf{\text{r}}\ \times \ \mathbf{\text{F}}} \right\| = \left\| \mathbf{\text{r}} \right\|\left\| \mathbf{\text{F}} \right\|\text{sin}\ \theta = (0.15\ \text{m})(6\ \text{N})\text{sin}\ 40\text{°} \approx 0.58\ \text{N} \cdot \text{m}.$$

Calculate the force required to produce $15\ \text{N} \cdot \text{m}$ torque at an angle of $30º$ from a 150-cm rod.

Section 2.4 Exercises

For the following exercises, the vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are given.

1. Find the cross product $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ of the vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}.$ Express the answer in component form.

2. Sketch the vectors $\mathbf{\text{u}},\mathbf{\text{v}},$ and $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}.$

183.

$\mathbf{\text{u}} = \left\langle {2,0,0} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {2,2,0} \right\rangle$

184\.

$\mathbf{\text{u}} = \left\langle {3,2,-1} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {1,1,0} \right\rangle$

185.

$\mathbf{\text{u}} = 2\mathbf{\text{i}} + 3\mathbf{\text{j}},$ $\mathbf{\text{v}} = \mathbf{\text{j}} + 2\mathbf{\text{k}}$

186\.

$\mathbf{\text{u}} = 2\mathbf{\text{j}} + 3\mathbf{\text{k}},$ $\mathbf{\text{v}} = 3\mathbf{\text{i}} + \mathbf{\text{k}}$

187.

Simplify $\left( {\mathbf{\text{i}}\ \times \ \mathbf{\text{i}} - 2\mathbf{\text{i}}\ \times \ \mathbf{\text{j}} - 4\mathbf{\text{i}}\ \times \ \mathbf{\text{k}} + 3\mathbf{\text{j}}\ \times \ \mathbf{\text{k}}} \right)\ \times \ \mathbf{\text{i}}.$

188\.

Simplify $\mathbf{\text{j}}\ \times \ \left( {\mathbf{\text{k}}\ \times \ \mathbf{\text{j}} + 2\mathbf{\text{j}}\ \times \ \mathbf{\text{i}} - 3\mathbf{\text{j}}\ \times \ \mathbf{\text{j}} + 5\mathbf{\text{i}}\ \times \ \mathbf{\text{k}}} \right).$

In the following exercises, vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are given. Find unit vector $\mathbf{\text{w}}$ in the direction of the cross product vector $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}.$ Express your answer using standard unit vectors.

189.

$\mathbf{\text{u}} = \left\langle {3,-1,2} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {-2,0,1} \right\rangle$

190\.

$\mathbf{\text{u}} = \left\langle {2,6,1} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {3,0,1} \right\rangle$

191.

$\mathbf{\text{u}} = \overset{\rightarrow}{AB},$ $\mathbf{\text{v}} = \overset{\rightarrow}{AC},$ where $A(1,0,1),$ $B(1,-1,3),$ and $C(0,0,5)$

192\.

$\mathbf{\text{u}} = \overset{\rightarrow}{OP},$ $\mathbf{\text{v}} = \overset{\rightarrow}{PQ},$ where $P(-1,1,0)$ and $Q(0,2,1)$

193.

Determine the real number $\alpha$ such that $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ and $\mathbf{\text{i}}$ are orthogonal, where $\mathbf{\text{u}} = 3\mathbf{\text{i}} + \mathbf{\text{j}} - 5\mathbf{\text{k}}$ and $\mathbf{\text{v}} = 4\mathbf{\text{i}} - 2\mathbf{\text{j}} + \alpha\mathbf{\text{k}}.$

194\.

Show that $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ and $2\mathbf{\text{i}} - 14\mathbf{\text{j}} + 2\mathbf{\text{k}}$ cannot be orthogonal for any $\alpha$ real number, where $\mathbf{\text{u}} = \mathbf{\text{i}} + 7\mathbf{\text{j}} - \mathbf{\text{k}}$ and $\mathbf{\text{v}} = \alpha\mathbf{\text{i}} + 5\mathbf{\text{j}} + \mathbf{\text{k}}.$

195\.

Show that $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ is orthogonal to $\mathbf{\text{u}} + \mathbf{\text{v}}$ and $\mathbf{\text{u}} - \mathbf{\text{v}},$ where $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are nonzero vectors.

196\.

Show that $\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}$ is orthogonal to $\left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right)\left( {\mathbf{\text{u}} + \mathbf{\text{v}}} \right) + \textbf{u},$ where $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are nonzero vectors.

197.

Calculate the determinant $\left| \begin{array}{lrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

1 & -1 & 7 \\

2 & 0 & 3

\end{array} \right|.$

198\.

Calculate the determinant $\left| \begin{array}{lrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

0 & 3 & -4 \\

1 & 6 & -1

\end{array} \right|.$

For the following exercises, the vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are given. Use determinant notation to find vector $\mathbf{\text{w}}$ orthogonal to vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}.$

199.

$\mathbf{\text{u}} = \left\langle {-1,0,e^{t}} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {1,e^{\text{−}t},0} \right\rangle,$ where $t$ is a real number

200\.

$\mathbf{\text{u}} = \left\langle {1,0,x} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {\frac{2}{x},1,0} \right\rangle,$ where $x$ is a nonzero real number

201.

Find vector $\left( {\mathbf{\text{a}} - 2\textbf{b}} \right)\ \times \ \textbf{c},$ where $\mathbf{\text{a}} = \left| \begin{array}{lrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

2 & -1 & 5 \\

0 & 1 & 8

\end{array} \right|,$ $\textbf{b} = \left| \begin{array}{lrr}

\textbf{i} & \textbf{j} & \textbf{k} \\

0 & 1 & 1 \\

2 & -1 & -2

\end{array} \right|,$ and $\textbf{c} = \mathbf{\text{i}} + \mathbf{\text{j}} + \textbf{k}.$

202\.

Find vector $\textbf{c}\ \times \ \left( {\mathbf{\text{a}} + 3\textbf{b}} \right),$ where $\mathbf{\text{a}} = \left| \begin{array}{lll}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

5 & 0 & 9 \\

0 & 1 & 0

\end{array} \right|,$ $\textbf{b} = \left| \begin{array}{lrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

0 & -1 & 1 \\

7 & 1 & -1

\end{array} \right|,$ and $\textbf{c} = \mathbf{\text{i}} - \mathbf{\text{k}}.$

203.

\[T\] Use the cross product $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ to find the acute angle between vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}},$ where $\mathbf{\text{u}} = \mathbf{\text{i}} + 2\mathbf{\text{j}}$ and $\mathbf{\text{v}} = \mathbf{\text{i}} + \mathbf{\text{k}}.$ Express the answer in degrees rounded to the nearest integer.

204\.

\[T\] Use the cross product $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ to find the obtuse angle between vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}},$ where $\mathbf{\text{u}} = \text{−}\mathbf{\text{i}} + 3\mathbf{\text{j}} + \mathbf{\text{k}}$ and $\mathbf{\text{v}} = \mathbf{\text{i}} - 2\mathbf{\text{j}}.$ Express the answer in degrees rounded to the nearest integer.

205\.

Use the sine and cosine of the angle between two nonzero vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ to prove Lagrange’s identity: $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|^{2} = \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2} - \left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right)^{2}.$

206\.

Verify Lagrange’s identity $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|^{2} = \left\| \mathbf{\text{u}} \right\|^{2}\left\| \mathbf{\text{v}} \right\|^{2} - \left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right)^{2}$ for vectors $\mathbf{\text{u}} = \text{−}\mathbf{\text{i}} + \mathbf{\text{j}} - 2\mathbf{\text{k}}$ and $\mathbf{\text{v}} = 2\mathbf{\text{i}} - \mathbf{\text{j}}.$

207\.

Nonzero vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are called *collinear* if there exists a nonzero scalar $\alpha$ such that $\mathbf{\text{v}} = \alpha\mathbf{\text{u}}.$ Show that $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are collinear if and only if $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \mathbf{0}.$

208\.

Nonzero vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are called *collinear* if there exists a nonzero scalar $\alpha$ such that $\mathbf{\text{v}} = \alpha\mathbf{\text{u}}.$ Show that vectors $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{AC}$ are collinear, where $A\left( {4,1,0} \right),$ $B\left( {6,5,-2} \right),$ and $C\left( {5,3,-1} \right).$

209.

Find the area of the parallelogram with adjacent sides $\mathbf{\text{u}} = \left\langle {3,2,0} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {0,2,1} \right\rangle.$

210\.

Find the area of the parallelogram with adjacent sides $\mathbf{\text{u}} = \mathbf{\text{i}} + \mathbf{\text{j}}$ and $\mathbf{\text{v}} = \mathbf{\text{i}} + \mathbf{\text{k}}.$

211.

Consider points $A\left( {3,-1,2} \right),B\left( {2,1,5} \right),$ and $C\left( {1,-2,-2} \right).$

1. Find the area of parallelogram $ABCD$ with adjacent sides $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{AC}.$

2. Find the area of triangle $ABC.$

3. Find the distance from point $A$ to line $BC.$

212\.

Consider points $A\left( {2,-3,4} \right),B\left( {0,1,2} \right),$ and $C(-1,2,0).$

1. Find the area of parallelogram $ABCD$ with adjacent sides $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{AC}.$

2. Find the area of triangle $ABC.$

3. Find the distance from point $B$ to line $AC.$

In the following exercises, vectors $\mathbf{\text{u}},\mathbf{\text{v}},\text{and}\ \mathbf{\text{w}}$ are given.

1. Find the triple scalar product $\mathbf{\text{u}} \cdot (\mathbf{\text{v}}\ \times \ \textbf{w}).$

2. Find the volume of the parallelepiped with the adjacent edges $\mathbf{\text{u}},\mathbf{\text{v}},\text{and}\ \mathbf{\text{w}}.$

213.

$\mathbf{\text{u}} = \mathbf{\text{i}} + \mathbf{\text{j}},$ $\mathbf{\text{v}} = \mathbf{\text{j}} + \mathbf{\text{k}},$ and $\mathbf{\text{w}} = \mathbf{\text{i}} + \mathbf{\text{k}}$

214\.

$\mathbf{\text{u}} = \left\langle {-3,5,-1} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {0,2,-2} \right\rangle,$ and $\mathbf{\text{w}} = \left\langle {3,1,1} \right\rangle$

215.

Calculate the triple scalar products $\mathbf{\text{v}} \cdot (\mathbf{\text{u}}\ \times \ \textbf{w})$ and $\mathbf{\text{w}} \cdot (\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}),$ where $\mathbf{\text{u}} = \left\langle {1,1,1} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {7,6,9} \right\rangle,$ and $\mathbf{\text{w}} = \left\langle {4,2,7} \right\rangle.$

216\.

Calculate the triple scalar products $\mathbf{\text{w}} \cdot (\mathbf{\text{v}}\ \times \ \mathbf{\text{u}})$ and $\mathbf{\text{u}} \cdot (\textbf{w}\ \times \ \mathbf{\text{v}}),$ where $\mathbf{\text{u}} = \left\langle {4,2,-1} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {2,5,-3} \right\rangle,$ and $\mathbf{\text{w}} = \left\langle {9,5,-10} \right\rangle.$

217.

Find vectors $\mathbf{\text{a}},\mathbf{\text{b}},\ \text{and}\ \mathbf{\text{c}}$ with a triple scalar product given by the determinant

$\left| \begin{array}{lll}

1 & 2 & 3 \\

0 & 2 & 5 \\

8 & 9 & 2

\end{array} \right|.$ Determine their triple scalar product.

218\.

The triple scalar product of vectors $\mathbf{\text{a}},\mathbf{\text{b}},\ \text{and}\ \mathbf{\text{c}}$ is given by the determinant

$\left| \begin{array}{rrr}

0 & -2 & 1 \\

0 & 1 & 4 \\

1 & -3 & 7

\end{array} \right|.$ Find vector $\mathbf{\text{a}} - \textbf{b} + \textbf{c}.$

219.

Consider the parallelepiped with edges $OA,OB,$ and $OC,$ where $A\left( {2,1,0} \right),B\left( {1,2,0} \right),$ and $C\left( {0,1,\alpha} \right).$

1. Find the real number $\alpha > 0$ such that the volume of the parallelepiped is $3$ units³.

2. For $\alpha = 1,$ find the height $h$ from vertex $C$ of the parallelepiped to the plane formed by the edges $OA$ and $OB$.

220\.

Consider points $A\left( {\alpha,0,0} \right),B\left( {0,\beta,0} \right),$ and $C\left( {0,0,\gamma} \right),$ with $\alpha,$ $\beta,$ and $\gamma$ positive real numbers.

1. Determine the volume of the parallelepiped with adjacent sides $\overset{\rightarrow}{OA},$ $\overset{\rightarrow}{OB},$ and $\overset{\rightarrow}{OC}.$

2. Find the volume of the tetrahedron with vertices $O,A,B,\ \text{and}\ C.$ (*Hint*: The volume of the tetrahedron is $1\text{/}6$ of the volume of the parallelepiped.)

3. Find the distance from the origin to the plane determined by $A,B,\ \text{and}\ C.$ Sketch the parallelepiped and tetrahedron.

221\.

Let $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ be three-dimensional vectors and $c$ be a real number. Prove the following properties of the cross product.

1. $\mathbf{\text{u}}\ \times \ \mathbf{\text{u}} = \mathbf{0}$

2. $\mathbf{\text{u}}\ \times \ \left( {\mathbf{\text{v}} + \mathbf{\text{w}}} \right) = \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) + \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{w}}} \right)$

3. $c\left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) = \left( {c\mathbf{\text{u}}} \right)\ \times \ \mathbf{\text{v}} = \mathbf{\text{u}}\ \times \ \left( {c\mathbf{\text{v}}} \right)$

4. $\mathbf{\text{u}} \cdot \mathbf{(}\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}\mathbf{) = 0}$

222\.

Show that vectors $\mathbf{\text{u}} = \left\langle {1,0,-8} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {0,1,6} \right\rangle,$ and $\mathbf{\text{w}} = \left\langle {-1,9,3} \right\rangle$ satisfy the following properties of the cross product.

1. $\mathbf{\text{u}}\ \times \ \mathbf{\text{u}} = \mathbf{0}$

2. $\mathbf{\text{u}}\ \times \ \left( {\mathbf{\text{v}} + \mathbf{\text{w}}} \right) = \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) + \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{w}}} \right)$

3. $c\left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) = \left( {c\mathbf{\text{u}}} \right)\ \times \ \mathbf{\text{v}} = \mathbf{\text{u}}\ \times \ \left( {c\mathbf{\text{v}}} \right)$

4. $\mathbf{\text{u}} \cdot \mathbf{(}\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}\mathbf{) = 0}$

223\.

Nonzero vectors $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ are said to be *linearly dependent* if one of the vectors is a linear combination of the other two. For instance, there exist two nonzero real numbers $\alpha$ and $\beta$ such that $\mathbf{\text{w}} = \alpha\mathbf{\text{u}} + \beta\mathbf{\text{v}}.$ Otherwise, the vectors are called *linearly independent*. Show that $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ are coplanar if and only if they are linear dependent.

224\.

Consider vectors $\mathbf{\text{u}} = \left\langle {1,4,-7} \right\rangle,$ $\mathbf{\text{v}} = \left\langle {2,-1,4} \right\rangle,$ $\mathbf{\text{w}} = \left\langle {0,-9,18} \right\rangle,$ and $\textbf{p} = \left\langle {0,-9,17} \right\rangle.$

1. Show that $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ are coplanar by using their triple scalar product

2. Show that $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ are coplanar, using the definition that there exist two nonzero real numbers $\alpha$ and $\beta$ such that $\mathbf{\text{w}} = \alpha\mathbf{\text{u}} + \beta\mathbf{\text{v}}.$

3. Show that $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{p}}$ are linearly independent—that is, none of the vectors is a linear combination of the other two.

225.

Consider points $A(0,0,2),$ $B\left( {1,0,2} \right),$ $C\left( {1,1,2} \right),$ and $D\left( {0,1,2} \right).$ Are vectors $\overset{\rightarrow}{AB},$ $\overset{\rightarrow}{AC},$ and $\overset{\rightarrow}{AD}$ linearly dependent (that is, one of the vectors is a linear combination of the other two)?

226\.

Show that vectors $\mathbf{\text{i}} + \mathbf{\text{j}},$ $\mathbf{\text{i}} - \mathbf{\text{j}},$ and $\mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}$ are linearly independent—that is, there do not exist two nonzero real numbers $\alpha$ and $\beta$ such that $\mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}} = \alpha\left( {\mathbf{\text{i}} + \mathbf{\text{j}}} \right) + \beta\left( {\mathbf{\text{i}} - \mathbf{\text{j}}} \right).$

227.

Let $\mathbf{\text{u}} = \left\langle {u_{1},u_{2}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2}} \right\rangle$ be two-dimensional vectors. The cross product of vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ is not defined. However, if the vectors are regarded as the three-dimensional vectors $\widetilde{\mathbf{\text{u}}} = \left\langle {u_{1},u_{2},0} \right\rangle$ and $\widetilde{\mathbf{\text{v}}} = \left\langle {v_{1},v_{2},0} \right\rangle,$ respectively, then, in this case, we can define the cross product of $\widetilde{\mathbf{\text{u}}}$ and $\widetilde{\mathbf{\text{v}}}.$ In particular, in determinant notation, the cross product of $\widetilde{\mathbf{\text{u}}}$ and $\widetilde{\mathbf{\text{v}}}$ is given by

$$\widetilde{\mathbf{\text{u}}}\ \times \ \widetilde{\mathbf{\text{v}}} = \left| \begin{array}{lll}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

u_{1} & u_{2} & 0 \\

v_{1} & v_{2} & 0

\end{array} \right|.$$

Use this result to compute $(\mathbf{\text{i}}\textit{cos}\ \theta + \mathbf{\text{j}}\textit{sin}\ \theta)\ \times \ (\mathbf{\text{i}}sin\theta - \mathbf{\text{j}}cos\theta),$ where $\theta$ is a real number.

228\.

Consider points $P\left( {2,1} \right),$ $Q\left( {4,2} \right),$ and $R\left( {1,2} \right).$

1. Find the area of triangle $P,Q,\ \text{and}\ R.$

2. Determine the distance from point $R$ to the line passing through $P\ \text{and}\ Q.$

229.

Determine a vector of magnitude $10$ perpendicular to the plane passing through the *x*-axis and point $P\left( {1,2,4} \right).$

230\.

Determine a unit vector perpendicular to the plane passing through the *z*-axis and point $A\left( {3,1,-2} \right).$

231\.

Consider $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ two three-dimensional vectors. If the magnitude of the cross product vector $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ is $k$ times larger than the magnitude of vector $\mathbf{\text{u}},$ show that the magnitude of $\mathbf{\text{v}}$ is greater than or equal to $k,$ where $k$ is a natural number.

232\.

\[T\] Assume that the magnitudes of two nonzero vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are known. The function $f(\theta) = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{sin}\ \theta$ defines the magnitude of the cross product vector $\mathbf{\text{u}}\ \times \ \textbf{v},$ where $\theta \in \lbrack 0,\pi\rbrack$ is the angle between $\mathbf{\text{u}}\ \text{and}\ \mathbf{\text{v}}.$

1. Graph the function $f.$

2. Find the absolute minimum and maximum of function $f.$ Interpret the results.

3. If $\left\| \mathbf{\text{u}} \right\| = 5$ and $\left\| \mathbf{\text{v}} \right\| = 2,$ find the angle between $\mathbf{\text{u}}\ \text{and}\ \mathbf{\text{v}}$ if the magnitude of their cross product vector is equal to $9.$

233.

Find all vectors $\mathbf{\text{w}} = \left\langle {w_{1},w_{2},w_{3}} \right\rangle$ that satisfy the equation $\left\langle {1,1,1} \right\rangle\ \times \ \mathbf{\text{w}} = \left\langle {-1,-1,2} \right\rangle.$

234\.

Solve the equation $\mathbf{\text{w}}\ \times \ \left\langle {1,0,-1} \right\rangle = \left\langle {3,0,3} \right\rangle,$ where $\mathbf{\text{w}} = \left\langle {w_{1},w_{2},w_{3}} \right\rangle$ is a nonzero vector with a magnitude of $3.$

235.

\[T\] A mechanic uses a 12-in. wrench to turn a bolt. The wrench makes a $30\text{°}$ angle with the horizontal. If the mechanic applies a vertical force of $10$ lb on the wrench handle, what is the magnitude of the torque at point $P$ (see the following figure)? Express the answer in foot-pounds rounded to two decimal places.

236\.

\[T\] A boy applies the brakes on a bicycle by applying a downward force of $20$ lb on the pedal when the 6-in. crank makes a $40\text{°}$ angle with the horizontal (see the following figure). Find the torque at point $P.$ Express your answer in foot-pounds rounded to two decimal places.

237.

\[T\] Find the magnitude of the force that needs to be applied to the end of a 20-cm wrench located on the positive direction of the *y*-axis if the force is applied in the direction $\left\langle {0,1,-2} \right\rangle$ and it produces a $100$ N·m torque to the bolt located at the origin.

238\.

\[T\] What is the magnitude of the force required to be applied to the end of a 1-ft wrench at an angle of $35\text{°}$ to produce a torque of $20$ ft-lbs?

239.

\[T\] The force vector $\mathbf{\text{F}}$ acting on a proton with an electric charge of $1.6\ \times \ 10^{-19}\text{C}$ (in coulombs) moving in a magnetic field $\mathbf{\text{B}}$ where the velocity vector $\mathbf{\text{v}}$ is given by $\mathbf{\text{F}} = 1.6\ \times \ 10^{-19}\left( {\mathbf{\text{v}}\ \times \ \textbf{B}} \right)$ (here, $\mathbf{\text{v}}$ is expressed in meters per second, $\mathbf{\text{B}}$ is in tesla \[T\], and $\mathbf{\text{F}}$ is in newtons \[N\]). Find the force that acts on a proton that moves in the *xy*-plane at velocity $\mathbf{\text{v}} = 10^{5}\mathbf{\text{i}} + 10^{5}\mathbf{\text{j}}$ (in meters per second) in a magnetic field given by $\textbf{B} = 0.3\mathbf{\text{j}}.$

240\.

\[T\] The force vector $\mathbf{\text{F}}$ acting on a proton with an electric charge of $1.6\ \times \ 10^{-19}\text{C}$ moving in a magnetic field $\mathbf{\text{B}}$ where the velocity vector v is given by $\mathbf{\text{F}} = 1.6\ \times \ 10^{-19}\left( {\mathbf{\text{v}}\ \times \ \textbf{B}} \right)$ (here, $\mathbf{\text{v}}$ is expressed in meters per second, $\mathbf{\text{B}}$ in $\text{T},$ and $\mathbf{\text{F}}$ in $\text{N}).$ If the magnitude of force $\mathbf{\text{F}}$ acting on a proton is $5.9\ \times \ 10^{-17}$ N and the proton is moving at the speed of 300 m/sec in magnetic field $\mathbf{\text{B}}$ of magnitude 2.4 T, find the angle between velocity vector $\mathbf{\text{v}}$ of the proton and magnetic field $\mathbf{\text{B}}.$ Express the answer in degrees rounded to the nearest integer.

241.

\[T\] Consider $\textbf{r}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t,2t} \right\rangle$ the position vector of a particle at time $t \in \lbrack 0,30\rbrack,$ where the components of $\mathbf{\text{r}}$ are expressed in centimeters and time in seconds. Let $\overset{\rightarrow}{OP}$ be the position vector of the particle after $1$ sec.

1. Determine unit vector $\textbf{B}(t)$ (called the *binormal unit vector*) that has the direction of cross product vector $\mathbf{\text{v}}(t)\ \times \ \textbf{a}(t),$ where $\mathbf{\text{v}}(t)$ and $\mathbf{\text{a}}(t)$ are the instantaneous velocity vector and, respectively, the acceleration vector of the particle after $t$ seconds.

2. Use a CAS to visualize vectors $\mathbf{\text{v}}(1),$ $\mathbf{\text{a}}(1),$ and $\textbf{B}(1)$ as vectors starting at point $P$ along with the path of the particle.

242\.

A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) $A(8,0,0),$ $B(8,18,0),$ $C(0,18,8),$ and $D(0,0,8)$ (see the following figure).

1. Find vector $\textbf{n} = \overset{\rightarrow}{AB}\ \times \ \overset{\rightarrow}{AD}$ perpendicular to the surface of the solar panels. Express the answer using standard unit vectors.

2. Assume unit vector $\textbf{s} = \frac{1}{\sqrt{3}}\mathbf{\text{i}} + \frac{1}{\sqrt{3}}\mathbf{\text{j}} + \frac{1}{\sqrt{3}}\mathbf{\text{k}}$ points toward the Sun at a particular time of the day and the flow of solar energy is $\mathbf{\text{F}} = 900\textbf{s}$ (in watts per square meter \[$\text{W/m}^{2}$\]). Find the predicted amount of electrical power the panel can produce, which is given by the dot product of vectors $\mathbf{\text{F}}$ and $\mathbf{\text{n}}$ (expressed in watts).

3. Determine the angle of elevation of the Sun above the solar panel. Express the answer in degrees rounded to the nearest whole number. (*Hint*: The angle between vectors $\mathbf{\text{n}}$ and $\mathbf{\text{s}}$ and the angle of elevation are complementary.)

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2.5 Equations of Lines and Planes in Space

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-5-equations-of-lines-and-planes-in-space

2.5 Equations of Lines and Planes in Space

• 2.5.1 Write the vector, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.
• 2.5.2 Find the distance from a point to a given line.
• 2.5.3 Write the vector and scalar equations of a plane through a given point with a given normal.
• 2.5.4 Find the distance from a point to a given plane.
• 2.5.5 Find the angle between two planes.

By now, we are familiar with writing equations that describe a line in two dimensions. To write an equation for a line, we must know two points on the line, or we must know the direction of the line and at least one point through which the line passes. In two dimensions, we use the concept of slope to describe the orientation, or direction, of a line. In three dimensions, we describe the direction of a line using a vector parallel to the line. In this section, we examine how to use equations to describe lines and planes in space.

Equations for a Line in Space

Let’s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, $\mathbf{\text{u}}$ and $\mathbf{\text{v}},$ are parallel, we claim there must be a scalar, $k,$ such that $\mathbf{\text{u}} = k\mathbf{\text{v}}.$ If $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have the same direction, simply choose $k = \frac{\left\| \mathbf{\text{u}} \right\|}{\left\| \mathbf{\text{v}} \right\|}.$ If $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have opposite directions, choose $k = - \frac{\left\| \mathbf{\text{u}} \right\|}{\left\| \mathbf{\text{v}} \right\|}.$ Note that the converse holds as well. If $\mathbf{\text{u}} = k\mathbf{\text{v}}$ for some scalar $\text{k},$ then either $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ have the same direction $\left( {k > 0} \right)$ or opposite directions $\left( {k < 0} \right),$ so $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are parallel. Therefore, two nonzero vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ are parallel if and only if $\mathbf{\text{u}} = k\mathbf{\text{v}}$ for some scalar $\text{k}.$ By convention, the zero vector $\mathbf{0}$ is considered to be parallel to all vectors.

As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the direction vector (Figure 2.63). Let $L$ be a line in space passing through point $P\left( {x_{0},y_{0},z_{0}} \right).$ Let $\mathbf{\text{v}} = \left\langle {a,b,c} \right\rangle$ be a vector parallel to $L.$ Then, for any point $Q\left( {x,y,z} \right)$ on line $L$, we know that $\overset{\rightarrow}{PQ}$ is parallel to $\mathbf{\text{v}}.$ Thus, as we just discussed, there is a scalar, $t,$ such that $\overset{\rightarrow}{PQ} = t\mathbf{\text{v}},$ which gives

$$\begin{array}{rll}

\overset{\rightarrow}{PQ} & = & {t\mathbf{\text{v}}} \\

\left\langle {x - x_{0},y - y_{0},z - z_{0}} \right\rangle & = & {t\left\langle {a,b,c} \right\rangle} \\

\left\langle {x - x_{0},y - y_{0},z - z_{0}} \right\rangle & = & {\left\langle {ta,tb,tc} \right\rangle.}

\end{array}$$ (2.11)

Using vector operations, we can rewrite Equation 2.11 as

$$\begin{array}{rll}

\left\langle {x - x_{0},y - y_{0},z - z_{0}} \right\rangle & = & \left\langle {ta,tb,tc} \right\rangle \\

{\left\langle {x,y,z} \right\rangle - \left\langle {x_{0},y_{0},z_{0}} \right\rangle} & = & {t\left\langle {a,b,c} \right\rangle} \\

\left\langle {x,y,z} \right\rangle & = & {\left\langle {x_{0},y_{0},z_{0}} \right\rangle + t\left\langle {a,b,c} \right\rangle.}

\end{array}$$

Setting $\mathbf{\text{r}} = \left\langle {x,y,z} \right\rangle$ and $\mathbf{\text{r}}_{0} = \left\langle {x_{0},y_{0},z_{0}} \right\rangle,$ we now have the vector equation of a line:

$$\mathbf{\text{r}} = \mathbf{\text{r}}_{0} + t\mathbf{\text{v}}.$$ (2.12)

Equating components, Equation 2.11 shows that the following equations are simultaneously true: $x - x_{0} = ta,$ $y - y_{0} = tb,$ and $z - z_{0} = tc.$ If we solve each of these equations for the component variables $x,y,\ \text{and}\ z,$ we get a set of equations in which each variable is defined in terms of the parameter *t* and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

$$x = x_{0} + ta\quad y = y_{0} + tb\quad z = z_{0} + tc.$$

If we solve each of the equations for $t$ assuming $a,b,\ \text{and}\ c$ are nonzero, we get a different description of the same line:

$$\frac{x - x_{0}}{a} = t\quad\frac{y - y_{0}}{b} = t\quad\frac{z - z_{0}}{c} = t.$$

Because each expression equals *t*, they all have the same value. We can set them equal to each other to create symmetric equations of a line:

$$\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}.$$

We summarize the results in the following theorem.

Parametric and Symmetric Equations of a Line

A line $L$ parallel to vector $\mathbf{\text{v}} = \left\langle {a,b,c} \right\rangle$ and passing through point $P\left( {x_{0},y_{0},z_{0}} \right)$ can be described by the following parametric equations:

$$x = x_{0} + ta,y = y_{0} + tb,\ \text{and}\ z = z_{0} + tc.$$ (2.13)

If the constants $a,b,\ \text{and}\ c$ are all nonzero, then $L$ can be described by the symmetric equation of the line:

$$\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}.$$ (2.14)

The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads to a different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is not unique either.

Equations of a Line in Space

Find parametric and symmetric equations of the line passing through points $\left( {1,4,-2} \right)$ and $(-3,5,0).$

Solution

First, identify a vector parallel to the line:

$$\mathbf{\text{v}} = \left\langle {-3 - 1,5 - 4,0 - (-2)} \right\rangle = \left\langle {-4,1,2} \right\rangle.$$

Use either of the given points on the line to complete the parametric equations:

$$x = 1 - 4t,y = 4 + t,\ \text{and}\ z = -2 + 2t.$$

Solve each equation for $t$ to create the symmetric equation of the line:

$$\frac{x - 1}{-4} = y - 4 = \frac{z + 2}{2}.$$

Find parametric and symmetric equations of the line passing through points $\left( {1,-3,2} \right)$ and $\left( {5,-2,8} \right).$

Sometimes we don’t want the equation of a whole line, just a line segment. In this case, we limit the values of our parameter $t.$ For example, let $P\left( {x_{0},y_{0},z_{0}} \right)$ and $Q\left( {x_{1},y_{1},z_{1}} \right)$ be points on a line, and let $\mathbf{\text{p}} = \left\langle {x_{0},y_{0},z_{0}} \right\rangle$ and $\mathbf{\text{q}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle$ be the associated position vectors. In addition, let $\mathbf{\text{r}} = \left\langle {x,y,z} \right\rangle.$ We want to find a vector equation for the line segment between $P$ and $Q.$ Using $P$ as our known point on the line, and $\overset{\rightarrow}{PQ} = \left\langle {x_{1} - x_{0},y_{1} - y_{0},z_{1} - z_{0}} \right\rangle$ as the direction vector equation, Equation 2.12 gives

$$\mathbf{\text{r}} = \mathbf{\text{p}} + t\left( \overset{\rightarrow}{PQ} \right).$$

Using properties of vectors, then

$$\begin{array}{cl}

\mathbf{\text{r}} & {= \mathbf{\text{p}} + t\left( \overset{\rightarrow}{PQ} \right)} \\

& {= \left\langle {x_{0},y_{0},z_{0}} \right\rangle + t\left\langle {x_{1} - x_{0},y_{1} - y_{0},z_{1} - z_{0}} \right\rangle} \\

& {= \left\langle {x_{0},y_{0},z_{0}} \right\rangle + t\left( {\left\langle {x_{1},y_{1},z_{1}} \right\rangle - \left\langle {x_{0},y_{0},z_{0}} \right\rangle} \right)} \\

& {= \left\langle {x_{0},y_{0},z_{0}} \right\rangle + t\left\langle {x_{1},y_{1},z_{1}} \right\rangle - t\left\langle {x_{0},y_{0},z_{0}} \right\rangle} \\

& {= \left( {1 - t} \right)\left\langle {x_{0},y_{0},z_{0}} \right\rangle + t\left\langle {x_{1},y_{1},z_{1}} \right\rangle} \\

& {= \left( {1 - t} \right)\mathbf{\text{p}} + t\mathbf{\text{q}}.}

\end{array}$$

Thus, the vector equation of the line passing through $P$ and $Q$ is

$$\mathbf{\text{r}} = \left( {1 - t} \right)\mathbf{\text{p}} + t\mathbf{\text{q}}.$$

Remember that we didn’t want the equation of the whole line, just the line segment between $P$ and $Q.$ Notice that when $t = 0,$ we have $\mathbf{r = p},$ and when $t = 1,$ we have $\mathbf{r = q}.$ Therefore, the vector equation of the line segment between $P$ and $Q$ is

$$\mathbf{\text{r}} = \left( {1 - t} \right)\mathbf{\text{p}} + t\mathbf{\text{q}},0 \leq t \leq 1.$$ (2.15)

Going back to Equation 2.12, we can also find parametric equations for this line segment. We have

$$\begin{array}{cll}

\mathbf{\text{r}} & = & {\mathbf{\text{p}} + t\left( \overset{\rightarrow}{PQ} \right)} \\

\left\langle {x,y,z} \right\rangle & = & {\left\langle {x_{0},y_{0},z_{0}} \right\rangle + t\left\langle {x_{1} - x_{0},y_{1} - y_{0},z_{1} - z_{0}} \right\rangle} \\

& = & {\left\langle {x_{0} + t\left( {x_{1} - x_{0}} \right),y_{0} + t\left( {y_{1} - y_{0}} \right),z_{0} + t\left( {z_{1} - z_{0}} \right)} \right\rangle.}

\end{array}$$

Then, the parametric equations are

$$x = x_{0} + t\left( {x_{1} - x_{0}} \right),y = y_{0} + t\left( {y_{1} - y_{0}} \right),z = z_{0} + t\left( {z_{1} - z_{0}} \right),0 \leq t \leq 1.$$ (2.16)

Parametric Equations of a Line Segment

Find parametric equations of the line segment between the points $P(2,1,4)$ and $Q\left( {3,-1,3} \right).$

Solution

By Equation 2.16, we have

$$x = x_{0} + t\left( {x_{1} - x_{0}} \right),y = y_{0} + t\left( {y_{1} - y_{0}} \right),z = z_{0} + t\left( {z_{1} - z_{0}} \right),0 \leq t \leq 1.$$

Working with each component separately, we get

$$\begin{array}{cl}

x & {= x_{0} + t\left( {x_{1} - x_{0}} \right)} \\

& {= 2 + t\left( {3 - 2} \right)} \\

& {= 2 + t,}

\end{array}$$ $$\begin{array}{cl}

y & {= y_{0} + t\left( {y_{1} - y_{0}} \right)} \\

& {= 1 + t\left( {-1 - 1} \right)} \\

& {= 1 - 2t,}

\end{array}$$

and

$$\begin{array}{cl}

z & {= z_{0} + t\left( {z_{1} - z_{0}} \right)} \\

& {= 4 + t\left( {3 - 4} \right)} \\

& {= 4 - t.}

\end{array}$$

Therefore, the parametric equations for the line segment are

$$x = 2 + t,y = 1 - 2t,z = 4 - t,0 \leq t \leq 1.$$

Find parametric equations of the line segment between points $P(-1,3,6)$ and $Q\left( {-8,2,4} \right).$

Distance between a Point and a Line

We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider “setback” requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.

Let $L$ be a line in the plane and let $M$ be any point not on the line. Then, we define distance $d$ from $M$ to $L$ as the length of line segment $\overset{—}{MP},$ where $P$ is a point on $L$ such that $\overset{—}{MP}$ is perpendicular to $L$ (Figure 2.64).

When we’re looking for the distance between a line and a point in space, Figure 2.64 still applies. We still define the distance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clear way to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the line and use properties of vectors to calculate the distance. Therefore, let $P$ be an arbitrary point on line $L$ and let $\mathbf{\text{v}}$ be a direction vector for $L$ (Figure 2.65).

By Area of a Parallelogram, vectors $\overset{\rightarrow}{PM}$ and $\mathbf{\text{v}}$ form two sides of a parallelogram with area $\left\| {\overset{\rightarrow}{PM}\ \times \ \mathbf{\text{v}}} \right\|.$ Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:

$$\left\| {\overset{\rightarrow}{PM}\ \times \ \mathbf{\text{v}}} \right\| = \left\| \mathbf{\text{v}} \right\| d.$$

We can use this formula to find a general formula for the distance between a line in space and any point not on the line.

Distance from a Point to a Line

Let $L$ be a line in space passing through point $P$ with direction vector $\mathbf{\text{v}}.$ If $M$ is any point not on $L,$ then the distance from $M$ to $L$ is

$$d = \frac{\left\| {\overset{\rightarrow}{PM}\ \times \ \mathbf{\text{v}}} \right\|}{\left\| \mathbf{\text{v}} \right\|}.$$

Calculating the Distance from a Point to a Line

Find the distance between the point $M = \left( {1,1,3} \right)$ and line $\frac{x - 3}{4} = \frac{y + 1}{2} = z - 3.$

Solution

From the symmetric equations of the line, we know that vector $\mathbf{\text{v}} = \left\langle {4,2,1} \right\rangle$ is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point $P\left( {3,-1,3} \right)$ lies on the line. Then,

$$\overset{\rightarrow}{PM} = \left\langle {1 - 3,1 - (-1),3 - 3} \right\rangle = \left\langle {-2,2,0} \right\rangle.$$

To calculate the distance, we need to find $\overset{\rightarrow}{PM}\ \times \ \mathbf{\text{v}}\text{:}$

$$\begin{array}{cl}

{\overset{\rightarrow}{PM}\ \times \ \mathbf{\text{v}}} & {= \left| \begin{array}{rcc}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

-2 & 2 & 0 \\

4 & 2 & 1

\end{array} \right|} \\

& {= \left( {2 - 0} \right)\mathbf{\text{i}} - \left( {-2 - 0} \right)\mathbf{\text{j}} + \left( {-4 - 8} \right)\mathbf{\text{k}}} \\

& {= 2\mathbf{\text{i}} + 2\mathbf{\text{j}} - 12\mathbf{\text{k}}.}

\end{array}$$

Therefore, the distance between the point and the line is (Figure 2.66)

$$\begin{array}{cl}

d & {= \frac{\left\| {\overset{\rightarrow}{PM}\ \times \ \mathbf{\text{v}}} \right\|}{\left\| \mathbf{\text{v}} \right\|}} \\

& {= \frac{\sqrt{2^{2} + 2^{2} + 12^{2}}}{\sqrt{4^{2} + 2^{2} + 1^{2}}}} \\

& {= \frac{2\sqrt{38}}{\sqrt{21}}.}

\end{array}$$

Find the distance between point $\left( {0,3,6} \right)$ and the line with parametric equations $x = 1 - t,y = 1 + 2t,z = 5 + 3t.$

Relationships between Lines

Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines (Figure 2.67).

To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the direction vectors are parallel and whether the lines share a point (Figure 2.68).

Classifying Lines in Space

For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.

1. $L_{1}:x = 2s - 1,y = s - 1,z = s - 4$

$L_{2}:x = t - 3,y = 3t + 8,z = 5 - 2t$

2. $L_{1}\text{:}$ $x = \text{−}y = z$

$L_{2}:\frac{x - 3}{2} = y = z - 2$

3. $L_{1}:x = 6s - 1,y = -2s,z = 3s + 1;s \neq 0$

$L_{2}:\frac{x - 4}{6} = \frac{y + 3}{-2} = \frac{z - 1}{3}$

Solution

1. Line $L_{1}$ has direction vector $\mathbf{\text{v}}_{\mathbf{1}} = \left\langle {2,1,1} \right\rangle;$ line $L_{2}$ has direction vector $\mathbf{\text{v}}_{\mathbf{2}} = \left\langle {1,3,-2} \right\rangle.$ Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To determine whether the lines intersect, we see if there is a point, $\left( {x,y,z} \right),$ that lies on both lines. To find this point, we use the parametric equations to create a system of equalities:

$$2s - 1 = t - 3;\quad s - 1 = 3t + 8;\quad s - 4 = 5 - 2t.$$

By the first equation, $t = 2s + 2.$ Substituting into the second equation yields

$$\begin{array}{rll}

{s - 1} & = & {3\left( {2s + 2} \right) + 8} \\

{s - 1} & = & {6s + 6 + 8} \\

{5s} & = & -15 \\

s & = & -3.

\end{array}$$

Substitution into the third equation, however, yields a contradiction:

$$\begin{array}{rll}

{s - 4} & = & {5 - 2\left( {2s + 2} \right)} \\

{s - 4} & = & {5 - 4s - 4} \\

{5s} & = & 5 \\

s & = & 1.

\end{array}$$

There is no single point that satisfies the parametric equations for $L_{1}\ \text{and}\ L_{2}$ simultaneously. These lines do not intersect, so they are skew (see the following figure).

2. Line *L₁* has direction vector $\mathbf{\text{v}}_{\mathbf{1}} = \left\langle {1,-1,1} \right\rangle$ and passes through the origin, $\left( {0,0,0} \right).$ Line $L_{2}$ has a different direction vector, $\mathbf{\text{v}}_{\mathbf{2}} = \left\langle {2,1,1} \right\rangle,$ so these lines are not parallel or equal. Let $r$ represent the parameter for line $L_{1}$ and let $s$ represent the parameter for $L_{2}\text{:}$

$$\begin{array}{lccl}

\begin{array}{ll}

x & {= r} \\

y & {= \text{−}r} \\

z & {= r}

\end{array} & & & \begin{array}{ll}

x & {= 2s + 3} \\

y & {= s} \\

z & {= s + 2.}

\end{array}

\end{array}$$

Solve the system of equations to find $r = 1$ and $s = - 1.$ If we need to find the point of intersection, we can substitute these parameters into the original equations to get $\left( {1,-1,1} \right)$ (see the following figure).

3. Lines $L_{1}$ and $L_{2}$ have equivalent direction vectors: $\mathbf{\text{v}} = \left\langle {6,-2,3} \right\rangle.$ These two lines are parallel (see the following figure).

Describe the relationship between the lines with the following parametric equations:

$$x = 1 - 4t,y = 3 + t,z = 8 - 6t$$ $$x = 2 + 3s,y = 2s,z = -1 - 3s.$$

Equations for a Plane

We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by three points. Perhaps the most surprising characterization of a plane is actually the most useful.

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let $\mathbf{\text{n}} = \left\langle {a,b,c} \right\rangle$ be a vector and $P = \left( {x_{0},y_{0},z_{0}} \right)$ be a point. Then the set of all points $Q = \left( {x,y,z} \right)$ such that $\overset{\rightarrow}{PQ}$ is orthogonal to $\mathbf{\text{n}}$ forms a plane (Figure 2.69). We say that $\mathbf{\text{n}}$ is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane: $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0.$ Rewriting this equation provides additional ways to describe the plane:

$$\begin{array}{rll}

& & \\

{\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ}} & = & 0 \\

{\left\langle {a,b,c} \right\rangle \cdot \left\langle {x - x_{0},y - y_{0},z - z_{0}} \right\rangle} & = & 0 \\

{a\left( {x - x_{0}} \right) + b\left( {y - y_{0}} \right) + c\left( {z - z_{0}} \right)} & = & 0.

\end{array}$$

Given a point $P$ and vector $\mathbf{\text{n}},$ the set of all points $Q$ satisfying the equation $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0$ forms a plane. The equation

$$\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0$$ (2.17)

is known as the vector equation of a plane.

The scalar equation of a plane containing point $P = \left( {x_{0},y_{0},z_{0}} \right)$ with normal vector $\mathbf{\text{n}} = \left\langle {a,b,c} \right\rangle$ is

$$a\left( {x - x_{0}} \right) + b\left( {y - y_{0}} \right) + c\left( {z - z_{0}} \right) = 0.$$ (2.18)

This equation can be expressed as $ax + by + cz + d = 0,$ where $d = \text{−}ax_{0} - by_{0} - cz_{0}.$ This form of the equation is sometimes called the general form of the equation of a plane.

As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.

Writing an Equation of a Plane Given Three Points in the Plane

Write an equation for the plane containing points $P = \left( {1,1,-2} \right),$ $Q = \left( {0,2,1} \right),$ and $R = \left( {-1,-1,0} \right)$ in both standard and general forms.

Solution

To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectors in the plane:

$$\begin{array}{rll}

\overset{\rightarrow}{PQ} & = & {\left\langle {0 - 1,2 - 1,1 - (-2)} \right\rangle = \left\langle {-1,1,3} \right\rangle} \\

\overset{\rightarrow}{QR} & = & {\left\langle {-1 - 0,-1 - 2,0 - 1} \right\rangle = \left\langle {-1,-3,-1} \right\rangle.}

\end{array}$$

The cross product $\overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{QR}$ is orthogonal to both $\overset{\rightarrow}{PQ}$ and $\overset{\rightarrow}{QR},$ so it is normal to the plane that contains these two vectors:

$$\begin{array}{cl}

\mathbf{\text{n}} & {= \overset{\rightarrow}{PQ}\ \times \ \overset{\rightarrow}{QR}} \\

& {= \left| \begin{array}{rrr}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

-1 & 1 & 3 \\

-1 & -3 & -1

\end{array} \right|} \\

& {= \left( {-1 + 9} \right)\mathbf{\text{i}} - \left( {1 + 3} \right)\mathbf{\text{j}} + \left( {3 + 1} \right)\mathbf{\text{k}}} \\

& {= 8\mathbf{\text{i}} - 4\mathbf{\text{j}} + 4\mathbf{\text{k}}.}

\end{array}$$

Thus, $\mathbf{\text{n}} = \left\langle {8,-4,4} \right\rangle,$ and we can choose any of the three given points to write an equation of the plane:

$$\begin{array}{rll}

{8(x - 1) - 4(y - 1) + 4(z + 2)} & = & 0 \\

{8x - 4y + 4z + 4} & = & 0.

\end{array}$$

The scalar equations of a plane vary depending on the normal vector and point chosen.

Writing an Equation for a Plane Given a Point and a Line

Find an equation of the plane that passes through point $\left( {1,4,3} \right)$ and contains the line given by $x = \frac{y - 1}{2} = z + 1.$

Solution

Symmetric equations describe the line that passes through point $\left( {0,1,\text{−}1} \right)$ parallel to vector $\mathbf{\text{v}}_{1} = \left\langle {1,2,1} \right\rangle$ (see the following figure). Use this point and the given point, $\left( {1,4,3} \right),$ to identify a second vector parallel to the plane:

$$\mathbf{\text{v}}_{2} = \left\langle {1 - 0,4 - 1,3 - (-1)} \right\rangle = \left\langle {1,3,4} \right\rangle.$$

Use the cross product of these vectors to identify a normal vector for the plane:

$$\begin{array}{cl}

\mathbf{\text{n}} & {= \mathbf{\text{v}}_{1}\ \times \ \mathbf{\text{v}}_{2}} \\

& {= \left| \begin{matrix}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

1 & 2 & 1 \\

1 & 3 & 4

\end{matrix} \right|} \\

& {= \left( {8 - 3} \right)\mathbf{\text{i}} - \left( {4 - 1} \right)\mathbf{\text{j}} + \left( {3 - 2} \right)\mathbf{\text{k}}} \\

& {= 5\mathbf{\text{i}} - 3\mathbf{\text{j}} + \mathbf{\text{k}}.}

\end{array}$$

The scalar equations for the plane are $5x - 3\left( {y - 1} \right) + \left( {z + 1} \right) = 0$ and $5x - 3y + z + 4 = 0.$

Find an equation of the plane containing the lines $L_{1}$ and $L_{2}\text{:}$

$$\begin{array}{l}

{L_{1}:x = \text{−}y = z} \\

{L_{2}:\frac{x - 3}{2} = y = z - 2.}

\end{array}$$

Now that we can write an equation for a plane, we can use the equation to find the distance $d$ between a point $P$ and the plane. It is defined as the shortest possible distance from $P$ to a point on the plane.

Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segment perpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a line segment perpendicular to the plane. Let $R$ be the point in the plane such that $\overset{\rightarrow}{RP}$ is orthogonal to the plane, and let $Q$ be an arbitrary point in the plane. Then the projection of vector $\overset{\rightarrow}{QP}$ onto the normal vector describes vector $\overset{\rightarrow}{RP},$ as shown in Figure 2.70.

The Distance between a Plane and a Point

Suppose a plane with normal vector $\mathbf{\text{n}}$ passes through point $Q.$ The distance $d$ from the plane to a point $P$ not in the plane is given by

$$d = \left\| {\text{proj}_{\mathbf{\text{n}}}\overset{\rightarrow}{QP}} \right\| = \left| {\text{comp}_{\mathbf{\text{n}}}\overset{\rightarrow}{QP}} \right| = \frac{\left| {\overset{\rightarrow}{QP} \cdot \mathbf{\text{n}}} \right|}{\left\| \mathbf{\text{n}} \right\|}.$$ (2.19)

Distance between a Point and a Plane

Find the distance between point $P = \left( {3,1,2} \right)$ and the plane given by $x - 2y + z = 5$ (see the following figure).

Solution

The coefficients of the plane’s equation provide a normal vector for the plane: $\mathbf{\text{n}} = \left\langle {1,-2,1} \right\rangle.$ To find vector $\overset{\rightarrow}{QP},$ we need a point in the plane. Any point will work, so set $y = z = 0$ to see that point $Q = \left( {5,0,0} \right)$ lies in the plane. Find the component form of the vector from $Q\ \text{to}\ P\text{:}$

$$\overset{\rightarrow}{QP} = \left\langle {3 - 5,1 - 0,2 - 0} \right\rangle = \left\langle {-2,1,2} \right\rangle.$$

Apply the distance formula from Equation 2.19:

$$\begin{array}{cl}

d & {= \frac{\left| {\overset{\rightarrow}{QP} \cdot \mathbf{\text{n}}} \right|}{\left\| \mathbf{\text{n}} \right\|}} \\

& {= \frac{\left| {\left\langle {-2,1,2} \right\rangle \cdot \left\langle {1,-2,1} \right\rangle} \right|}{\sqrt{1^{2} + (-2)^{2} + 1^{2}}}} \\

& {= \frac{\left| {-2 - 2 + 2} \right|}{\sqrt{6}}} \\

& {= \frac{2}{\sqrt{6}}.}

\end{array}$$

Find the distance between point $P = \left( {5,-1,0} \right)$ and the plane given by $4x + 2y - z = 3.$

Parallel and Intersecting Planes

We have discussed the various possible relationships between two lines in two dimensions and three dimensions. When we describe the relationship between two planes in space, we have only two possibilities: the two distinct planes are parallel or they intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection is a line (Figure 2.71).

We can use the equations of the two planes to find parametric equations for the line of intersection.

Finding the Line of Intersection for Two Planes

Find parametric and symmetric equations for the line formed by the intersection of the planes given by $x + y + z = 0$ and $2x - y + z = 0$ (see the following figure).

Solution

Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, $y\text{:}$

$$\begin{matrix}

\underset{\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}}{\begin{array}{rcccccc}

x & + & y & + & z & = & 0 \\

{2x} & - & y & + & z & = & 0

\end{array}} \\

\\

{\mspace{11mu} 3x\mspace{57mu} + 2z\mspace{8mu} = \mspace{7mu} 0.}

\end{matrix}$$

This gives us $x = - \frac{2}{3}z.$ We substitute this value into the first equation to express $y$ in terms of $z\text{:}$

$$\begin{array}{rll}

{x + y + z} & = & 0 \\

{- \frac{2}{3}z + y + z} & = & 0 \\

{y + \frac{1}{3}z} & = & 0 \\

y & = & {- \frac{1}{3}z.}

\end{array}$$

We now have the first two variables, $x$ and $y,$ in terms of the third variable, $z.$ Now we define $z$ in terms of $t.$ To eliminate the need for fractions, we choose to define the parameter $t$ as $t = - \frac{1}{3}z.$ Then, $z = -3t.$ Substituting the parametric representation of $z$ back into the other two equations, we see that the parametric equations for the line of intersection are $x = 2t,y = t,z = -3t.$ The symmetric equations for the line are $\frac{x}{2} = y = \frac{z}{-3}.$

Find parametric equations for the line formed by the intersection of planes $x + y - z = 3$ and $3x - y + 3z = 5.$

In addition to finding the equation of the line of intersection between two planes, we may need to find the angle formed by the intersection of two planes. For example, builders constructing a house need to know the angle where different sections of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate the angle between the two planes. We can do this because the angle between the normal vectors is the same as the angle between the planes. Figure 2.72 shows why this is true.

We can find the measure of the angle *θ* between two intersecting planes by first finding the cosine of the angle, using the following equation:

$$\text{cos}\ \theta = \frac{\left| {\mathbf{\text{n}}_{1} \cdot \mathbf{\text{n}}_{2}} \right|}{\left\| \mathbf{\text{n}}_{1} \right\|\left\| \mathbf{\text{n}}_{2} \right\|}.$$

We can then use the angle to determine whether two planes are parallel or orthogonal or if they intersect at some other angle.

Finding the Angle between Two Planes

Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.

1. $x + 2y - z = 8\ \text{and}\ 2x + 4y - 2z = 10$

2. $2x - 3y + 2z = 3\ \text{and}\ 6x + 2y - 3z = 1$

3. $x + y + z = 4\ \text{and}\ x - 3y + 5z = 1$

Solution

1. The normal vectors for these planes are $\mathbf{\text{n}}_{1} = \left\langle {1,2,-1} \right\rangle$ and $\mathbf{\text{n}}_{2} = \left\langle {2,4,-2} \right\rangle.$ These two vectors are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.

2. The normal vectors for these planes are $\mathbf{\text{n}}_{1} = \left\langle {2,-3,2} \right\rangle$ and $\mathbf{\text{n}}_{2} = \left\langle {6,2,-3} \right\rangle.$ Taking the dot product of these vectors, we have

$$\mathbf{\text{n}}_{1} \cdot \mathbf{\text{n}}_{2} = \left\langle {2,-3,2} \right\rangle \cdot \left\langle {6,2,-3} \right\rangle = 2(6) - 3(2) + 2(-3) = 0.$$

The normal vectors are orthogonal, so the corresponding planes are orthogonal as well.

3. The normal vectors for these planes are $\mathbf{\text{n}}_{1} = \left\langle {1,1,1} \right\rangle$ and $\mathbf{\text{n}}_{2} = \left\langle {1,-3,5} \right\rangle\text{:}$

$$\begin{array}{cl}

{\text{cos}\ \theta} & {= \frac{\left| {\mathbf{\text{n}}_{1} \cdot \mathbf{\text{n}}_{2}} \right|}{\left\| \mathbf{\text{n}}_{1} \right\|\left\| \mathbf{\text{n}}_{2} \right\|}} \\

& {= \frac{\left| {\left\langle {1,1,1} \right\rangle \cdot \left\langle {1,-3,5} \right\rangle} \right|}{\sqrt{1^{2} + 1^{2} + 1^{2}}\ \sqrt{1^{2} + {(-3)}^{2} + 5^{2}}}} \\

& {= \frac{3}{\sqrt{105}}.}

\end{array}$$

The angle between the two planes is $1.27$ rad, or approximately $73\text{°}.$

Find the measure of the angle between planes $x + y - z = 3$ and $3x - y + 3z = 5.$ Give the answer in radians and round to two decimal places.

When we find that two planes are parallel, we may need to find the distance between them. To find this distance, we simply select a point in one of the planes. The distance from this point to the other plane is the distance between the planes.

Previously, we introduced the formula for calculating this distance in Equation 2.19:

$$d = \frac{\overset{\rightarrow}{QP} \cdot \mathbf{\text{n}}}{\left\| \mathbf{\text{n}} \right\|},$$

where $Q$ is a point on the plane, $P$ is a point not on the plane, and $\mathbf{\text{n}}$ is the normal vector that passes through point $Q.$ Consider the distance from point $\left( {x_{0},y_{0},z_{0}} \right)$ to plane $ax + by + cz + k = 0.$ Let $\left( {x_{1},y_{1},z_{1}} \right)$ be any point in the plane. Substituting into the formula yields

$$\begin{array}{cl}

d & {= \frac{\left| {a\left( {x_{0} - x_{1}} \right) + b\left( {y_{0} - y_{1}} \right) + c\left( {z_{0} - z_{1}} \right)} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}} \\

& {= \frac{\left| {ax_{0} + by_{0} + cz_{0} + k} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}.}

\end{array}$$

We state this result formally in the following theorem.

Distance from a Point to a Plane

Let $P\left( {x_{0},y_{0},z_{0}} \right)$ be a point. The distance from $P$ to plane $ax + by + cz + k = 0$ is given by

$$d = \frac{\left| {ax_{0} + by_{0} + cz_{0} + k} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}.$$

Finding the Distance between Parallel Planes

Find the distance between the two parallel planes given by $2x + y - z = 2$ and $2x + y - z = 8.$

Solution

Point $\left( {1,0,0} \right)$ lies in the first plane. The desired distance, then, is

$$\begin{array}{cl}

d & {= \frac{\left| {ax_{0} + by_{0} + cz_{0} + k} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}} \\

& {= \frac{\left| {2(1) + 1(0) + (-1)(0) + (-8)} \right|}{\sqrt{2^{2} + 1^{2} + (-1)^{2}}}} \\

& {= \frac{6}{\sqrt{6}} = \sqrt{6}.}

\end{array}$$

Find the distance between parallel planes $5x - 2y + z = 6$ and $5x - 2y + z = -3.$

Distance between Two Skew Lines

Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product.

The symmetric forms of two lines, $L_{1}$ and $L_{2},$ are

$$\begin{array}{l}

\\

\\

{L_{1}:\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}} \\

{L_{2}:\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}.}

\end{array}$$

You are to develop a formula for the distance $d$ between these two lines, in terms of the values $a_{1},b_{1},c_{1};a_{2},b_{2},c_{2};x_{1},y_{1},z_{1};\ \text{and}\ x_{2},y_{2},z_{2}.$ The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.

1. First, write down two vectors, $\mathbf{\text{v}}_{1}$ and $\mathbf{\text{v}}_{2},$ that lie along $L_{1}$ and $L_{2},$ respectively.

2. Find the cross product of these two vectors and call it $\mathbf{\text{N}}.$ This vector is perpendicular to $\mathbf{\text{v}}_{1}\ \text{and}\ \mathbf{\text{v}}_{2},$ and hence is perpendicular to both lines.

3. From vector $\mathbf{\text{N}},$ form a unit vector $\mathbf{\text{n}}$ in the same direction.

4. Use symmetric equations to find a convenient vector $\mathbf{\text{v}}_{12}$ that lies between any two points, one on each line. Again, this can be done directly from the symmetric equations.

5. The dot product of two vectors is the magnitude of the projection of one vector onto the other times the magnitude of the other vector—that is, $\mathbf{\text{A}} \cdot \mathbf{\text{B}} = \left\| \mathbf{\text{A}} \right\|\left\| \mathbf{\text{B}} \right\|\text{cos}\ \theta,$ where $\theta$ is the angle between the vectors. Using the dot product, find the projection of vector $\mathbf{\text{v}}_{12}$ found in step $4$ onto unit vector $\mathbf{\text{n}}$ found in step 3. This projection is perpendicular to both lines, and hence its length must be the perpendicular distance $d$ between them. Note that the value of $d$ may be negative, depending on your choice of vector $\mathbf{\text{v}}_{12}$ or the order of the cross product, so use absolute value signs around the numerator.

6. Check that your formula gives the correct distance of $|-25|\text{/}\sqrt{198} \approx 1.78$ between the following two lines:

$$\begin{array}{l}

\\

\\

{L_{1}:\frac{x - 5}{2} = \frac{y - 3}{4} = \frac{z - 1}{3}} \\

{L_{2}:\frac{x - 6}{3} = \frac{y - 1}{5} = \frac{z}{7}.}

\end{array}$$

7. Is your general expression valid when the lines are parallel? If not, why not? (*Hint:* What do you know about the value of the cross product of two parallel vectors? Where would that result show up in your expression for $d?)$

8. Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersect if they are not parallel and they are in the same plane. Hence, consider the direction of $\mathbf{\text{n}}$ and $\mathbf{\text{v}}_{12}.$ What is the result of their dot product?

9. Consider the following application. Engineers at a refinery have determined they need to install support struts between many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these struts at the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure, they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute them to the installation crews without spending valuable time making measurements.

The rectangular frame structure has the dimensions $4.0\ \times \ 15.0\ \times \ 10.0\ \text{m}$ (height, width, and depth). One sector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposed corner (the one farthest away at the top); call this $L_{1}.$ A second pipe enters and exits at the two different opposite lower corners; call this $L_{2}$ (Figure 2.74).

Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector $\mathbf{\text{n}},$ define a vector that spans two points on each line, and finally determine the minimum distance between the lines. (Take the origin to be at the lower corner of the first pipe.) Similarly, you may also develop the symmetric equations for each line and substitute directly into your formula.

Section 2.5 Exercises

In the following exercises, points $P$ and $Q$ are given. Let $L$ be the line passing through points $P$ and $Q.$

1. Find the vector equation of line $L.$

2. Find parametric equations of line $L.$

3. Find symmetric equations of line $L.$

4. Find parametric equations of the line segment determined by $P$ and $Q.$

243.

$P\left( {-3,5,9} \right),$ $Q\left( {4,-7,2} \right)$

244\.

$P\left( {4,0,5} \right),Q\left( {2,3,1} \right)$

245.

$P\left( {-1,0,5} \right),$ $Q\left( {4,0,3} \right)$

246\.

$P\left( {7,-2,6} \right),$ $Q\left( {-3,0,6} \right)$

For the following exercises, point $P$ and vector $\mathbf{\text{v}}$ are given. Let $L$ be the line passing through point $P$ with direction $\mathbf{\text{v}}.$

1. Find parametric equations of line $L.$

2. Find symmetric equations of line $L.$

3. Find the intersection of the line with the *xy*-plane.

247.

$P\left( {1,-2,3} \right),$ $\mathbf{\text{v}} = \left\langle {1,2,3} \right\rangle$

248\.

$P(3,1,5),$ $\mathbf{\text{v}} = \left\langle {1,1,1} \right\rangle$

249.

$P(3,1,5),$ $\mathbf{\text{v}} = \overset{\rightarrow}{QR},$ where $Q(2,2,3)$ and $R(3,2,3)$

250\.

$P(2,3,0),$ $\mathbf{\text{v}} = \overset{\rightarrow}{QR},$ where $Q(0,4,5)$ and $R(0,4,6)$

For the following exercises, line $L$ is given.

1. Find point $P$ that belongs to the line and direction vector $\mathbf{\text{v}}$ of the line. Express $\mathbf{\text{v}}$ in component form.

2. Find the distance from the origin to line $L.$

251.

$x = 1 + t,y = 3 + t,z = 5 + 4t,$ $t \in \mathbb{R}$

252\.

$\text{−}x = y + 1,z = 2$

253.

Find the distance between point $A\left( {-3,1,1} \right)$ and the line of symmetric equations

$x = \text{−}y = \text{−}z.$

254\.

Find the distance between point $A\left( {4,2,5} \right)$ and the line of parametric equations

$x = -1 - t,y = \text{−}t,z = 2,$ $t \in \mathbb{R}.$

For the following exercises, lines $L_{1}$ and $L_{2}$ are given.

1. Verify whether lines $L_{1}$ and $L_{2}$ are parallel.

2. If the lines $L_{1}$ and $L_{2}$ are parallel, then find the distance between them.

255.

$L_{1}:x = 1 + t,y = t,z = 2 + t,$ $t \in \mathbb{R},$ $L_{2}:x - 3 = y - 1 = z - 3$

256\.

$L_{1}:x = 2,y = 1,z = t,$ $L_{2}:x = 1,y = 1,z = 2 - 3t,$ $t \in \mathbb{R}$

257\.

Show that the line passing through points $P\left( {3,1,0} \right)$ and $Q\left( {1,4,-3} \right)$ is perpendicular to the line with equations $x~ = ~3~ + ~3t,~y~ = ~1~ + ~8t,~z~ = ~6t,$ $t \in \mathbb{R}.$

258\.

Are the lines of equations $x = -2 + 2t,y = -6,z = 2 + 6t$ and $x = -1 + t,y = 1 + t,z = t,$ $t \in \mathbb{R},$ perpendicular to each other?

259.

Find the point of intersection of the lines of equations $x = -2y = 3z$ and $x = -5 - t,y = -1 + t,z = t - 11,$ $t \in \mathbb{R}.$

260\.

Find the intersection point of the *x*-axis with the line of parametric equations

$x = 10 + t,y = 2 - 2t,z = -3 + 3t,$ $t \in \mathbb{R}.$

For the following exercises, lines $L_{1}$ and $L_{2}$ are given. Determine whether the lines are equal, parallel but not equal, skew, or intersecting.

261.

$L_{1}:x = y - 1 = \text{−}z$ and $L_{2}:x - 2 = \text{−}y = \frac{z}{2}$

262\.

$L_{1}:x = 2t,y = 0,z = 3,$ $t \in \mathbb{R}$ and $L_{2}:x = 0,y = 8 + s,z = 7 + s,$ $s \in \mathbb{R}$

263.

$L_{1}:x = -1 + 2t,y = 1 + 3t,z = 7t,$ $t \in \mathbb{R}$ and $L_{2}:x - 1 = \frac{2}{3}\left( {y - 4} \right) = \frac{2}{7}z - 2$

264\.

$L_{1}:3x = y + 1 = 2z$ and $L_{2}:x = 6 + 2t,y = 17 + 6t,z = 9 + 3t,$ $t \in \mathbb{R}$

265.

Consider line $L$ of symmetric equations $x - 2 = \text{−}y = \frac{z}{2}$ and point $A(1,1,1).$

1. Find parametric equations for a line parallel to $L$ that passes through point $A.$

2. Find symmetric equations of a line skew to $L$ and that passes through point $A.$

3. Find symmetric equations of a line that intersects $L$ and passes through point $A.$

266\.

Consider line $L$ of parametric equations $x = t,y = 2t,z = 3,$ $t \in \mathbb{R}.$

1. Find parametric equations for a line parallel to $L$ that passes through the origin.

2. Find parametric equations of a line skew to $L$ that passes through the origin.

3. Find symmetric equations of a line that intersects $L$ and passes through the origin.

For the following exercises, point $P$ and vector $\mathbf{\text{n}}$ are given.

1. Find the scalar equation of the plane that passes through $P$ and has normal vector $\mathbf{\text{n}}.$

2. Find the general form of the equation of the plane that passes through $P$ and has normal vector $\mathbf{\text{n}}.$

267.

$P(0,0,0),$ $\mathbf{\text{n}} = 3\mathbf{\text{i}} - 2\mathbf{\text{j}} + 4\mathbf{\text{k}}$

268\.

$P\left( {3,2,2} \right),$ $\mathbf{\text{n}} = 2\mathbf{\text{i}} + 3\mathbf{\text{j}} - \mathbf{\text{k}}$

269.

$P\left( {1,2,3} \right),$ $\mathbf{\text{n}} = \left\langle {1,2,3} \right\rangle$

270\.

$P(0,0,0),$ $\mathbf{\text{n}} = \left\langle {-3,2,-1} \right\rangle$

For the following exercises, the equation of a plane is given.

1. Find normal vector $\mathbf{\text{n}}$ to the plane. Express $\mathbf{\text{n}}$ using standard unit vectors.

2. Find the intersections of the plane with the coordinate axes.

3. Sketch the plane.

271.

\[T\] $4x + 5y + 10z - 20 = 0$

272\.

$3x + 4y - 12 = 0$

273.

$3x - 2y + 4z = 0$

274\.

$x + z = 0$

275.

Given point $P\left( {1,2,3} \right)$ and vector $\mathbf{\text{n}} = \mathbf{\text{i}} + \mathbf{\text{j}},$ find point $Q$ on the *x*-axis such that $\overset{\rightarrow}{PQ}$ and $\mathbf{\text{n}}$ are orthogonal.

276\.

Show there is no plane perpendicular to $\mathbf{\text{n}} = \mathbf{\text{i}} + \mathbf{\text{j}}$ that passes through points $P\left( {1,2,3} \right)$ and $Q\left( {2,3,4} \right).$

277.

Find parametric equations of the line passing through point $P(-2,1,3)$ that is perpendicular to the plane of equation $2x - 3y + z = 7.$

278\.

Find symmetric equations of the line passing through point $P(2,5,4)$ that is perpendicular to the plane of equation $2x + 3y - 5z = 0.$

279\.

Show that line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ is parallel to plane $x - 2y + z = 6.$

280\.

Find the real number $\alpha$ such that the line of parametric equations $x = t,y = 2 - t,z = 3 + t,$ $t \in \mathbb{R}$ is parallel to the plane of equation $\alpha x + 5y + z - 10 = 0.$

For the following exercises, points $P,Q,\ \text{and}\ R$ are given.

1. Find the general equation of the plane passing through $P,Q,\ \text{and}\ R.$

2. Write the vector equation $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PS} = 0$ of the plane at a., where $S\left( {x,y,z} \right)$ is an arbitrary point of the plane.

3. Find parametric equations of the line passing through the origin that is perpendicular to the plane passing through $P,Q,\ \text{and}\ R.$

281.

$P(1,1,1),Q(2,4,3),$ and $R(-1,-2,-1)$

282\.

$P\left( {-2,1,4} \right),Q\left( {3,1,3} \right),$ and $R\left( {-2,1,0} \right)$

283.

Consider the planes of equations $x + y + z = 1$ and $x + z = 0.$

1. Show that the planes intersect.

2. Find symmetric equations of the line passing through point $P(1,4,6)$ that is parallel to the line of intersection of the planes.

284\.

Consider the planes of equations $\text{−}y + z - 2 = 0$ and $x - y = 0.$

1. Show that the planes intersect.

2. Find parametric equations of the line passing through point $P(-8,0,2)$ that is parallel to the line of intersection of the planes.

285.

Find the scalar equation of the plane that passes through point $P(-1,2,1)$ and is perpendicular to the line of intersection of planes $x + y - z - 2 = 0$ and $2x - y + 3z - 1 = 0.$

286\.

Find the general equation of the plane that passes through the origin and is perpendicular to the line of intersection of planes $\text{−}x + y + 2 = 0$ and $z - 3 = 0.$

287.

Determine whether the line of parametric equations $x = 1 + 2t,y = -2t,z = 2 + t,$ $t \in \mathbb{R}$ intersects the plane with equation $3x + 4y + 6z - 7 = 0.$ If it does intersect, find the point of intersection.

288\.

Determine whether the line of parametric equations $x = 5,y = 4 - t,z = 2t,$ $t \in \mathbb{R}$ intersects the plane with equation $2x - y + z = 5.$ If it does intersect, find the point of intersection.

289.

Find the distance from point $P\left( {1,5,-4} \right)$ to the plane of equation $3x - y + 2z - 6 = 0.$

290\.

Find the distance from point $P\left( {1,-2,3} \right)$ to the plane of equation $\left( {x - 3} \right) + 2\left( {y + 1} \right) - 4z = 0.$

For the following exercises, the equations of two planes are given.

1. Determine whether the planes are parallel, orthogonal, or neither.

2. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer.

291.

\[T\] $x + y + z = 0,$ $2x - y + z - 7 = 0$

292\.

$5x - 3y + z = 4,$ $x + 4y + 7z = 1$

293.

$x - 5y - z = 1,$ $5x - 25y - 5z = -3$

294\.

\[T\] $x - 3y + 6z = 4,$ $5x + y - z = 4$

295.

Show that the lines of equations $x = t,y = 1 + t,z = 2 + t,$ $t \in \mathbb{R}\text{,}$ and $\frac{x}{2} = \frac{y - 1}{3} = z - 3$ are skew, and find the distance between them.

296\.

Show that the lines of equations $x = -1 + t,y = -2 + t,z = 3t,$ $t \in \mathbb{R},$ and $x = 5 + s,y = -8 + 2s,z = 7s,$ $s \in \mathbb{R}$ are skew, and find the distance between them.

297.

Consider point $C\left( {-3,2,4} \right)$ and the plane of equation $2x + 4y - 3z = 8.$

1. Find the radius of the sphere with center $C$ tangent to the given plane.

2. Find point *P* of tangency.

298\.

Consider the plane of equation $x - y - z - 8 = 0.$

1. Find an equation of the sphere with center $C$ at the origin that is tangent to the given plane.

2. Find parametric equations of the line passing through the origin and the point of tangency.

299.

Two children are playing with a ball. The girl throws the ball to the boy. The ball travels in

the air, curves $3$ ft to the right, and falls $5$ ft away from the girl (see the following figure). If the plane that contains the trajectory of the ball is perpendicular to the ground, find its equation.

300\.

\[T\] John allocates $d$ dollars to consume monthly three goods of prices $a,b,\ \text{and}\ c.$ In this context, the budget equation is defined as $ax + by + cz = d,$ where $x \geq 0,y \geq 0,$ and $z \geq 0$ represent the number of items bought from each of the goods. The budget set is given by $\left\{ \left( {x,y,z} \right) \middle| ax + by + cz \leq d,x \geq 0,y \geq 0,z \geq 0 \right\},$ and the budget plane is the part of the plane of equation $ax + by + cz = d$ for which $x \geq 0,y \geq 0,$ and $z \geq 0.$ Consider $a = \text{\$}8,$ $b = \text{\$}5,$ $c = \text{\$}10,$ and $d = \text{\$}500.$

1. Use a CAS to graph the budget set and budget plane.

2. For $z = 25,$ find the new budget equation and graph the budget set in the same system of coordinates.

301.

\[T\] Consider $\mathbf{\text{r}}(t) = \left\langle {\text{sin}\ t,\text{cos}\ t,2t} \right\rangle$ the position vector of a particle at time $t \in \lbrack 0,3\rbrack,$ where the components of r are expressed in centimeters and time is measured in seconds. Let $\overset{\rightarrow}{OP}$ be the position vector of the particle after $1$ sec.

1. Determine the velocity vector $\mathbf{\text{v}}(1)$ of the particle after $1$ sec.

2. Find the scalar equation of the plane that is perpendicular to $\mathbf{\text{v}}(1)$ and passes through point $P.$ This plane is called the *normal plane* to the path of the particle at point $P.$

3. Use a CAS to visualize the path of the particle along with the velocity vector and normal plane at point $P.$

302\.

\[T\] A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) $A(8,0,0),$ $B(8,18,0),$ $C(0,18,8),$ and $D(0,0,8)$ (see the following figure).

1. Find the general form of the equation of the plane that contains the solar panel by using points $A,B,\ \text{and}\ C,$ and show that its normal vector is equivalent to $\overset{\rightarrow}{AB}\ \times \ \overset{\rightarrow}{AD}.$

2. Find parametric equations of line $L_{1}$ that passes through the center of the solar panel and has direction vector $\mathbf{\text{s}} = \frac{1}{\sqrt{3}}\mathbf{\text{i}} + \frac{1}{\sqrt{3}}\mathbf{\text{j}} + \frac{1}{\sqrt{3}}\mathbf{\text{k}},$ which points toward the position of the Sun at a particular time of day.

3. Find symmetric equations of line $L_{2}$ that passes through the center of the solar panel and is perpendicular to it.

4. Determine the angle of elevation of the Sun above the solar panel by using the angle between lines $L_{1}$ and $L_{2}.$

---

2.6 Quadric Surfaces

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-6-quadric-surfaces

2.6 Quadric Surfaces

• 2.6.1 Identify a cylinder as a type of three-dimensional surface.
• 2.6.2 Recognize the main features of ellipsoids, paraboloids, and hyperboloids.
• 2.6.3 Use traces to draw the intersections of quadric surfaces with the coordinate planes.

We have been exploring vectors and vector operations in three-dimensional space, and we have developed equations to describe lines, planes, and spheres. In this section, we use our knowledge of planes and spheres, which are examples of three-dimensional figures called *surfaces*, to explore a variety of other surfaces that can be graphed in a three-dimensional coordinate system.

Identifying Cylinders

The first surface we’ll examine is the cylinder. Although most people immediately think of a hollow pipe or a soda straw when they hear the word *cylinder*, here we use the broad mathematical meaning of the term. As we have seen, cylindrical surfaces don’t have to be circular. A rectangular heating duct is a cylinder, as is a rolled-up yoga mat, the cross-section of which is a spiral shape.

In the two-dimensional coordinate plane, the equation $x^{2} + y^{2} = 9$ describes a circle centered at the origin with radius $3.$ In three-dimensional space, this same equation represents a surface. Imagine copies of a circle stacked on top of each other centered on the *z*-axis (Figure 2.75), forming a hollow tube. We can then construct a cylinder from the set of lines parallel to the *z*-axis passing through circle $x^{2} + y^{2} = 9$ in the *xy*-plane, as shown in the figure. In this way, any curve in one of the coordinate planes can be extended to become a surface.

A set of lines parallel to a given line passing through a given curve is known as a cylindrical surface, or cylinder. The parallel lines are called rulings.

From this definition, we can see that we still have a cylinder in three-dimensional space, even if the curve is not a circle. Any curve can form a cylinder, and the rulings that compose the cylinder may be parallel to any given line (Figure 2.76).

Graphing Cylindrical Surfaces

Sketch the graphs of the following cylindrical surfaces.

1. $x^{2} + z^{2} = 25$

2. $z = 2x^{2} - y$

3. $y = \text{sin}\ x$

Solution

1. The variable $y$ can take on any value without limit. Therefore, the lines ruling this surface are parallel to the *y*-axis. The intersection of this surface with the *xz*-plane forms a circle centered at the origin with radius $5$ (see the following figure).

2. In this case, the equation contains all three variables $—x,y,$ and $z—$ so none of the variables can vary arbitrarily. The easiest way to visualize this surface is to use a computer graphing utility (see the following figure).

3. In this equation, the variable *z* can take on any value without limit. Therefore, the lines composing this surface are parallel to the *z*-axis. The intersection of this surface with the *xy*-plane outlines curve $y = \ \text{sin}\ x$ (see the following figure).

Sketch or use a graphing tool to view the graph of the cylindrical surface defined by equation $z = y^{2}.$

When sketching surfaces, we have seen that it is useful to sketch the intersection of the surface with a plane parallel to one of the coordinate planes. These curves are called traces. We can see them in the plot of the cylinder in Figure 2.80.

The traces of a surface are the cross-sections created when the surface intersects a plane parallel to one of the coordinate planes.

Traces are useful in sketching cylindrical surfaces. For a cylinder in three dimensions, though, only one set of traces is useful. Notice, in Figure 2.80, that the trace of the graph of $z = \text{sin}\ x$ in the *xz*-plane is useful in constructing the graph. The trace in the *xy*-plane, though, is just a series of parallel lines, and the trace in the *yz*-plane is simply one line.

Cylindrical surfaces are formed by a set of parallel lines. Not all surfaces in three dimensions are constructed so simply, however. We now explore more complex surfaces, and traces are an important tool in this investigation.

Quadric Surfaces

We have learned about surfaces in three dimensions described by first-order equations; these are planes. Some other common types of surfaces can be described by second-order equations. We can view these surfaces as three-dimensional extensions of the conic sections we discussed earlier: the ellipse, the parabola, and the hyperbola. We call these graphs quadric surfaces.

Quadric surfaces are the graphs of equations that can be expressed in the form

$$Ax^{2} + By^{2} + Cz^{2} + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0.$$

When a quadric surface intersects a coordinate plane, the trace is a conic section.

An ellipsoid is a surface described by an equation of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1.$ Set $x = 0$ to see the trace of the ellipsoid in the *yz*-plane. To see the traces in the *xy*- and *xz*-planes, set $z = 0$ and $y = 0,$ respectively. Notice that, if $a = b,$ the trace in the *xy*-plane is a circle. Similarly, if $a = c,$ the trace in the *xz*-plane is a circle and, if $b = c,$ then the trace in the *yz*-plane is a circle. A sphere, then, is an ellipsoid with $a = b = c.$

Sketching an Ellipsoid

Sketch the ellipsoid $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{3^{2}} + \frac{z^{2}}{5^{2}} = 1.$

Solution

Start by sketching the traces. To find the trace in the *xy*-plane, set $z = 0\text{:}$ $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{3^{2}} = 1$ (see Figure 2.81). To find the other traces, first set $y = 0$ and then set $x = 0.$

Now that we know what traces of this solid look like, we can sketch the surface in three dimensions (Figure 2.82).

The trace of an ellipsoid is an ellipse in each of the coordinate planes. However, this does not have to be the case for all quadric surfaces. Many quadric surfaces have traces that are different kinds of conic sections, and this is usually indicated by the name of the surface. For example, if a surface can be described by an equation of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{z}{c},$ then we call that surface an elliptic paraboloid. The trace in the *xy*-plane is an ellipse, but the traces in the *xz*-plane and *yz*-plane are parabolas (Figure 2.83). Other elliptic paraboloids can have other orientations simply by interchanging the variables to give us a different variable in the linear term of the equation $\frac{x^{2}}{a^{2}} + \frac{z^{2}}{c^{2}} = \frac{y}{b}$ or $\frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = \frac{x}{a}.$

Identifying Traces of Quadric Surfaces

Describe the traces of the elliptic paraboloid $x^{2} + \frac{y^{2}}{2^{2}} = \frac{z}{5}.$

Solution

To find the trace in the *xy*-plane, set $z = 0\text{:}$ $x^{2} + \frac{y^{2}}{2^{2}} = 0.$ The trace in the plane $z = 0$ is simply one point, the origin. Since a single point does not tell us what the shape is, we can move up the *z*-axis to an arbitrary plane to find the shape of other traces of the figure.

The trace in plane $z = 5$ is the graph of equation $x^{2} + \frac{y^{2}}{2^{2}} = 1,$ which is an ellipse. In the *xz*-plane, the equation becomes $z = 5x^{2}.$ The trace is a parabola in this plane and in any plane with the equation $y = b.$

In planes parallel to the *yz*-plane, the traces are also parabolas, as we can see in the following figure.

A hyperboloid of one sheet is any surface that can be described with an equation of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1.$ Describe the traces of the hyperboloid of one sheet given by equation $\frac{x^{2}}{3^{2}} + \frac{y^{2}}{2^{2}} - \frac{z^{2}}{5^{2}} = 1.$

Hyperboloids of one sheet have some fascinating properties. For example, they can be constructed using straight lines, such as in the sculpture in Figure 2.85(a). In fact, cooling towers for nuclear power plants are often constructed in the shape of a hyperboloid. The builders are able to use straight steel beams in the construction, which makes the towers very strong while using relatively little material (Figure 2.85(b)).

Chapter Opener: Finding the Focus of a Parabolic Reflector

Energy hitting the surface of a parabolic reflector is concentrated at the focal point of the reflector (Figure 2.86). If the surface of a parabolic reflector is described by equation $\frac{x^{2}}{100} + \frac{y^{2}}{100} = \frac{z}{4},$ where is the focal point of the reflector?

Solution

Since *z* is the first-power variable, the axis of the reflector corresponds to the *z*-axis. The coefficients of $x^{2}$ and $y^{2}$ are equal, so the cross-section of the paraboloid perpendicular to the *z*-axis is a circle. We can consider a trace in the *xz*-plane or the *yz*-plane; the result is the same. Setting $y = 0,$ the trace is a parabola opening up along the *z*-axis, with standard equation $x^{2} = 4pz,$ where $p$ is the focal length of the parabola. In this case, this equation becomes $x^{2} = 100 \cdot \frac{z}{4} = 4pz$ or $25 = 4p.$ So *p* is $6.25$ m, which tells us that the focus of the paraboloid is $6.25$ m up the axis from the vertex. Because the vertex of this surface is the origin, the focal point is $\left( {0,0,6.25} \right).$

Seventeen standard quadric surfaces can be derived from the general equation

$$Ax^{2} + By^{2} + Cz^{2} + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0.$$

The following figures summarizes the most important ones.

In the following two figures, the “axis” of a quadric surface may or may not be an axis of symmetry. However, all traces on the surface formed by any plane perpendicular to an “axis” will be of the same conic section type.

Identifying Equations of Quadric Surfaces

Identify the surfaces represented by the given equations.

1. $16x^{2} + 9y^{2} + 16z^{2} = 144$

2. $9x^{2} - 18x + 4y^{2} + 16y - 36z + 25 = 0$

Solution

1. The $x,y,$ and $z$ terms are all squared, and are all positive, so this is probably an ellipsoid. However, let’s put the equation into the standard form for an ellipsoid just to be sure. We have

$$16x^{2} + 9y^{2} + 16z^{2} = 144.$$

Dividing through by 144 gives

$$\frac{x^{2}}{9} + \frac{y^{2}}{16} + \frac{z^{2}}{9} = 1.$$

So, this is, in fact, an ellipsoid, centered at the origin.

2. We first notice that the $z$ term is raised only to the first power, so this is either an elliptic paraboloid or a hyperbolic paraboloid. We also note there are $x$ terms and $y$ terms that are not squared, so this quadric surface is not centered at the origin. We need to complete the square to put this equation in one of the standard forms. We have

$$\begin{array}{rll}

& & \\

{9x^{2} - 18x + 4y^{2} + 16y - 36z + 25} & = & 0 \\

{9x^{2} - 18x + 4y^{2} + 16y + 25} & = & {36z} \\

{9\left( {x^{2} - 2x} \right) + 4\left( {y^{2} + 4y} \right) + 25} & = & {36z} \\

{9\left( {x^{2} - 2x + 1 - 1} \right) + 4\left( {y^{2} + 4y + 4 - 4} \right) + 25} & = & {36z} \\

{9\left( {x - 1} \right)^{2} - 9 + 4\left( {y + 2} \right)^{2} - 16 + 25} & = & {36z} \\

{9\left( {x - 1} \right)^{2} + 4\left( {y + 2} \right)^{2}} & = & {36z} \\

{\frac{\left( {x - 1} \right)^{2}}{4} + \frac{\left( {y + 2} \right)^{2}}{9}} & = & {z.}

\end{array}$$

This is an elliptic paraboloid centered at $\left( {1,2,0} \right).$

Identify the surface represented by equation $9x^{2} + y^{2} - z^{2} + 2z - 10 = 0.$

Section 2.6 Exercises

For the following exercises, sketch and describe the cylindrical surface of the given equation.

303.

\[T\] $x^{2} + z^{2} = 1$

304\.

\[T\] $x^{2} + y^{2} = 9$

305.

\[T\] $z = \text{cos}\left( {\frac{\pi}{2} + x} \right)$

306\.

\[T\] $z = e^{x}$

307.

\[T\] $z = 9 - y^{2}$

308\.

\[T\] $z = \text{ln}(x)$

For the following exercises, the graph of a quadric surface is given.

1. Specify the name of the quadric surface.

2. Determine the axis of the quadric surface.

309. 310. 311. 312.

For the following exercises, match the given quadric surface with its corresponding equation in standard form.

1. $\frac{x^{2}}{4} + \frac{y^{2}}{9} - \frac{z^{2}}{12} = 1$

2. $\frac{x^{2}}{4} - \frac{y^{2}}{9} - \frac{z^{2}}{12} = 1$

3. $\frac{x^{2}}{4} + \frac{y^{2}}{9} + \frac{z^{2}}{12} = 1$

4. $z = 4x^{2} + 3y^{2}$

5. $z = 4x^{2} - y^{2}$

6. $4x^{2} + y^{2} - z^{2} = 0$

313.

Hyperboloid of two sheets

314\.

Ellipsoid

315.

Elliptic paraboloid

316\.

Hyperbolic paraboloid

317.

Hyperboloid of one sheet

318\.

Elliptic cone

For the following exercises, rewrite the given equation of the quadric surface in standard form. Identify the surface.

319.

$\text{−}x^{2} + 36y^{2} + 36z^{2} = 9$

320\.

$-4x^{2} + 25y^{2} + z^{2} = 100$

321.

$-3x^{2} + 5y^{2} - z^{2} = 10$

322\.

$3x^{2} - y^{2} - 6z^{2} = 18$

323.

$5y = x^{2} - z^{2}$

324\.

$8x^{2} - 5y^{2} - 10z = 0$

325.

$x^{2} + 5y^{2} + 3z^{2} - 15 = 0$

326\.

$63x^{2} + 7y^{2} + 9z^{2} - 63 = 0$

327.

$x^{2} + 5y^{2} - 8z^{2} = 0$

328\.

$5x^{2} - 4y^{2} + 20z^{2} = 0$

329.

$6x = 3y^{2} + 2z^{2}$

330\.

$49y = x^{2} + 7z^{2}$

For the following exercises, find the trace of the given quadric surface in the specified plane of coordinates and sketch it.

331.

\[T\] $x^{2} + z^{2} + 4y = 0,z = 0$

332\.

\[T\] $x^{2} + z^{2} + 4y = 0,x = 0$

333.

\[T\] $-4x^{2} + 25y^{2} + z^{2} = 100,x = 0$

334\.

\[T\] $-4x^{2} + 25y^{2} + z^{2} = 100,y = 0$

335.

\[T\] $x^{2} + \frac{y^{2}}{4} + \frac{z^{2}}{100} = 1,x = 0$

336\.

\[T\] $x^{2} - y - z^{2} = 1,y = 0$

337.

Use the graph of the given quadric surface to answer the questions.

1. Specify the name of the quadric surface.

2. Which of the equations—$16x^{2} + 9y^{2} + 36z^{2} = 3600,9x^{2} + 36y^{2} + 16z^{2} = 3600,$ or $36x^{2} + 9y^{2} + 16z^{2} = 3600$—corresponds to the graph?

3. Use b. to write the equation of the quadric surface in standard form.

338\.

Use the graph of the given quadric surface to answer the questions.

1. Specify the name of the quadric surface.

2. Which of the equations—$36z = 9x^{2} + y^{2},9x^{2} + 4y^{2} = 36z,\ \text{or}\ - 36z = -81x^{2} + 4y^{2}$—corresponds to the graph above?

3. Use b. to write the equation of the quadric surface in standard form.

For the following exercises, the equation of a quadric surface is given.

1. Use the method of completing the square to write the equation in standard form.

2. Identify the surface.

339.

$x^{2} + 2z^{2} + 6x - 8z + 1 = 0$

340\.

$4x^{2} - y^{2} + z^{2} - 8x + 2y + 2z + 3 = 0$

341.

$x^{2} + 4y^{2} - 4z^{2} - 6x - 16y - 16z + 5 = 0$

342\.

$x^{2} + z^{2} - 4y + 4 = 0$

343.

$x^{2} + \frac{y^{2}}{4} - \frac{z^{2}}{3} + 6x + 9 = 0$

344\.

$x^{2} - y^{2} + z^{2} - 12z + 2x + 37 = 0$

345.

Write the standard form of the equation of the ellipsoid centered at the origin that passes through points $A\left( {2,0,0} \right),B\left( {0,0,1} \right),$ and $C\left( {\frac{1}{2},\sqrt{11},\frac{1}{2}} \right).$

346\.

Write the standard form of the equation of the ellipsoid centered at point $P\left( {1,1,0} \right)$ that passes through points $A\left( {6,1,0} \right),B\left( {4,2,0} \right)$ and $C\left( {1,2,1} \right).$

347.

Determine the intersection points of elliptic cone $x^{2} - y^{2} - z^{2} = 0$ with the line of symmetric equations $\frac{x - 1}{2} = \frac{y + 1}{3} = z.$

348\.

Determine the intersection points of parabolic hyperboloid $z = 3x^{2} - 2y^{2}$ with the line of parametric equations $x = 3t,y = 2t,z = 19t,$ where $t \in \mathbb{R}.$

349.

Find an equation of the quadric surface with points $P\left( {x,y,z} \right)$ that are equidistant from point $Q\left( {0,-1,0} \right)$ and plane of equation $y = 1.$ Identify the surface.

350\.

Find an equation of the quadric surface with points $P\left( {x,y,z} \right)$ that are equidistant from point $Q\left( {0,2,0} \right)$ and plane of equation $y = -2.$ Identify the surface.

351.

If the surface of a parabolic reflector is described by equation $400z = x^{2} + y^{2},$ find the focal point of the reflector.

352\.

Consider the parabolic reflector described by equation $z = 20x^{2} + 20y^{2}.$ Find its focal point.

353\.

Show that quadric surface $x^{2} + y^{2} + z^{2} + 2xy + 2xz + 2yz + x + y + z = 0$ reduces to two parallel planes.

354\.

Show that quadric surface $x^{2} + y^{2} + z^{2} - 2xy - 2xz + 2yz - 1 = 0$ reduces to two parallel planes passing.

355.

\[T\] The intersection between cylinder $\left( {x - 1} \right)^{2} + y^{2} = 1$ and sphere $x^{2} + y^{2} + z^{2} = 4$ is called a *Viviani curve*.

1. Solve the system consisting of the equations of the surfaces to find the equations of the intersection curve. (*Hint:* Find $x$ and $y$ in terms of $z.)$

2. Use a computer algebra system (CAS) to visualize the intersection curve on sphere $x^{2} + y^{2} + z^{2} = 4.$

356\.

Hyperboloid of one sheet $25x^{2} + 25y^{2} - z^{2} = 25$ and elliptic cone $-25x^{2} + 75y^{2} + z^{2} = 0$ are represented in the following figure along with their intersection curves. Identify the intersection curves and find their equations (*Hint:* Find *y* from the system consisting of the equations of the surfaces.)

357.

\[T\] Use a CAS to create the intersection between cylinder $9x^{2} + 4y^{2} = 18$ and ellipsoid $36x^{2} + 16y^{2} + 9z^{2} = 144,$ and find the equations of the intersection curves.

358\.

\[T\] A spheroid is an ellipsoid with two equal semiaxes. For instance, the equation of a spheroid with the *z*-axis as its axis of symmetry is given by $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} + \frac{z^{2}}{c^{2}} = 1,$ where $a$ and $c$ are positive real numbers. The spheroid is called *oblate* if $c < a,$ and *prolate* for $c > a.$

1. The eye cornea is approximated as a prolate spheroid with an axis that is the eye, where $a = 8.7\ \text{mm and}\ c = 9.6\ \text{mm}.$ Write the equation of the spheroid that models the cornea and sketch the surface.

2. Give two examples of objects with prolate spheroid shapes.

359.

\[T\] In cartography, Earth is approximated by an oblate spheroid rather than a sphere. The radii at the equator and poles are approximately $3963$ mi and $3950$ mi, respectively.

1. Write the equation in standard form of the ellipsoid that represents the shape of Earth. Assume the center of Earth is at the origin and that the trace formed by plane $z = 0$ corresponds to the equator.

2. Sketch the graph.

3. Find an equation of the intersection curve of the surface with plane $z = 1000$ that is parallel to the *xy*-plane. The intersection curve is called a *parallel*.

4. Find an equation of the intersection curve of the surface with plane $x + y = 0$ that passes through the *z*-axis. The intersection curve is called a *meridian*.

360\.

\[T\] A set of buzzing stunt magnets (or “rattlesnake eggs”) includes two sparkling, polished, superstrong spheroid-shaped magnets well-known for children’s entertainment. Each magnet is $1.625$ in. long and $0.5$ in. wide at the middle. While tossing them into the air, they create a buzzing sound as they attract each other.

1. Write the equation of the prolate spheroid centered at the origin that describes the shape of one of the magnets.

2. Write the equations of the prolate spheroids that model the shape of the buzzing stunt magnets. Use a CAS to create the graphs.

361.

\[T\] A heart-shaped surface is given by equation $\left( {x^{2} + \frac{9}{4}y^{2} + z^{2} - 1} \right)^{3} - x^{2}z^{3} - \frac{9}{80}y^{2}z^{3} = 0.$

1. Use a CAS to graph the surface that models this shape.

2. Determine and sketch the trace of the heart-shaped surface on the *xz*-plane.

362\.

\[T\] The ring torus symmetric about the *z*-axis is a special type of surface in topology and its equation is given by $\left( {x^{2} + y^{2} + z^{2} + R^{2} - r^{2}} \right)^{2} = 4R^{2}\left( {x^{2} + y^{2}} \right),$ where $R > r > 0.$ The numbers $R$ and $r$ are called are the major and minor radii, respectively, of the surface. The following figure shows a ring torus for which $R = 2\ \text{and}\ r = 1.$

1. Write the equation of the ring torus with $R = 2\ \text{and}\ r = 1,$ and use a CAS to graph the surface. Compare the graph with the figure given.

2. Determine the equation and sketch the trace of the ring torus from a. on the *xy*-plane.

3. Give two examples of objects with ring torus shapes.

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2.7 Cylindrical and Spherical Coordinates

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-7-cylindrical-and-spherical-coordinates

2.7 Cylindrical and Spherical Coordinates

• 2.7.1 Convert from cylindrical to rectangular coordinates.
• 2.7.2 Convert from rectangular to cylindrical coordinates.
• 2.7.3 Convert from spherical to rectangular coordinates.
• 2.7.4 Convert from rectangular to spherical coordinates.

The Cartesian coordinate system provides a straightforward way to describe the location of points in space. Some surfaces, however, can be difficult to model with equations based on the Cartesian system. This is a familiar problem; recall that in two dimensions, polar coordinates often provide a useful alternative system for describing the location of a point in the plane, particularly in cases involving circles. In this section, we look at two different ways of describing the location of points in space, both of them based on extensions of polar coordinates. As the name suggests, cylindrical coordinates are useful for dealing with problems involving cylinders, such as calculating the volume of a round water tank or the amount of oil flowing through a pipe. Similarly, spherical coordinates are useful for dealing with problems involving spheres, such as finding the volume of domed structures.

Cylindrical Coordinates

When we expanded the traditional Cartesian coordinate system from two dimensions to three, we simply added a new axis to model the third dimension. Starting with polar coordinates, we can follow this same process to create a new three-dimensional coordinate system, called the cylindrical coordinate system. In this way, cylindrical coordinates provide a natural extension of polar coordinates to three dimensions.

In the cylindrical coordinate system, a point in space (Figure 2.89) is represented by the ordered triple $\left( {r,\theta,z} \right),$ where

• $\left( {r,\theta} \right)$ are the polar coordinates of the point’s projection in the *xy*-plane
• $z$ is the usual $z\text{-coordinate}$ in the Cartesian coordinate system

In the *xy*-plane, the right triangle shown in Figure 2.89 provides the key to transformation between cylindrical and Cartesian, or rectangular, coordinates.

Conversion between Cylindrical and Cartesian Coordinates

The rectangular coordinates $\left( {x,y,z} \right)$ and the cylindrical coordinates $\left( {r,\theta,z} \right)$ of a point are related as follows:

$$\begin{matrix}

x & = & {r\ \text{cos}\ \theta} & & & \text{These equations are used to convert from} \\

y & = & {r\ \text{sin}\ \theta} & & & \text{cylindrical coordinates to rectangular} \\

z & = & z & & & \text{coordinates.} \\

& \text{and} & & & & \\

r^{2} & = & {x^{2} + y^{2}} & & & \text{These equations are used to convert from} \\

{\text{tan}\ \theta} & = & \frac{y}{x} & & & \text{rectangular coordinates to cylindrical} \\

z & = & z & & & \text{coordinates.}

\end{matrix}$$

As when we discussed conversion from rectangular coordinates to polar coordinates in two dimensions, it should be noted that the equation $\text{tan}\ \theta = \frac{y}{x}$ has an infinite number of solutions. However, if we restrict $\theta$ to values between $0$ and $2\pi,$ then we can find a unique solution based on the quadrant of the *xy*-plane in which original point $\left( {x,y,z} \right)$ is located. Note that if $x = 0,$ then the value of $\theta$ is either $\frac{\pi}{2},\frac{3\pi}{2},$ or $0,$ depending on the value of $y.$

Notice that these equations are derived from properties of right triangles. To make this easy to see, consider point $P$ in the *xy*-plane with rectangular coordinates $\left( {x,y,0} \right)$ and with cylindrical coordinates $\left( {r,\theta,0} \right),$ as shown in the following figure.

Let’s consider the differences between rectangular and cylindrical coordinates by looking at the surfaces generated when each of the coordinates is held constant. If $c$ is a constant, then in rectangular coordinates, surfaces of the form $x = c,$ $y = c,$ or $z = c$ are all planes. Planes of these forms are parallel to the *yz*-plane, the *xz*-plane, and the *xy*-plane, respectively. When we convert to cylindrical coordinates, the *z*-coordinate does not change. Therefore, in cylindrical coordinates, surfaces of the form $z = c$ are planes parallel to the *xy*-plane. Now, let’s think about surfaces of the form $r = c.$ The points on these surfaces are at a fixed distance from the *z*-axis. In other words, these surfaces are vertical circular cylinders. Last, what about $\theta = c?$ The points on a surface of the form $\theta = c$ are at a fixed angle from the *x*-axis, which gives us a half-plane that starts at the *z*-axis (Figure 2.91 and Figure 2.92).

Converting from Cylindrical to Rectangular Coordinates

Plot the point with cylindrical coordinates $\left( {4,\frac{2\pi}{3},-2} \right)$ and express its location in rectangular coordinates.

Solution

Conversion from cylindrical to rectangular coordinates requires a simple application of the equations listed in Conversion between Cylindrical and Cartesian Coordinates:

$$\begin{array}{rll}

x & = & {r\ \text{cos}\ \theta = 4\ \text{cos}\ \frac{2\pi}{3} = -2} \\

y & = & {r\ \text{sin}\ \theta = 4\ \text{sin}\ \frac{2\pi}{3} = 2\sqrt{3}} \\

z & = & -2.

\end{array}$$

The point with cylindrical coordinates $\left( {4,\frac{2\pi}{3},-2} \right)$ has rectangular coordinates $\left( {-2,2\sqrt{3},-2} \right)$ (see the following figure).

Point $R$ has cylindrical coordinates $\left( {5,\frac{\pi}{6},4} \right)$. Plot $R$ and describe its location in space using rectangular, or Cartesian, coordinates.

If this process seems familiar, it is with good reason. This is exactly the same process that we followed in Introduction to Parametric Equations and Polar Coordinates to convert from polar coordinates to two-dimensional rectangular coordinates.

Converting from Rectangular to Cylindrical Coordinates

Convert the rectangular coordinates $(1,-3,5)$ to cylindrical coordinates.

Solution

Use the second set of equations from Conversion between Cylindrical and Cartesian Coordinates to translate from rectangular to cylindrical coordinates:

$$\begin{array}{rll}

r^{2} & = & {x^{2} + y^{2}} \\

r & = & {\text{±}\sqrt{1^{2} + (-3)^{2}} = \text{±}\sqrt{10}.}

\end{array}$$

We choose the positive square root, so $r = \sqrt{10}.$ Now, we apply the formula to find $\theta.$ In this case, $y$ is negative and $x$ is positive, which means we must select the value of $\theta$ between $\frac{3\pi}{2}$ and $2\pi\text{:}$

$$\begin{array}{rll}

{\text{tan}\ \theta} & = & {\frac{y}{x} = \frac{-3}{1}} \\

\theta & = & {\text{arctan}{(-3) + 2\pi} \approx 5.03\ \text{rad}.}

\end{array}$$

In this case, the *z*-coordinates are the same in both rectangular and cylindrical coordinates:

$$z = 5.$$

The point with rectangular coordinates $(1,-3,5)$ has cylindrical coordinates approximately equal to $\left( {\sqrt{10},5.03,5} \right).$

Convert point $\left( {-8,8,-7} \right)$ from Cartesian coordinates to cylindrical coordinates.

The use of cylindrical coordinates is common in fields such as physics. Physicists studying electrical charges and the capacitors used to store these charges have discovered that these systems sometimes have a cylindrical symmetry. These systems have complicated modeling equations in the Cartesian coordinate system, which make them difficult to describe and analyze. The equations can often be expressed in more simple terms using cylindrical coordinates. For example, the cylinder described by equation $x^{2} + y^{2} = 25$ in the Cartesian system can be represented by cylindrical equation $r = 5.$

Identifying Surfaces in the Cylindrical Coordinate System

Describe the surfaces with the given cylindrical equations.

1. $\theta = \frac{\pi}{4}$

2. $r^{2} + z^{2} = 9$

3. $z = r$

Solution

1. When the angle $\theta$ is held constant while $r$ and $z$ are allowed to vary, the result is a half-plane (see the following figure).

2. Substitute $r^{2} = x^{2} + y^{2}$ into equation $r^{2} + z^{2} = 9$ to express the rectangular form of the equation: $x^{2} + y^{2} + z^{2} = 9.$ This equation describes a sphere centered at the origin with radius $3$ (see the following figure).

3. To describe the surface defined by equation $z = r,$ is it useful to examine traces parallel to the *xy*-plane. For example, the trace in plane $z = 1$ is circle $r = 1,$ the trace in plane $z = 3$ is circle $r = 3,$ and so on. Each trace is a circle. As the value of $z$ increases, the radius of the circle also increases. The resulting surface is a cone (see the following figure).

Describe the surface with cylindrical equation $r = 6.$

Spherical Coordinates

In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a distance. In the cylindrical coordinate system, location of a point in space is described using two distances $\left( {r\ \text{and}\ z} \right)$ and an angle measure $(\theta).$ In the spherical coordinate system, we again use an ordered triple to describe the location of a point in space. In this case, the triple describes one distance and two angles. Spherical coordinates make it simple to describe a sphere, just as cylindrical coordinates make it easy to describe a cylinder. Grid lines for spherical coordinates are based on angle measures, like those for polar coordinates.

In the spherical coordinate system, a point $P$ in space (Figure 2.97) is represented by the ordered triple $(\rho,\theta,\varphi)$ where

• $\rho$ (the Greek letter rho) is the distance between $P$ and the origin $\left( {\rho \neq 0} \right);$
• $\theta$ is the same angle used to describe the location in cylindrical coordinates;
• $\varphi$ (the Greek letter phi) is the angle formed by the positive *z*-axis and line segment $\overset{—}{OP},$ where $O$ is the origin and $0 \leq \varphi \leq \pi.$

By convention, the origin is represented as $\left( {0,0,0} \right)$ in spherical coordinates.

Converting among Spherical, Cylindrical, and Rectangular Coordinates

Rectangular coordinates $\left( {x,y,z} \right)$ and spherical coordinates $\left( {\rho,\theta,\varphi} \right)$ of a point are related as follows:

$$\begin{matrix}

x & = & {\rho\ \text{sin}\ \varphi\ \text{cos}\ \theta} & & & \text{These equations are used to convert from} \\

y & = & {\rho\ \text{sin}\ \varphi\ \text{sin}\ \theta} & & & \text{spherical coordinates to rectangular} \\

z & = & {\rho\ \text{cos}\ \varphi} & & & \text{coordinates.} \\

& \text{and} & & & & \\

\rho^{2} & = & {x^{2} + y^{2} + z^{2}} & & & \text{These equations are used to convert from} \\

{\text{tan}\ \theta} & = & \frac{y}{x} & & & \text{rectangular coordinates to spherical} \\

\varphi & = & {\text{arccos}{\left( \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right).}} & & & \text{coordinates.}

\end{matrix}$$

If a point has cylindrical coordinates $\left( {r,\theta,z} \right),$ then these equations define the relationship between cylindrical and spherical coordinates.

$$\begin{matrix}

r & = & {\rho\ \text{sin}\ \varphi} & & & \text{These equations are used to convert from} \\

\theta & = & \theta & & & \text{spherical coordinates to cylindrical} \\

z & = & {\rho\ \text{cos}\ \varphi} & & & \text{coordinates.} \\

& \text{and} & & & & \\

\rho & = & \sqrt{r^{2} + z^{2}} & & & \text{These equations are used to convert from} \\

\theta & = & \theta & & & \text{cylindrical coordinates to spherical} \\

\varphi & = & {\text{arccos}\left( \frac{z}{\sqrt{r^{2} + z^{2}}} \right)} & & & \text{coordinates.}

\end{matrix}$$

The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry. Looking at Figure 2.98, it is easy to see that $r = \rho\ \text{sin}\ \varphi.$ Then, looking at the triangle in the *xy*-plane with $r$ as its hypotenuse, we have $x = r\ \text{cos}\ \theta = \rho\ \text{sin}\ \varphi\ \text{cos}\ \theta.$ The derivation of the formula for $y$ is similar. Figure 2.96 also shows that $\rho^{2} = r^{2} + z^{2} = x^{2} + y^{2} + z^{2}$ and $z = \rho\ \text{cos}\ \varphi.$ Solving this last equation for $\varphi$ and then substituting $\rho = \sqrt{r^{2} + z^{2}}$ (from the first equation) yields $\varphi = \text{arccos}\left( \frac{z}{\sqrt{r^{2} + z^{2}}} \right).$ Also, note that, as before, we must be careful when using the formula $\text{tan}\ \theta = \frac{y}{x}$ to choose the correct value of $\theta.$

As we did with cylindrical coordinates, let’s consider the surfaces that are generated when each of the coordinates is held constant. Let $c$ be a constant, and consider surfaces of the form $\rho = c.$ Points on these surfaces are at a fixed distance from the origin and form a sphere. The coordinate $\theta$ in the spherical coordinate system is the same as in the cylindrical coordinate system, so surfaces of the form $\theta = c$ are half-planes, as before. Last, consider surfaces of the form $\varphi = c.$ The points on these surfaces are at a fixed angle from the *z*-axis and form a half-cone (Figure 2.99).

Converting from Spherical Coordinates

Plot the point with spherical coordinates $\left( {8,\frac{\pi}{3},\frac{\pi}{6}} \right)$ and express its location in both rectangular and cylindrical coordinates.

Solution

Use the equations in Converting among Spherical, Cylindrical, and Rectangular Coordinates to translate between spherical and cylindrical coordinates (Figure 2.100):

$$\begin{array}{l}

\\

\\

{x = \rho\ \text{sin}\ \varphi\ \text{cos}\ \theta = 8\ \text{sin}\left( \frac{\pi}{6} \right)\text{cos}\left( \frac{\pi}{3} \right) = 8\left( \frac{1}{2} \right)\frac{1}{2} = 2} \\

{y = \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta = 8\ \text{sin}\left( \frac{\pi}{6} \right)\text{sin}\left( \frac{\pi}{3} \right) = 8\left( \frac{1}{2} \right)\frac{\sqrt{3}}{2} = 2\sqrt{3}} \\

{z = \rho\ \text{cos}\ \varphi = 8\ \text{cos}\left( \frac{\pi}{6} \right) = 8\left( \frac{\sqrt{3}}{2} \right) = 4\sqrt{3}.}

\end{array}$$

The point with spherical coordinates $\left( {8,\frac{\pi}{3},\frac{\pi}{6}} \right)$ has rectangular coordinates $\left( {2,2\sqrt{3},4\sqrt{3}} \right).$

Finding the values in cylindrical coordinates is equally straightforward:

$$\begin{array}{rll}

& & \\

r & = & {\rho\ \text{sin}\ \varphi = 8\ \text{sin}\ \frac{\pi}{6} = 4} \\

\theta & = & \theta \\

z & = & {\rho\ \text{cos}\ \varphi = 8\ \text{cos}\ \frac{\pi}{6} = 4\sqrt{3}.}

\end{array}$$

Thus, cylindrical coordinates for the point are $\left( {4,\frac{\pi}{3},4\sqrt{3}} \right).$

Plot the point with spherical coordinates $\left( {2, - \frac{5\pi}{6},\frac{\pi}{6}} \right)$ and describe its location in both rectangular and cylindrical coordinates.

Converting from Rectangular Coordinates

Convert the rectangular coordinates $\left( {-1,1,\sqrt{6}} \right)$ to both spherical and cylindrical coordinates.

Solution

Start by converting from rectangular to spherical coordinates:

$$\begin{array}{lccl}

\begin{array}{rll}

\rho^{2} & = & {x^{2} + y^{2} + z^{2} = (-1)^{2} + 1^{2} + \left( \sqrt{6} \right)^{2} = 8} \\

\rho & = & {2\sqrt{2}}

\end{array} & & & \begin{array}{rll}

{\text{tan}\ \theta} & = & \frac{1}{-1} \\

\theta & = & {\text{arctan}(-1) = \frac{3\pi}{4}.}

\end{array}

\end{array}$$

Because $\left( {x,y} \right) = \left( {-1,1} \right),$ then the correct choice for $\theta$ is $\frac{3\pi}{4}.$

There are actually two ways to identify $\varphi.$ We can use the equation $\varphi = \text{arccos}\left( \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right).$ A more simple approach, however, is to use equation $z = \rho\ \text{cos}\ \varphi.$ We know that $z = \sqrt{6}$ and $\rho = 2\sqrt{2},$ so

$$\sqrt{6} = 2\sqrt{2}\ \text{cos}\ \varphi,\ \text{so}\ \text{cos}\ \varphi = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$$

and therefore $\varphi = \frac{\pi}{6}.$ The spherical coordinates of the point are $\left( {2\sqrt{2},\frac{3\pi}{4},\frac{\pi}{6}} \right).$

To find the cylindrical coordinates for the point, we need only find $r\text{:}$

$$r = \rho\ \text{sin}\ \varphi = 2\sqrt{2}\ \text{sin}\left( \frac{\pi}{6} \right) = \sqrt{2}.$$

The cylindrical coordinates for the point are $\left( {\sqrt{2},\frac{3\pi}{4},\sqrt{6}} \right).$

Identifying Surfaces in the Spherical Coordinate System

Describe the surfaces with the given spherical equations.

1. $\theta = \frac{\pi}{3}$

2. $\varphi = \frac{5\pi}{6}$

3. $\rho = 6$

4. $\rho = \text{sin}\ \theta\ \text{sin}\ \varphi$

Solution

1. The variable $\theta$ represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates $\left( {\rho,\frac{\pi}{3},\varphi} \right)$ lie on the plane that forms angle $\theta = \frac{\pi}{3}$ with the positive *x*-axis. Because $\rho > 0,$ the surface described by equation $\theta = \frac{\pi}{3}$ is the half-plane shown in Figure 2.101.

2. Equation $\varphi = \frac{5\pi}{6}$ describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring $\frac{5\pi}{6}$ rad with the positive *z*-axis. These points form a half-cone (Figure 2.102). Because there is only one value for $\varphi$ that is measured from the positive *z*-axis, we do not get the full cone (with two pieces).

To find the equation in rectangular coordinates, use equation $\varphi = \text{arccos}\left( \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right).$

$$\begin{array}{rll}

\frac{5\pi}{6} & = & {\text{arccos}\left( \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right)} \\

{\text{cos}\ \frac{5\pi}{6}} & = & \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \\

{- \frac{\sqrt{3}}{2}} & = & \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \\

\frac{3}{4} & = & \frac{z^{2}}{x^{2} + y^{2} + z^{2}} \\

{\frac{3x^{2}}{4} + \frac{3y^{2}}{4} + \frac{3z^{2}}{4}} & = & z^{2} \\

{\frac{3x^{2}}{4} + \frac{3y^{2}}{4} - \frac{z^{2}}{4}} & = & 0.

\end{array}$$

This is the equation of a cone centered on the *z*-axis.

3. Equation $\rho = 6$ describes the set of all points $6$ units away from the origin—a sphere with radius $6$ (Figure 2.103).

4. To identify this surface, convert the equation from spherical to rectangular coordinates, using equations $y = \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta$ and $\rho^{2} = x^{2} + y^{2} + z^{2}\text{:}$

$$\begin{array}{rllccc}

\rho & = & {\text{sin}\ \theta\ \text{sin}\ \varphi} & & & \\

\rho^{2} & = & {\rho\ \text{sin}\ \theta\ \text{sin}\ \varphi} & & & {\text{Multiply both sides of the equation by}\ \rho.} \\

{x^{2} + y^{2} + z^{2}} & = & y & & & \text{Substitute rectangular variables using the equations above.} \\

{x^{2} + y^{2} - y + z^{2}} & = & 0 & & & {\text{Subtract}\ y\ \text{from both sides of the equation.}} \\

{x^{2} + y^{2} - y + \frac{1}{4} + z^{2}} & = & \frac{1}{4} & & & \text{Complete the square.} \\

{x^{2} + \left( {y - \frac{1}{2}} \right)^{2} + z^{2}} & = & {\frac{1}{4}.} & & & \text{Rewrite the middle terms as a perfect square.}

\end{array}$$

The equation describes a sphere centered at point $\left( {0,\frac{1}{2},0} \right)$ with radius $\frac{1}{2}.$

Describe the surfaces defined by the following equations.

1. $\rho = 13$

2. $\theta = \frac{2\pi}{3}$

3. $\varphi = \frac{\pi}{4}$

Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as the volume of the space inside a domed stadium or wind speeds in a planet’s atmosphere. A sphere that has Cartesian equation $x^{2} + y^{2} + z^{2} = c^{2}$ has the simple equation $\rho = c$ in spherical coordinates.

In geography, latitude and longitude are used to describe locations on Earth’s surface, as shown in Figure 2.104. Although the shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth. Let’s assume Earth has the shape of a sphere with radius $4000$ mi. We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees.

Let the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing the positive *z*-axis. The prime meridian represents the trace of the surface as it intersects the *xz*-plane. The equator is the trace of the sphere intersecting the *xy*-plane.

Converting Latitude and Longitude to Spherical Coordinates

The latitude of Columbus, Ohio, is $40\text{°}$ N and the longitude is $83\text{°}$ W, which means that Columbus is $40\text{°}$ north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earth through the equator directly south of Columbus. The measure of the angle formed by the rays is $40\text{°}.$ In the same way, measuring from the prime meridian, Columbus lies $83\text{°}$ to the west. Express the location of Columbus in spherical coordinates.

Solution

The radius of Earth is $4000$ mi, so $\rho = 4000.$ The intersection of the prime meridian and the equator lies on the positive *x*-axis. Movement to the west is then described with negative angle measures, which shows that $\theta = -83\text{°},$ Because Columbus lies $40\text{°}$ north of the equator, it lies $50\text{°}$ south of the North Pole, so $\varphi = 50\text{°}.$ In spherical coordinates, Columbus lies at point $\left( {4000,-83\text{°},50\text{°}} \right).$

Sydney, Australia is at $34\text{°}\text{S}$ and $151\text{°}\text{E}.$ Express Sydney’s location in spherical coordinates.

Cylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand. A thoughtful choice of coordinate system can make a problem much easier to solve, whereas a poor choice can lead to unnecessarily complex calculations. In the following example, we examine several different problems and discuss how to select the best coordinate system for each one.

Choosing the Best Coordinate System

In each of the following situations, we determine which coordinate system is most appropriate and describe how we would orient the coordinate axes. There could be more than one right answer for how the axes should be oriented, but we select an orientation that makes sense in the context of the problem. *Note*: There is not enough information to set up or solve these problems; we simply select the coordinate system (Figure 2.105).

1. Find the center of gravity of a bowling ball.

2. Determine the velocity of a submarine subjected to an ocean current.

3. Calculate the pressure in a conical water tank.

4. Find the volume of oil flowing through a pipeline.

5. Determine the amount of leather required to make a football.

Solution

1. Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The origin should be located at the physical center of the ball. There is no obvious choice for how the *x*-, *y*- and *z*-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choice is to align the *z*-axis with the axis of symmetry of the weight block.

2. A submarine generally moves in a straight line. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice. The *z*-axis should probably point upward. The *x*- and *y*-axes could be aligned to point east and north, respectively. The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port.

3. A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation $z = kr,$ where $k$ is a constant. In spherical coordinates, we have seen that surfaces of the form $\varphi = c$ are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form $z^{2} = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}.$ In this case, we could choose any of the three. However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might want to avoid that choice. In addition, we are talking about a water tank, and the depth of the water might come into play at some point in our calculations, so it might be nice to have a component that represents height and depth directly. Based on this reasoning, cylindrical coordinates might be the best choice. Choose the *z*-axis to align with the axis of the cone. The orientation of the other two axes is arbitrary. The origin should be the bottom point of the cone.

4. A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, we would likely choose to orient our *z*-axis with the center axis of the pipeline. The *x*-axis could be chosen to point straight downward or to some other logical direction. The origin should be chosen based on the problem statement. Note that this puts the *z*-axis in a horizontal orientation, which is a little different from what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem.

5. A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. The *z*-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one of the ends. The position of the *x*-axis is arbitrary.

Which coordinate system is most appropriate for creating a star map, as viewed from Earth (see the following figure)?

How should we orient the coordinate axes?

Section 2.7 Exercises

Use the following figure as an aid in identifying the relationship between the rectangular, cylindrical, and spherical coordinate systems.

For the following exercises, the cylindrical coordinates $\left( {r,\theta,z} \right)$ of a point are given. Find the rectangular coordinates $\left( {x,y,z} \right)$ of the point.

363.

$\left( {4,\frac{\pi}{6},3} \right)$

364\.

$\left( {3,\frac{\pi}{3},5} \right)$

365.

$\left( {4,\frac{7\pi}{6},3} \right)$

366\.

$\left( {2,\pi,-4} \right)$

For the following exercises, the rectangular coordinates $\left( {x,y,z} \right)$ of a point are given. Find the cylindrical coordinates $\left( {r,\theta,z} \right)$ of the point.

367.

$\left( {1,\sqrt{3},2} \right)$

368\.

$\left( {1,1,5} \right)$

369.

$\left( {3,-3,7} \right)$

370\.

$\left( {-2\sqrt{2},2\sqrt{2},4} \right)$

For the following exercises, the equation of a surface in cylindrical coordinates is given.

Find an equation of the surface in rectangular coordinates. Identify and graph the surface.

371.

\[T\] $r = 4$

372\.

\[T\] $z = r^{2}\text{cos}^{2}\theta$

373.

\[T\] $r^{2}\text{cos}(2\theta) + z^{2} + 1 = 0$

374\.

\[T\] $r = 3\ \text{sin}\ \theta$

375.

\[T\] $r = 2\ \text{cos}\ \theta$

376\.

\[T\] $r^{2} + z^{2} = 5$

377.

\[T\] $r = 2\ \text{sec}\ \theta$

378\.

\[T\] $r = 3\ \text{csc}\ \theta$

For the following exercises, the equation of a surface in rectangular coordinates is given. Find an equation of the surface in cylindrical coordinates.

379.

$z = 3$

380\.

$x = 6$

381.

$x^{2} + y^{2} + z^{2} = 9$

382\.

$y = 2x^{2}$

383.

$x^{2} + y^{2} - 16x = 0$

384\.

$x^{2} + y^{2} - 3\sqrt{x^{2} + y^{2}} + 2 = 0$

For the following exercises, the spherical coordinates $\left( {\rho,\theta,\varphi} \right)$ of a point are given. Find the rectangular coordinates $\left( {x,y,z} \right)$ of the point.

385.

$\left( {3,0,\pi} \right)$

386\.

$\left( {1,\frac{\pi}{6},\frac{\pi}{6}} \right)$

387.

$\left( {12, - \frac{\pi}{4},\frac{\pi}{4}} \right)$

388\.

$\left( {3,\frac{\pi}{4},\frac{\pi}{6}} \right)$

For the following exercises, the rectangular coordinates $\left( {x,y,z} \right)$ of a point are given. Find the spherical coordinates $\left( {\rho,\theta,\varphi} \right)$ of the point. Express the measure of the angles in degrees rounded to the nearest integer.

389.

$\left( {4,0,0} \right)$

390\.

$\left( {-1,2,1} \right)$

391.

$\left( {0,3,0} \right)$

392\.

$\left( {-2,2\sqrt{3},4} \right)$

For the following exercises, the equation of a surface in spherical coordinates is given. Find an equation of the surface in rectangular coordinates. Identify and graph the surface.

393.

\[T\] $\rho = 3$

394\.

\[T\] $\varphi = \frac{\pi}{3}$

395.

\[T\] $\rho = 2\ \text{cos}\ \varphi$

396\.

\[T\] $\rho = 4\ \text{csc}\ \varphi$

397.

\[T\] $\varphi = \frac{\pi}{2}$

398\.

\[T\] $\rho = 6\ \text{csc}\ \varphi\ \text{sec}\ \theta$

For the following exercises, the equation of a surface in rectangular coordinates is given. Find an equation of the surface in spherical coordinates. Identify the surface.

399.

$x^{2} + y^{2} - 3z^{2} = 0,$ $z \neq 0$

400\.

$x^{2} + y^{2} + z^{2} - 4z = 0$

401.

$z = 6$

402\.

$x^{2} + y^{2} = 9$

For the following exercises, the cylindrical coordinates of a point are given. Find its associated spherical coordinates, with the measure of the angle $\varphi$ in radians rounded to four decimal places.

403.

\[T\] $\left( {1,\frac{\pi}{4},3} \right)$

404\.

\[T\] $\left( {5,\pi,12} \right)$

405.

$\left( {3,\frac{\pi}{2},3} \right)$

406\.

$\left( {3, - \frac{\pi}{6},3} \right)$

For the following exercises, the spherical coordinates of a point are given. Find its associated cylindrical coordinates.

407.

$\left( {2, - \frac{\pi}{4},\frac{\pi}{2}} \right)$

408\.

$\left( {4,\frac{\pi}{4},\frac{\pi}{6}} \right)$

409.

$\left( {8,\frac{\pi}{3},\frac{\pi}{2}} \right)$

410\.

$\left( {9, - \frac{\pi}{6},\frac{\pi}{3}} \right)$

For the following exercises, find the most suitable system of coordinates to describe the solids.

411.

The solid situated in the first octant with a vertex at the origin and enclosed by a cube of edge length $a,$ where $a > 0$

412\.

A spherical shell determined by the region between two concentric spheres centered at the origin, of radii of $a$ and $b,$ respectively, where $b > a > 0$

413.

A solid inside sphere $x^{2} + y^{2} + z^{2} = 9$ and outside cylinder $\left( {x - \frac{3}{2}} \right)^{2} + y^{2} = \frac{9}{4}$

414\.

A cylindrical shell of height $10$ determined by the region between two cylinders with the same center, parallel rulings, and radii of $2$ and $5,$ respectively

415.

\[T\] Use a CAS to graph the region between elliptic paraboloid $z = x^{2} + y^{2}$ and cone $x^{2} + y^{2} - z^{2} = 0.$ Then describe the region in cylindrical coordinates.

416\.

\[T\] Use a CAS to graph in spherical coordinates the “ice cream-cone region” situated above the *xy*-plane between sphere $x^{2} + y^{2} + z^{2} = 4$ and elliptical cone $x^{2} + y^{2} - z^{2} = 0.$

417.

Washington, DC, is located at $39\text{°}$ N and $77\text{°}$ W (see the following figure). Assume the radius of Earth is $4000$ mi. Express the location of Washington, DC, in spherical coordinates.

418\.

San Francisco is located at $37.78\text{°}\text{N}$ and $122.42\text{°}\text{W}.$ Assume the radius of Earth is $4000$ mi. Express the location of San Francisco in spherical coordinates.

419.

Find the latitude and longitude of Rio de Janeiro if its spherical coordinates are $\left( {4000,\text{−}43.17\text{°},102.91\text{°}} \right).$

420\.

Find the latitude and longitude of Berlin if its spherical coordinates are $\left( {4000,13.38\text{°},37.48\text{°}} \right).$

421.

\[T\] Consider the torus of equation $\left( {x^{2} + y^{2} + z^{2} + R^{2} - r^{2}} \right)^{2} = 4R^{2}\left( {x^{2} + y^{2}} \right),$ where $R \geq r > 0.$

1. Write the equation of the torus in spherical coordinates.

2. If $R = r,$ the surface is called a *horn torus*. Show that the equation of a horn torus in spherical coordinates is $\rho = 2R\ \text{sin}\ \varphi.$

3. Use a CAS to graph the horn torus with $R = r = 2$ in spherical coordinates.

422\.

\[T\] The “bumpy sphere” with an equation in spherical coordinates is $\rho = a + b\ \text{cos}(m\theta)\text{sin}(n\varphi),$ with $\theta \in \lbrack 0,2\pi\rbrack$ and $\varphi \in \lbrack 0,\pi\rbrack,$ where $a$ and $b$ are positive numbers and $m$ and $n$ are positive integers, may be used in applied mathematics to model tumor growth.

1. Show that the “bumpy sphere” is contained inside a sphere of equation $\rho = a + b.$ Find the values of $\theta$ and $\varphi$ at which the two surfaces intersect.

2. Use a CAS to graph the surface for $a = 14,$ $b = 2,$ $m = 4,$ and $n = 6$ along with sphere $\rho = a + b.$

3. Find an equation of the intersection curve of the surface at b. with the cone $\varphi = \frac{\pi}{12}.$ Graph the intersection curve in the plane of intersection.

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Chapter Review

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-chapter-review

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Key Terms

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-key-terms

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Key Terms

Calculus Volume 3Key Terms

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Key Terms

component

a scalar that describes either the vertical or horizontal direction of a vector

coordinate plane

a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the *xy*-plane, *xz*-plane, or the *yz*-plane

cross product

$\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \left( {u_{2}v_{3} - u_{3}v_{2}} \right)\mathbf{\text{i}} - \left( {u_{1}v_{3} - u_{3}v_{1}} \right)\mathbf{\text{j}} + \left( {u_{1}v_{2} - u_{2}v_{1}} \right)\mathbf{\text{k}},$ where $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$

cylinder

a set of lines parallel to a given line passing through a given curve

cylindrical coordinate system

a way to describe a location in space with an ordered triple $\left( {r,\theta,z} \right),$ where $\left( {r,\theta} \right)$ represents the polar coordinates of the point’s projection in the *xy*-plane, and $z$ represents the point’s projection onto the *z*-axis

determinant

a real number associated with a square matrix

direction angles

the angles formed by a nonzero vector and the coordinate axes

direction cosines

the cosines of the angles formed by a nonzero vector and the coordinate axes

direction vector

a vector parallel to a line that is used to describe the direction, or orientation, of the line in space

dot product or scalar product

$\mathbf{\text{u}} \cdot \mathbf{\text{v}} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}$ where $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$

ellipsoid

a three-dimensional surface described by an equation of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1;$ all traces of this surface are ellipses

elliptic cone

a three-dimensional surface described by an equation of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 0;$ traces of this surface include ellipses and intersecting lines

elliptic paraboloid

a three-dimensional surface described by an equation of the form $z = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}};$ traces of this surface include ellipses and parabolas

equivalent vectors

vectors that have the same magnitude and the same direction

general form of the equation of a plane

an equation in the form $ax + by + cz + d = 0,$ where $\mathbf{\text{n}} = \left\langle {a,b,c} \right\rangle$ is a normal vector of the plane, $P = \left( {x_{0},y_{0},z_{0}} \right)$ is a point on the plane, and $d = \text{−}ax_{0} - by_{0} - cz_{0}$

hyperboloid of one sheet

a three-dimensional surface described by an equation of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1;$ traces of this surface include ellipses and hyperbolas

hyperboloid of two sheets

a three-dimensional surface described by an equation of the form $\frac{z^{2}}{c^{2}} - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1;$ traces of this surface include ellipses and hyperbolas

initial point

the starting point of a vector

magnitude

the length of a vector

normal vector

a vector perpendicular to a plane

normalization

using scalar multiplication to find a unit vector with a given direction

octants

the eight regions of space created by the coordinate planes

orthogonal vectors

vectors that form a right angle when placed in standard position

parallelepiped

a three-dimensional prism with six faces that are parallelograms

parallelogram method

a method for finding the sum of two vectors; position the vectors so they share the same initial point; the vectors then form two adjacent sides of a parallelogram; the sum of the vectors is the diagonal of that parallelogram

parametric equations of a line

the set of equations $x = x_{0} + ta,$ $y = y_{0} + tb,$ and $z = z_{0} + tc$ describing the line with direction vector $\mathbf{\text{v}} = \left\langle {a,b,c} \right\rangle$ passing through point $\left( {x_{0},y_{0},z_{0}} \right)$

quadric surfaces

surfaces in three dimensions having the property that the traces of the surface are conic sections (ellipses, hyperbolas, and parabolas)

right-hand rule

a common way to define the orientation of the three-dimensional coordinate system; when the right hand is curved around the *z*-axis in such a way that the fingers curl from the positive *x*-axis to the positive *y*-axis, the thumb points in the direction of the positive *z*-axis

rulings

parallel lines that make up a cylindrical surface

scalar

a real number

scalar equation of a plane

the equation $a\left( {x - x_{0}} \right) + b\left( {y - y_{0}} \right) + c\left( {z - z_{0}} \right) = 0$ used to describe a plane containing point $P = \left( {x_{0},y_{0},z_{0}} \right)$ with normal vector $\mathbf{\text{n}} = \left\langle {a,b,c} \right\rangle$ or its alternate form $ax + by + cz + d = 0,$ where $d = \text{−}ax_{0} - by_{0} - cz_{0}$

scalar multiplication

a vector operation that defines the product of a scalar and a vector

scalar projection

the magnitude of the vector projection of a vector

skew lines

two lines that are not parallel but do not intersect

sphere

the set of all points equidistant from a given point known as the *center*

spherical coordinate system

a way to describe a location in space with an ordered triple $\left( {\rho,\theta,\varphi} \right),$ where $\rho$ is the distance between $P$ and the origin $\left( {\rho \neq 0} \right),$ $\theta$ is the same angle used to describe the location in cylindrical coordinates, and $\varphi$ is the angle formed by the positive *z*-axis and line segment $\overset{—}{OP},$ where $O$ is the origin and $0 \leq \varphi \leq \pi$

standard equation of a sphere

$\left( {x - a} \right)^{2} + \left( {y - b} \right)^{2} + \left( {z - c} \right)^{2} = r^{2}$ describes a sphere with center $\left( {a,b,c} \right)$ and radius $r$

standard unit vectors

unit vectors along the coordinate axes: $\mathbf{\text{i}} = \left\langle {1,0} \right\rangle,\mathbf{\text{j}} = \left\langle {0,1} \right\rangle$

standard-position vector

a vector with initial point $\left( {0,0} \right)$

symmetric equations of $\mathbf{\text{a}}$ line

the equations $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ describing the line with direction vector $\mathbf{\text{v}} = \left\langle {a,b,c} \right\rangle$ passing through point $\left( {x_{0},y_{0},z_{0}} \right)$

terminal point

the endpoint of a vector

three-dimensional rectangular coordinate system

a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple $\left( {x,y,z} \right)$ that plots its location relative to the defining axes

torque

the effect of a force that causes an object to rotate

trace

the intersection of a three-dimensional surface with a coordinate plane

triangle inequality

the length of any side of a triangle is less than the sum of the lengths of the other two sides

triangle method

a method for finding the sum of two vectors; position the vectors so the terminal point of one vector is the initial point of the other; these vectors then form two sides of a triangle; the sum of the vectors is the vector that forms the third side; the initial point of the sum is the initial point of the first vector; the terminal point of the sum is the terminal point of the second vector

triple scalar product

the dot product of a vector with the cross product of two other vectors: $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)$

unit vector

a vector with margnitude $1$

vector

a mathematical object that has both magnitude and direction

vector addition

a vector operation that defines the sum of two vectors

vector difference

the vector difference $\mathbf{\text{v}} - \mathbf{\text{w}}$ is defined as $\mathbf{v +}\left( {\text{−}\mathbf{\text{w}}} \right) = \mathbf{v +}(-1)\mathbf{\text{w}}$

vector equation of a line

the equation $\mathbf{\text{r}} = \mathbf{\text{r}}_{0} + t\mathbf{\text{v}}$ used to describe a line with direction vector $\mathbf{\text{v}} = \left\langle {a,b,c} \right\rangle$ passing through point $P = \left( {x_{0},y_{0},z_{0}} \right),$ where $\mathbf{\text{r}}_{0} = \left\langle {x_{0},y_{0},z_{0}} \right\rangle,$ is the position vector of point $P$

vector equation of a plane

the equation $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0,$ where $P$ is a given point in the plane, $Q$ is any point in the plane, and $\mathbf{\text{n}}$ is a normal vector of the plane

vector product

the cross product of two vectors

vector projection

the component of a vector that follows a given direction

vector sum

the sum of two vectors, $\mathbf{\text{v}}$ and $\mathbf{\text{w}},$ can be constructed graphically by placing the initial point of $\mathbf{\text{w}}$ at the terminal point of $\mathbf{\text{v}};$ then the vector sum $\mathbf{v + w}$ is the vector with an initial point that coincides with the initial point of $\mathbf{\text{v}},$ and with a terminal point that coincides with the terminal point of $\mathbf{\text{w}}$

work done by a force

work is generally thought of as the amount of energy it takes to move an object; if we represent an applied force by a vector F and the displacement of an object by a vector s, then the work done by the force is the dot product of F and s.

zero vector

the vector with both initial point and terminal point $\left( {0,0} \right)$

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/2-key-terms

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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Key Equations

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-key-equations

Skip to ContentGo to accessibility page\Calculus Volume 3

Key Equations

Calculus Volume 3Key Equations

------------------------------------------------------------------------

Key Equations

| | |

|-----------------------------------------------------------------|---------------------------------------------------------------------------------------------------------------------------|

| Distance between two points in space: | $d = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2} + \left( {z_{2} - z_{1}} \right)^{2}}$ |

| **Sphere with center $\left( {a,b,c} \right)$ and radius *r*:** | $\left( {x - a} \right)^{2} + \left( {y - b} \right)^{2} + \left( {z - c} \right)^{2} = r^{2}$ |

| | |

|------------------------------------------------------------------------------------------------------|----------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Dot product of u and v | $\begin{array}{cl}

{\mathbf{\text{u}} \cdot \mathbf{\text{v}}} & {= u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}} \\

& {= \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta}

\end{array}$ |

| Cosine of the angle formed by $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ | $\text{cos}\ \theta = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|}$ |

| Vector projection of $\mathbf{\text{v}}$ onto $\mathbf{\text{u}}$ | $\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}} = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}$ |

| Scalar projection of $\mathbf{\text{v}}$ onto $\mathbf{\text{u}}$ | $\text{comp}_{\mathbf{\text{u}}}\mathbf{\text{v}} = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|}$ |

| Work done by a force F to move an object through displacement vector $\overset{\rightarrow}{PQ}$ | $W = \mathbf{\text{F}} \cdot \overset{\rightarrow}{PQ} = \left\| \mathbf{\text{F}} \right\|\left\| \overset{\rightarrow}{PQ} \right\|\text{cos}\ \theta$ |

| | |

|-------------------------------------------------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| The cross product of two vectors in terms of the unit vectors | $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = (u_{2}v_{3} - u_{3}v_{2})\mathbf{\text{i}} - (u_{1}v_{3} - u_{3}v_{1})\mathbf{\text{j}} + (u_{1}v_{2} - u_{2}v_{1})\mathbf{\text{k}}$ |

| | |

|------------------------------------------|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Vector Equation of a Line | $\mathbf{\text{r}} = \mathbf{\text{r}}_{0} + t\mathbf{\text{v}}$ |

| Parametric Equations of a Line | $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ |

| Vector Equation of a Plane | $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0$ |

| Scalar Equation of a Plane | $a\left( {x - x_{0}} \right) + b\left( {y - y_{0}} \right) + c\left( {z - z_{0}} \right) = 0$ |

| Distance between a Plane and a Point | $d = \left\| {\text{proj}_{\mathbf{\text{n}}}\overset{\rightarrow}{QP}} \right\| = \left| {\text{comp}_{\mathbf{\text{n}}}\overset{\rightarrow}{QP}} \right| = \frac{\left| {\overset{\rightarrow}{QP} \cdot \mathbf{\text{n}}} \right|}{\left\| \mathbf{\text{n}} \right\|}$ |

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Citation information

• Use the information below to generate a citation. We recommend using a citation tool such as this one.
• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/2-key-equations

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

---

Key Concepts

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-key-concepts

Skip to ContentGo to accessibility page\Calculus Volume 3

Key Concepts

Calculus Volume 3Key Concepts

------------------------------------------------------------------------

Key Concepts

2.1 Vectors in the Plane

• Vectors are used to represent quantities that have both magnitude and direction.
• We can add vectors by using the parallelogram method or the triangle method to find the sum. We can multiply a vector by a scalar to change its length or give it the opposite direction.
• Subtraction of vectors is defined in terms of adding the negative of the vector.
• A vector is written in component form as $\mathbf{\text{v}} = \left\langle {x,y} \right\rangle.$
• The magnitude of a vector is a scalar: $\left\| \mathbf{\text{v}} \right\| = \sqrt{x^{2} + y^{2}}.$
• A unit vector $\mathbf{\text{u}}$ has magnitude $1$ and can be found by dividing a vector by its magnitude: $\mathbf{\text{u}} = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\mathbf{\text{v}}.$ The standard unit vectors are $\mathbf{\text{i}} = \left\langle {1,0} \right\rangle\ \text{and}\ \mathbf{\text{j}} = \left\langle {0,1} \right\rangle.$ A vector $\mathbf{\text{v}} = \left\langle {x,y} \right\rangle$ can be expressed in terms of the standard unit vectors as $\mathbf{\text{v}} = x\mathbf{\text{i}} + y\mathbf{\text{j}}.$
• Vectors are often used in physics and engineering to represent forces and velocities, among other quantities.

2.2 Vectors in Three Dimensions

• The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples $\left( {x,y,z} \right)$ are used to describe the location of a point in space.
• The distance $d$ between points $\left( {x_{1},y_{1},z_{1}} \right)$ and $\left( {x_{2},y_{2},z_{2}} \right)$ is given by the formula

$$d = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2} + \left( {z_{2} - z_{1}} \right)^{2}}.$$

• In three dimensions, the equations $x = a,y = b,\text{and}\ z = c$ describe planes that are parallel to the coordinate planes.
• The standard equation of a sphere with center $\left( {a,b,c} \right)$ and radius $r$ is

$$\left( {x - a} \right)^{2} + \left( {y - b} \right)^{2} + \left( {z - c} \right)^{2} = r^{2}.$$

• In three dimensions, as in two, vectors are commonly expressed in component form, $\mathbf{\text{v}} = \left\langle {x,y,z} \right\rangle,$ or in terms of the standard unit vectors, $x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}.$
• Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let $\mathbf{\text{v}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle$ and $\mathbf{\text{w}} = \left\langle {x_{2},y_{2},z_{2}} \right\rangle$ be vectors, and let $k$ be a scalar.
• Scalar multiplication: $k\mathbf{\text{v}} = \left\langle {kx_{1},ky_{1},kz_{1}} \right\rangle$
• Vector addition: $\mathbf{\text{v}} + \mathbf{\text{w}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle + \left\langle {x_{2},y_{2},z_{2}} \right\rangle = \left\langle {x_{1} + x_{2},y_{1} + y_{2},z_{1} + z_{2}} \right\rangle$
• Vector subtraction: $\mathbf{\text{v}} - \mathbf{\text{w}} = \left\langle {x_{1},y_{1},z_{1}} \right\rangle - \left\langle {x_{2},y_{2},z_{2}} \right\rangle = \left\langle {x_{1} - x_{2},y_{1} - y_{2},z_{1} - z_{2}} \right\rangle$
• Vector magnitude: $\left\| \mathbf{\text{v}} \right\| = \sqrt{x_{1}{}^{2} + y_{1}{}^{2} + z_{1}{}^{2}}$
• Unit vector in the direction of v: $\frac{\mathbf{\text{v}}}{\left\| \mathbf{\text{v}} \right\|} = \frac{1}{\left\| \mathbf{\text{v}} \right\|}\left\langle {x_{1},y_{1},z_{1}} \right\rangle = \left\langle {\frac{x_{1}}{\left\| \mathbf{\text{v}} \right\|},\frac{y_{1}}{\left\| \mathbf{\text{v}} \right\|},\frac{z_{1}}{\left\| \mathbf{\text{v}} \right\|}} \right\rangle,$ $\mathbf{\text{v}} \neq \mathbf{0}$

2.3 The Dot Product

• The dot product, or scalar product, of two vectors $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ is $\mathbf{\text{u}} \cdot \mathbf{\text{v}} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}.$
• The dot product satisfies the following properties:
• $\mathbf{\text{u}} \cdot \mathbf{\text{v}} = \mathbf{\text{v}} \cdot \mathbf{\text{u}}$
• $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}} + \mathbf{\text{w}}} \right) = \mathbf{\text{u}} \cdot \mathbf{\text{v}} + \mathbf{\text{u}} \cdot \mathbf{\text{w}}$
• $c\left( {\mathbf{\text{u}} \cdot \mathbf{\text{v}}} \right) = \left( {c\mathbf{\text{u}}} \right) \cdot \mathbf{\text{v}} = \mathbf{\text{u}} \cdot \left( {c\mathbf{\text{v}}} \right)$
• $\mathbf{\text{v}} \cdot \mathbf{\text{v}} = \left\| \mathbf{\text{v}} \right\|^{2}$
• The dot product of two vectors can be expressed, alternatively, as $\mathbf{\text{u}} \cdot \mathbf{\text{v}} = \left\| \mathbf{\text{u}} \right\|\left\| \mathbf{\text{v}} \right\|\text{cos}\ \theta.$ This form of the dot product is useful for finding the measure of the angle formed by two vectors.
• Vectors $\textbf{u}$ and $\textbf{v}$ are orthogonal if $\mathbf{\text{u}} \cdot \mathbf{\text{v}} = 0.$
• The angles formed by a nonzero vector and the coordinate axes are called the *direction angles* for the vector. The cosines of these angles are known as the *direction cosines*.
• The vector projection of $\textbf{v}$ onto $\textbf{u}$ is the vector $\text{proj}_{\mathbf{\text{u}}}\mathbf{\text{v}} = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|^{2}}\mathbf{\text{u}}.$ The magnitude of this vector is known as the *scalar projection* of $\textbf{v}$ onto $\textbf{u}$, given by $\text{comp}_{\mathbf{\text{u}}}\mathbf{\text{v}} = \frac{\mathbf{\text{u}} \cdot \mathbf{\text{v}}}{\left\| \mathbf{\text{u}} \right\|}.$
• Work is done when a force is applied to an object, causing displacement. When the force is represented by the vector F and the displacement is represented by the vector s, then the work done *W* is given by the formula $W = \mathbf{\text{F}} \cdot s = \left\| \mathbf{\text{F}} \right\|\left\| \mathbf{\text{s}} \right\|\text{cos}\ \theta.$

2.4 The Cross Product

• The cross product $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}$ of two vectors $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ is a vector orthogonal to both $\mathbf{\text{u}}$ and $\mathbf{\text{v}}.$ Its length is given by $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\| = \left\| \mathbf{\text{u}} \right\| \cdot \left\| \mathbf{\text{v}} \right\| \cdot \text{sin}\ \theta,$ where $\theta$ is the angle between $\mathbf{\text{u}}$ and $\mathbf{\text{v}}.$ Its direction is given by the right-hand rule.
• The algebraic formula for calculating the cross product of two vectors,

$\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle\ \text{and}\ \mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle,$ is

$\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \left( {u_{2}v_{3} - u_{3}v_{2}} \right)\mathbf{\text{i}} - \left( {u_{1}v_{3} - u_{3}v_{1}} \right)\mathbf{\text{j}} + \left( {u_{1}v_{2} - u_{2}v_{1}} \right)\mathbf{\text{k}}.$

• The cross product satisfies the following properties for vectors $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}},$ and scalar $c\text{:}$
• $\mathbf{\text{u}}\ \times \ \mathbf{\text{v}} = \text{−}\left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{u}}} \right)$
• $\mathbf{\text{u}}\ \times \ \left( {\mathbf{\text{v}} + \mathbf{\text{w}}} \right) = \mathbf{\text{u}}\ \times \ \mathbf{\text{v}} + \mathbf{\text{u}}\ \times \ \mathbf{\text{w}}$
• $c\left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) = \left( {c\mathbf{\text{u}}} \right)\ \times \ \mathbf{\text{v}} = \mathbf{\text{u}}\ \times \ \left( {c\mathbf{\text{v}}} \right)$
• $\mathbf{\text{u}}\ \times \ \mathbf{0} = \mathbf{0}\ \times \ \mathbf{\text{u}} = \mathbf{0}$
• $\mathbf{\text{v}}\ \times \ \mathbf{\text{v}} = 0$
• $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right) = \left( {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right) \cdot \mathbf{\text{w}}$
• The cross product of vectors $\mathbf{\text{u}} = \left\langle {u_{1},u_{2},u_{3}} \right\rangle$ and $\mathbf{\text{v}} = \left\langle {v_{1},v_{2},v_{3}} \right\rangle$ is the determinant $\left| \begin{array}{lll}

\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\

u_{1} & u_{2} & u_{3} \\

v_{1} & v_{2} & v_{3}

\end{array} \right|.$

• If vectors $\mathbf{\text{u}}$ and $\mathbf{\text{v}}$ form adjacent sides of a parallelogram, then the area of the parallelogram is given by $\left\| {\mathbf{\text{u}}\ \times \ \mathbf{\text{v}}} \right\|.$
• The triple scalar product of vectors $\mathbf{\text{u}},$ $\mathbf{\text{v}},$ and $\mathbf{\text{w}}$ is $\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right).$
• The volume of a parallelepiped with adjacent edges given by vectors $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ is $V = \left| {\mathbf{\text{u}} \cdot \left( {\mathbf{\text{v}}\ \times \ \mathbf{\text{w}}} \right)} \right|.$
• If the triple scalar product of vectors $\mathbf{\text{u}},\mathbf{\text{v}},\ \text{and}\ \mathbf{\text{w}}$ is zero, then the vectors are coplanar. The converse is also true: If the vectors are coplanar, then their triple scalar product is zero.
• The cross product can be used to identify a vector orthogonal to two given vectors or to a plane.
• Torque $\tau$ measures the tendency of a force to produce rotation about an axis of rotation. If force $\mathbf{\text{F}}$ is acting at a distance $\mathbf{\text{r}}$ from the axis, then torque is equal to the cross product of $\mathbf{\text{r}}$ and $\mathbf{\text{F}}\text{:}$ $\tau = \mathbf{\text{r}}\ \times \ \mathbf{\text{F}}.$

2.5 Equations of Lines and Planes in Space

• In three dimensions, the direction of a line is described by a direction vector. The vector equation of a line with direction vector $\mathbf{\text{v}} = \left\langle {a,b,c} \right\rangle$ passing through point $P = \left( {x_{0},y_{0},z_{0}} \right)$ is $\mathbf{\text{r}} = \mathbf{\text{r}}_{0} + t\mathbf{\text{v}},$ where $\mathbf{\text{r}}_{0} = \left\langle {x_{0},y_{0},z_{0}} \right\rangle$ is the position vector of point $P.$ This equation can be rewritten to form the parametric equations of the line: $x = x_{0} + ta,$ $y = y_{0} + tb,$ and $z = z_{0} + tc.$ The line can also be described with the symmetric equations $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}.$
• Let $L$ be a line in space passing through point $P$ with direction vector $\mathbf{\text{v}}.$ If $Q$ is any point not on $L,$ then the distance from $Q$ to $L$ is $d = \frac{\left\| {\overset{\rightarrow}{PQ}\ \times \ \mathbf{\text{v}}} \right\|}{\left\| \mathbf{\text{v}} \right\|}.$
• In three dimensions, two lines may be parallel but not equal, equal, intersecting, or skew.
• Given a point $P$ and vector $\mathbf{\text{n}},$ the set of all points $Q$ satisfying equation $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0$ forms a plane. Equation $\mathbf{\text{n}} \cdot \overset{\rightarrow}{PQ} = 0$ is known as the *vector equation of a plane.*
• The scalar equation of a plane containing point $P = \left( {x_{0},y_{0},z_{0}} \right)$ with normal vector $\mathbf{\text{n}} = \left\langle {a,b,c} \right\rangle$ is $a\left( {x - x_{0}} \right) + b\left( {y - y_{0}} \right) + c\left( {z - z_{0}} \right) = 0.$ This equation can be expressed as $ax + by + cz + d = 0,$ where $d = \text{−}ax_{0} - by_{0} - cz_{0}.$ This form of the equation is sometimes called the *general form of the equation of a plane*.
• Suppose a plane with normal vector n passes through point $Q.$ The distance $D$ from the plane to point $P$ not in the plane is given by

$$D = \left\| {\text{proj}_{\mathbf{\text{n}}}\overset{\rightarrow}{QP}} \right\| = \left| {\text{comp}_{\mathbf{\text{n}}}\overset{\rightarrow}{QP}} \right| = \frac{\left| {\overset{\rightarrow}{QP} \cdot \mathbf{\text{n}}} \right|}{\left\| \mathbf{\text{n}} \right\|}.$$

• The normal vectors of parallel planes are parallel. When two planes intersect, they form a line.
• The measure of the angle $\theta$ between two intersecting planes can be found using the equation: $\text{cos}\ \theta = \frac{\left| {\mathbf{\text{n}}_{1} \cdot \mathbf{\text{n}}_{2}} \right|}{\left\| \mathbf{\text{n}}_{1} \right\|\left\| \mathbf{\text{n}}_{2} \right\|},$ where $\mathbf{\text{n}}_{1}$ and $\mathbf{\text{n}}_{2}$ are normal vectors to the planes.
• The distance $D$ from point $\left( {x_{0},y_{0},z_{0}} \right)$ to plane $ax + by + cz + d = 0$ is given by

$$D = \frac{\left| {a\left( {x_{0} - x_{1}} \right) + b\left( {y_{0} - y_{1}} \right) + c\left( {z_{0} - z_{1}} \right)} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{\left| {ax_{0} + by_{0} + cz_{0} + d} \right|}{\sqrt{a^{2} + b^{2} + c^{2}}}.$$

2.6 Quadric Surfaces

• A set of lines parallel to a given line passing through a given curve is called a *cylinder*, or a *cylindrical surface*. The parallel lines are called *rulings*.
• The intersection of a three-dimensional surface and a plane is called a *trace*. To find the trace in the *xy*-, *yz*-, or *xz*-planes, set $z = 0,x = 0,\ \text{or}\ y = 0,$ respectively.
• Quadric surfaces are three-dimensional surfaces with traces composed of conic sections. Every quadric surface can be expressed with an equation of the form $Ax^{2} + By^{2} + Cz^{2} + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0.$
• To sketch the graph of a quadric surface, start by sketching the traces to understand the framework of the surface.
• Important quadric surfaces are summarized in Figure 2.87 and Figure 2.88.

2.7 Cylindrical and Spherical Coordinates

• In the cylindrical coordinate system, a point in space is represented by the ordered triple $\left( {r,\theta,z} \right),$ where $\left( {r,\theta} \right)$ represents the polar coordinates of the point’s projection in the *xy*-plane and $z$ represents the point’s projection onto the *z*-axis.
• To convert a point from cylindrical coordinates to Cartesian coordinates, use equations $x = r\ \text{cos}\ \theta,$ $y = r\ \text{sin}\ \theta,$ and $z = z.$
• To convert a point from Cartesian coordinates to cylindrical coordinates, use equations $r^{2} = x^{2} + y^{2},$ $\text{tan}\ \theta = \frac{y}{x},$ and $z = z.$
• In the spherical coordinate system, a point $P$ in space is represented by the ordered triple $\left( {\rho,\theta,\varphi} \right),$ where $\rho$ is the distance between $P$ and the origin $\left( {\rho \neq 0} \right),$ $\theta$ is the same angle used to describe the location in cylindrical coordinates, and $\varphi$ is the angle formed by the positive *z*-axis and line segment $\overset{—}{OP},$ where $O$ is the origin and $0 \leq \varphi \leq \pi.$
• To convert a point from spherical coordinates to Cartesian coordinates, use equations $x = \rho\ \text{sin}\ \varphi\ \text{cos}\ \theta,$ $y = \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta,$ and $z = \rho\ \text{cos}\ \varphi.$
• To convert a point from Cartesian coordinates to spherical coordinates, use equations $\rho^{2} = x^{2} + y^{2} + z^{2},$ $\text{tan}\ \theta = \frac{y}{x},$ and $\varphi = \text{arccos}\left( \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right).$
• To convert a point from spherical coordinates to cylindrical coordinates, use equations $r = \rho\ \text{sin}\ \varphi,$ $\theta = \theta,$ and $z = \rho\ \text{cos}\ \varphi.$
• To convert a point from cylindrical coordinates to spherical coordinates, use equations $\rho = \sqrt{r^{2} + z^{2}},$ $\theta = \theta,$ and $\varphi = \text{arccos}\left( \frac{z}{\sqrt{r^{2} + z^{2}}} \right).$

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/2-key-concepts

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Review Exercises

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/2-review-exercises

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Review Exercises

Calculus Volume 3Review Exercises

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Review Exercises

For the following exercises, determine whether the statement is *true or false*. Justify the answer with a proof or a counterexample.

423.

For vectors $\mathbf{\text{a}}$ and $\mathbf{\text{b}}$ and any given scalar $c,$ $c\left( {\mathbf{\text{a}} \cdot \mathbf{\text{b}}} \right) = \left( {c\mathbf{\text{a}}} \right) \cdot \mathbf{\text{b}}.$

424\.

For vectors $\mathbf{\text{a}}$ and $\mathbf{\text{b}}$ and any given scalar $c,$ $c\left( {\mathbf{\text{a}}\ \times \ \mathbf{\text{b}}} \right) = \left( {c\mathbf{\text{a}}} \right)\ \times \ \mathbf{\text{b}}.$

425.

The symmetric equation for the line of intersection between two planes $x + y + z = 2$ and $x + 2y - 4z = 5$ is given by $- \frac{x - 1}{6} = \frac{y - 1}{5} = z.$

426\.

If $\mathbf{\text{a}} \cdot \mathbf{\text{b}} = 0,$ then $\mathbf{\text{a}}$ is perpendicular to $\mathbf{\text{b}}.$

For the following exercises, use the given vectors to find the quantities.

427.

$\mathbf{\text{a}} = 9\mathbf{\text{i}} - 2\mathbf{\text{j}},\mathbf{\text{b}} = -3\mathbf{\text{i}} + \mathbf{\text{j}}$

1. $3\mathbf{\text{a}} + \mathbf{\text{b}}$

2. $\left\| \mathbf{\text{a}} \right\|$

3. $\left. \left. {\mathbf{\text{a}}\ \times \ } \right\|{\mathbf{\text{b}}\ \times \ }\mathbf{\text{c}} \right\|$

4. ${\mathbf{\text{b}}\ \cdot \ }\mathbf{\text{a}}$

428\.

$\mathbf{\text{a}} = 2\mathbf{\text{i}} + \mathbf{\text{j}} - 9\mathbf{\text{k}},\mathbf{\text{b}} = \text{−}\mathbf{\text{i}} + 2\mathbf{\text{k}},\mathbf{\text{c}} = 4\mathbf{\text{i}} - 2\mathbf{\text{j}} + \mathbf{\text{k}}$

1. $2\mathbf{\text{a}} - \mathbf{\text{b}}$

2. $\mathbf{\text{b}}\ \times \ \mathbf{\text{c}}$

3. $\mathbf{\text{b}}\ \times \ \left\| {\mathbf{\text{b}}\ \times \ \mathbf{\text{c}}} \right\|$

4. $\mathbf{\text{c}}\ \times \ \left\| {\mathbf{\text{b}}\ \times \ \mathbf{\text{a}}} \right\|$

5. $\text{proj}_{\mathbf{\text{a}}}\mathbf{\text{b}}$

429.

Find the values of $a$ such that vectors $\left\langle {2,4,a} \right\rangle$ and $\left\langle {0,-1,a} \right\rangle$ are orthogonal.

For the following exercises, find the unit vectors.

430\.

Find the unit vector that has the same direction as vector $\mathbf{\text{v}}$ that begins at $\left( {0,-3} \right)$ and ends at $\left( {4,10} \right).$

431.

Find the unit vector that has the same direction as vector $\mathbf{\text{v}}$ that begins at $\left( {1,4,10} \right)$ and ends at $\left( {3,0,4} \right).$

For the following exercises, find the area or volume of the given shapes.

432\.

The parallelogram spanned by vectors $\mathbf{\text{a}} = \left\langle {1,13} \right\rangle\ \text{and}\ \mathbf{\text{b}} = \left\langle {3,21} \right\rangle$

433.

The parallelepiped formed by $\mathbf{\text{a}} = \left\langle {1,4,1} \right\rangle\ \text{and}\ \mathbf{\text{b}} = \left\langle {3,6,2} \right\rangle,$ and $\mathbf{\text{c}} = \left\langle {-2,1,-5} \right\rangle$

For the following exercises, find the vector and parametric equations of the line with the given properties.

434\.

The line that passes through point $\left( {2,-3,7} \right)$ that is parallel to vector $\left\langle {1,3,-2} \right\rangle$

435.

The line that passes through points $\left( {1,3,5} \right)$ and $\left( {-2,6,-3} \right)$

For the following exercises, find an equation of the plane with the given properties.

436\.

The plane that passes through point $\left( {4,7,-1} \right)$ and has normal vector $\mathbf{\text{n}} = \left\langle {3,4,2} \right\rangle$

437.

The plane that passes through points $\left( {0,1,5} \right),\left( {2,-1,6} \right),\ \text{and}\ \left( {3,2,5} \right).$

For the following exercises, find the traces for the surfaces in planes $x = k,y = k,\ \text{and}\ z = k.$ Then, describe and draw the surfaces.

438\.

$9x^{2} + 4y^{2} - 16y + 36z^{2} = 20$

439.

$x^{2} = y^{2} + z^{2}$

For the following exercises, write the given equation in cylindrical coordinates and spherical coordinates.

440\.

$x^{2} + y^{2} + z^{2} = 144$

441.

$z = x^{2} + y^{2} - 1$

For the following exercises, convert the given equations from cylindrical or spherical coordinates to rectangular coordinates. Identify the given surface.

442\.

$\rho^{2}\left( {\text{sin}^{2}(\varphi) - \text{cos}^{2}(\varphi)} \right) = 1$

443.

$r^{2} - 2r\ \text{cos}(\theta) + z^{2} = 1$

For the following exercises, consider a small boat crossing a river.

444\.

If the boat velocity is $5$ km/h due north in still water and the water has a current of $2$ km/h due west (see the following figure), what is the velocity of the boat relative to shore? What is the angle $\theta$ that the boat is actually traveling?

445.

When the boat reaches the shore, two ropes are thrown to people to help pull the boat ashore. One rope is at an angle of $25\text{°}$ and the other is at $35\text{°}.$ If the boat must be pulled straight and at a force of $500\text{N},$ find the magnitude of force for each rope (see the following figure).

446\.

An airplane is flying in the direction of 52° east of north with a speed of 450 mph. A strong wind has a bearing 33° east of north with a speed of 50 mph. What is the resultant ground speed and bearing of the airplane?

447.

Calculate the work done by moving a particle from position $(1,2,0)$ to $(8,4,5)$ along a straight line with a force $\mathbf{\text{F}} = 2\mathbf{\text{i}} + 3\mathbf{\text{j}} - \mathbf{\text{k}}.$

The following problems consider your unsuccessful attempt to take the tire off your car using a wrench to loosen the bolts. Assume the wrench is $0.3$ m long and you are able to apply a 200-N force.

448\.

Because your tire is flat, you are only able to apply your force at a $60\text{°}$ angle. What is the torque at the center of the bolt? Assume this force is not enough to loosen the bolt.

449.

Someone lends you a tire jack and you are now able to apply a 200-N force at an $80\text{°}$ angle. Is your resulting torque going to be more or less? What is the new resulting torque at the center of the bolt? Assume this force is not enough to loosen the bolt.

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Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction

Citation information

• Use the information below to generate a citation. We recommend using a citation tool such as this one.
• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 3
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-3/pages/2-review-exercises

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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