Vectors in R^n及向量应用 ℝⁿ 中的向量

In order to satisfy $\vec{x} \cdot \vec{u}_{2} = 0$, the following equation must hold. $x_2 + 2x_3 = 0$

Both of these equations must be satisfied, so we have the following system of equations.

$$\begin{array}{c} x_1 - x_3 = 0 \\ x_2 + 2x_3 = 0 \end{array}$$

To solve, set up the augmented matrix.

$$\left[ \begin{array}{rrr|r} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \end{array} \right]$$

Using Gaussian Elimination, we find that $W^{\perp} = \mbox{span} \left\{ \left[ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right] \right\}$, and hence $\left\{ \left[ \begin{array}{r} 1 \\ -2 \\ 1 \end{array} \right] \right\}$ is a basis for $W^{\perp}$.

The following results summarize the important properties of the orthogonal projection.

定理 4：正交投影 Orthogonal Projection

Let $W$ be a subspace of $\mathbb{R}^n$, $Y$ be any point in $\mathbb{R}^n$, and let $Z$ be the point in $W$ closest to $Y$. Then,

1. The position vector $\vec{z}$ of the point $Z$ is given by $\vec{z} = \mathrm{proj}_{W}\left( \vec{y}\right)$

2. $\vec{z} \in W$ and $\vec{y} - \vec{z} \in W^{\perp}$

3. $\| Y - Z \| < \| Y - Z_1 \|$ for all $Z_1 \neq Z \in W$

Consider the following example of this concept.

示例 13：求离已知向量最近的向量 Find a Vector Closest to a Given Vector

Let $\vec{x}_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array}\right], \vec{x}_{2}=\left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array}\right], \vec{x}_{3}=\left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array}\right], \mbox{ and } \vec{v}=\left[\begin{array}{c} 4 \\ 3 \\ -2 \\ 5 \end{array}\right]$. We want to find the vector in $W =\mathrm{span}\{\vec{x}_{1}, \vec{x}_{2},\vec{x}_{3}\}$ closest to $\vec{y}$.

解 Solution

We will first use the Gram-Schmidt Process to construct the orthogonal basis, $B$, of $W$: $B=\left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array}\right], \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}\right], \left[\begin{array}{r} 1 \\ 2 \\ -1 \\ 0 \end{array}\right] \right\}.$

By Theorem 4, $\mathrm{proj}_{U}(\vec{v}) = \frac{2}{2} \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array}\right] + \frac{5}{1}\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}\right] + \frac{12}{6}\left[\begin{array}{r} 1 \\ 2 \\ -1 \\ 0 \end{array}\right] = \left[\begin{array}{r} 3 \\ 4 \\ -1 \\ 5 \end{array}\right]$ is the vector in $U$ closest to $\vec{y}$.

Consider the next example.

示例 14：把向量写成两个向量之和 Vector Written as a Sum of Two Vectors

Let $W$ be a subspace given by $W = \mbox{span} \left\{ \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right], \left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 2 \end{array} \right] \right\}$, and $Y = (1,2,3,4)$. Find the point $Z$ in $W$ closest to $Y$, and moreover write $\vec{y}$ as the sum of a vector in $W$ and a vector in $W^{\perp}$.

解 Solution

From Theorem 3 the point $Z$ in $W$ closest to $Y$ is given by $\vec{z} = \mathrm{proj}_{W}\left( \vec{y}\right)$.

Notice that since the above vectors already give an orthogonal basis for $W$, we have:

$$\begin{aligned} \vec{z} &= \mathrm{proj}_{W}\left( \vec{y}\right) \\ &= \left( \frac{\vec{y} \cdot \vec{w}_{1}}{ \vec{w}_{1}^2}\right) \vec{w}_{1} + \left( \frac{\vec{y} \cdot \vec{w}_{2}}{ \vec{w}_{2}^2}\right) \vec{w}_{2} \\ &= \left( \frac{4}{2} \right) \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] + \left( \frac{10}{5} \right) \left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 2 \end{array} \right] \\ &= \left[ \begin{array}{c} 2 \\ 2 \\ 2 \\ 4 \end{array} \right] \end{aligned}$$

Therefore the point in $W$ closest to $Y$ is $Z = (2,2,2,4)$. Now, we need to write $\vec{y}$ as the sum of a vector in $W$ and a vector in $W^{\perp}$. This can easily be done as follows: $\vec{y} = \vec{z} + (\vec{y} - \vec{z})$ since $\vec{z}$ is in $W$ and as we have seen $\vec{y} - \vec{z}$ is in $W^{\perp}$. The vector $\vec{y} - \vec{z}$ is given by $\vec{y} - \vec{z} = \left[ \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \end{array} \right] - \left[ \begin{array}{c} 2 \\ 2 \\ 2 \\ 4 \end{array} \right] = \left[ \begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array} \right]$ Therefore, we can write $\vec{y}$ as $\left[ \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 2 \\ 2 \\ 4 \end{array} \right] + \left[ \begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array} \right]$

Example 15: Point in a Plane Closest to a Given Point

Find the point $Z$ in the plane $3x+y-2z=0$ that is closest to the point $Y=(1,1,1)$.

解 Solution

The solution will proceed as follows.

1. Find a basis $X$ of the subspace $W$ of $\mathbb{R}^3$ defined by the equation $3x+y-2z=0$. 2. Orthogonalize the basis $X$ to get an orthogonal basis $B$ of $W$. 3. Find the projection on $W$ of the position vector of the point $Y$.

We now begin the solution.

1. $3x+y-2z=0$ is a system of one equation in three variables. Putting the augmented matrix in reduced row-echelon form: $\left[\begin{array}{rrr|r} 3 & 1 & -2 & 0 \end{array}\right] \rightarrow \left[\begin{array}{rrr|r} 1 & \frac{1}{3} & -\frac{2}{3} & 0 \end{array}\right]$ gives general solution $x=\frac{1}{3}s+\frac{2}{3}t$, $y=s$, $z=t$ for any $s,t\in\mathbb{R}$. Then $W=\mathrm{span} \left\{ \left[\begin{array}{r} -\frac{1}{3} \\ 1 \\ 0 \end{array}\right], \left[\begin{array}{r} \frac{2}{3} \\ 0 \\ 1 \end{array}\right]\right\}$. Let $X=\left\{ \left[\begin{array}{r} -1 \\ 3 \\ 0 \end{array}\right], \left[\begin{array}{r} 2 \\ 0 \\ 3 \end{array}\right]\right\}$. Then $X$ is linearly independent and $\mathrm{span}(X)=W$, so $X$ is a basis of $W$.

2. Use the Gram-Schmidt Process to get an orthogonal basis of $W$: $\vec{f}_{1}=\left[\begin{array}{r} -1 \\ 3 \\ 0\end{array}\right]\mbox{ and }\vec{f}_{2}=\left[\begin{array}{r}2 \\ 0 \\ 3\end{array}\right]-\frac{-2}{10}\left[\begin{array}{r} -1 \\ 3 \\ 0\end{array}\right]=\frac{1}{5}\left[\begin{array}{r}9 \\ 3 \\ 15\end{array}\right].$ Therefore $B=\left\{\left[\begin{array}{r}-1\\3\\0\end{array}\right], \left[\begin{array}{r}3\\1\\5 \end{array}\right]\right\}$ is an orthogonal basis of $W$.

3. To find the point $Z$ on $W$ closest to $Y=(1,1,1)$, compute $\begin{aligned} \mathrm{proj}_{W}\left[\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right] & = \frac{2}{10} \left[\begin{array}{r} -1 \\ 3 \\ 0 \end{array}\right] + \frac{9}{35}\left[\begin{array}{r} 3 \\ 1 \\ 5 \end{array}\right] \\ & = \frac{1}{7}\left[\begin{array}{r} 4 \\ 6 \\ 9 \end{array}\right].\end{aligned}$ Therefore, $Z=\left( \frac{4}{7}, \frac{6}{7}, \frac{9}{7}\right)$.

4.12 最小二乘逼近 Least Squares Approximation

It should not be surprising to hear that many problems do not have a perfect solution, and in these cases the objective is always to try to do the best possible. For example what does one do if there are no solutions to a system of linear equations $A\vec{x}=\vec{b}$? It turns out that what we do is find $\vec{x}$ such that $A\vec{x}$ is as close to $\vec{b}$ as possible. A very important technique that follows from orthogonal projections is that of the least square approximation, and allows us to do exactly that.

We begin with a lemma.

Recall that we can form the image of an $m \times n$ matrix $A$ by $\mathrm{im}\left( A\right) = \left\{ A\vec{x} : \vec{x} \in \mathbb{R}^n \right\}$. Rephrasing Theorem 4 using the subspace $W=\mathrm{im}\left( A\right)$ gives the equivalence of an orthogonality condition with a minimization condition. The following picture illustrates this orthogonality condition and geometric meaning of this theorem.

定理 5：极小元的存在性 Existence of Minimizers

Let $\vec{y}\in \mathbb{R}^{m}$ and let $A$ be an $m\times n$ matrix.

Choose $\vec{z}\in W= \mathrm{im}\left( A\right)$ given by $\vec{z} = \mathrm{proj}_{W}\left( \vec{y}\right)$, and let $\vec{x} \in \mathbb{R}^{n}$ such that $\vec{z}=A\vec{x}$.

Then

1. $\vec{y} - A\vec{x} \in W^{\perp}$

2. $\|\vec{y} - A\vec{x}\| < \|\vec{y} - \vec{u}\|$ for all $\vec{u} \neq \vec{z} \in W$

We note a simple but useful observation.

引理 1：转置与点积 Transpose and Dot Product

Let $A$ be an $m\times n$ matrix. Then $A\vec{x} \cdot \vec{y} = \vec{x}\cdot A^T\vec{y}$

证明

Proof. This follows from the definitions: $A\vec{x} \cdot \vec{y}=\sum_{i,j}a_{ij}x_{j} y_{i} =\sum_{i,j}x_{j} a_{ji} y_{i}= \vec{x} \cdot A^T\vec{y}$

The next corollary gives the technique of least squares.

推论 1：最小二乘与正规方程 Least Squares and Normal Equation

A specific value of $\vec{x}$ which solves the problem of Theorem 5 is obtained by solving the equation $A^TA\vec{x}=A^T\vec{y}$ Furthermore, there always exists a solution to this system of equations.

证明

Proof. For $\vec{x}$ the minimizer of Theorem 5, $\left( \vec{y}-A\vec{x}\right) \cdot A \vec{u} =0$ for all $\vec{u} \in \mathbb{R}^{n}$ and from Lemma 1, this is the same as saying $A^T\left( \vec{y}-A\vec{x}\right) \cdot \vec{u}=0$ for all $u \in \mathbb{R}^{n}.$ This implies $A^T\vec{y}-A^TA\vec{x}=\vec{0}.$ Therefore, there is a solution to the equation of this corollary, and it solves the minimization problem of Theorem 5.

Note that $\vec{x}$ might not be unique but $A\vec{x}$, the closest point of $A\left(\mathbb{R}^{n}\right)$ to $\vec{y}$, is unique as was shown in the above argument.

Consider the following example.

示例 16：方程组的最小二乘解 Least Squares Solution to a System

Find a least squares solution to the system $\left[ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right]$

解 Solution

First, consider whether there exists a real solution. To do so, set up the augmented matrix given by $\left[ \begin{array}{rr|r} 2 & 1 & 2 \\ -1 & 3 & 1 \\ 4 & 5 & 1 \end{array} \right]$ The reduced row-echelon form of this augmented matrix is $\left[ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$

It follows that there is no real solution to this system. Therefore we wish to find the least squares solution. The normal equations are $\begin{aligned} A^T A \vec{x} &= A^T \vec{y} \\ \left[ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right] \left[ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] &=\left[ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right] \left[ \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right]\end{aligned}$ and so we need to solve the system $\left[ \begin{array}{rr} 21 & 19 \\ 19 & 35 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{r} 7 \\ 10 \end{array} \right]$ This is a familiar exercise and the solution is $\left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{c} \frac{5}{34} \\ \frac{7}{34} \end{array} \right]$

Consider another example.

示例 17：方程组的最小二乘解 Least Squares Solution to a System

Find a least squares solution to the system $\left[ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{c} 3 \\ 2 \\ 9 \end{array} \right]$

解 Solution

First, consider whether there exists a real solution. To do so, set up the augmented matrix given by $\left[ \begin{array}{rr|r} 2 & 1 & 3 \\ -1 & 3 & 2 \\ 4 & 5 & 9 \end{array} \right]$ The reduced row-echelon form of this augmented matrix is $\left[ \begin{array}{rr|r} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right]$

It follows that the system has a solution given by $x=y=1$. However we can also use the normal equations and find the least squares solution. $\left[ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right] \left[ \begin{array}{rr} 2 & 1 \\ -1 & 3 \\ 4 & 5 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{rrr} 2 & -1 & 4 \\ 1 & 3 & 5 \end{array} \right] \left[ \begin{array}{r} 3 \\ 2 \\ 9 \end{array} \right]$ Then $\left[ \begin{array}{rr} 21 & 19 \\ 19 & 35 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{c} 40 \\ 54 \end{array} \right]$

The least squares solution is $\left[ \begin{array}{c} x \\ y \end{array} \right] =\left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$ which is the same as the solution found above.

An important application of Corollary 1 is the problem of finding the least squares regression line in statistics. Suppose you are given points in the $xy$ plane $\left\{ \left( x_{1},y_{1}\right), \left( x_{2},y_{2}\right), \cdots, \left( x_{n},y_{n}\right) \right\}$ and you would like to find constants $m$ and $b$ such that the line $\vec{y}=m\vec{x}+b$ goes through all these points. Of course this will be impossible in general. Therefore, we try to find $m,b$ such that the line will be as close as possible. The desired system is

$$\left[ \begin{array}{c} y_{1} \\ \vdots \\ y_{n} \end{array} \right] =\left[ \begin{array}{cc} x_{1} & 1 \\ \vdots & \vdots \\ x_{n} & 1 \end{array} \right] \left[ \begin{array}{c} m \\ b \end{array} \right]$$

which is of the form $\vec{y}=A\vec{x}$. It is desired to choose $m$ and $b$ to make

$$\left\| A\left[ \begin{array}{c} m \\ b \end{array} \right] -\left[ \begin{array}{c} y_{1} \\ \vdots \\ y_{n} \end{array} \right] \right\|^{2}$$

as small as possible. According to Theorem 5 and Corollary 1, the best values for $m$ and $b$ occur as the solution to

$$A^{T}A\left[ \begin{array}{c} m \\ b \end{array} \right] =A^{T}\left[ \begin{array}{c} y_{1} \\ \vdots \\ y_{n} \end{array} \right] ,\\ \mbox{where}\\ A=\left[ \begin{array}{cc} x_{1} & 1 \\ \vdots & \vdots \\ x_{n} & 1 \end{array} \right]$$

Thus, computing $A^{T}A,$

$$\left[ \begin{array}{cc} \sum_{i=1}^{n}x_{i}^{2} & \sum_{i=1}^{n}x_{i} \\ \sum_{i=1}^{n}x_{i} & n \end{array} \right] \left[ \begin{array}{c} m \\ b \end{array} \right] =\left[ \begin{array}{c} \sum_{i=1}^{n}x_{i}y_{i} \\ \sum_{i=1}^{n}y_{i} \end{array} \right]$$

Solving this system of equations for $m$ and $b$ (using Cramer's rule for example) yields:

$$m= \frac{-\left( \sum_{i=1}^{n}x_{i}\right) \left( \sum_{i=1}^{n}y_{i}\right) +\left( \sum_{i=1}^{n}x_{i}y_{i}\right) n}{\left( \sum_{i=1}^{n}x_{i}^{2}\right) n-\left( \sum_{i=1}^{n}x_{i}\right)^{2}} ,\quad b=\frac{-\left( \sum_{i=1}^{n}x_{i}\right) \sum_{i=1}^{n}x_{i}y_{i}+\left( \sum_{i=1}^{n}y_{i}\right) \sum_{i=1}^{n}x_{i}^{2}}{\left( \sum_{i=1}^{n}x_{i}^{2}\right) n-\left( \sum_{i=1}^{n}x_{i}\right)^{2}}$$

Consider the following example.

示例 18：最小二乘回归 Least Squares Regression

Find the least squares regression line $\vec{y}=m\vec{x}+b$ for the following set of data points: $\left\{ (0,1), (1,2), (2,2), (3,4), (4,5) \right\}$

解 Solution

In this case we have $n=5$ data points and we obtain: $\begin{array}{ll} \sum_{i=1}^{5}x_{i} = 10 & \sum_{i=1}^{5}y_{i} = 14 \\ \sum_{i=1}^{5}x_{i}y_{i} = 38 & \sum_{i=1}^{5}x_{i}^{2} = 30 \end{array}$ and hence $\begin{aligned} m &= \frac{- 10 \cdot 14 + 5 \cdot 38}{5\cdot 30-10^2} = 1.00 \\ b &= \frac{- 10 \cdot 38 + 14\cdot 30}{5\cdot 30-10^2} = 0.80 \end{aligned}$

The least squares regression line for the set of data points is: $\vec{y} = \vec{x}+.8$

One could use this line to approximate other values for the data. For example for $x=6$ one could use $y(6)=6+.8=6.8$ as an approximate value for the data.

The following diagram shows the data points and the corresponding regression line.

One could clearly do a least squares fit for curves of the form $y=ax^{2}+bx+c$ in the same way. In this case you want to solve as well as possible for $a,b,$ and $c$ the system $\left[ \begin{array}{ccc} x_{1}^{2} & x_{1} & 1 \\ \vdots & \vdots & \vdots \\ x_{n}^{2} & x_{n} & 1 \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] =\left[ \begin{array}{c} y_{1} \\ \vdots \\ y_{n} \end{array} \right]$ and one would use the same technique as above. Many other similar problems are important, including many in higher dimensions and they are all solved the same way.

4.12 应用 Applications

学习目标 Outcomes

1. Apply the concepts of vectors in $\mathbb{R}^n$ to the applications of physics and work.

向量与物理 Vectors and Physics

Suppose you push on something. Then, your push is made up of two components, how hard you push and the direction you push. This illustrates the concept of force.

定义 1：力 Force

Force is a vector. The magnitude of this vector is a measure of how hard it is pushing. It is measured in units such as Newtons or pounds or tons. The direction of this vector is the direction in which the push is taking place.

Vectors are used to model force and other physical vectors like velocity. As with all vectors, a vector modeling force has two essential ingredients, its magnitude and its direction.

Recall the special vectors which point along the coordinate axes. These are given by $\vec{e}_{i} = \left[ 0 \cdots 0 \ 1 \ 0 \cdots 0 \right]^T$ where the $1$ is in the $i^{th}$ slot and there are zeros in all the other spaces. The direction of $\vec{e}_{i}$ is referred to as the $i^{th}$ direction.

Consider the following picture which illustrates the case of $\mathbb{R}^{3}.$ Recall that in $\mathbb{R}^3$, we may refer to these vectors as $\vec{i}, \vec{j},$ and $\vec{k}$.

Given a vector $\vec{u}=\left[ u_{1} \cdots u_{n}\right]^T ,$ it follows that $\vec{u}=u_{1}\vec{e}_{1}+\cdots +u_{n}\vec{e}_{n}= \sum_{k=1}^{n}u_{i}\vec{e}_{i}$

What does addition of vectors mean physically? Suppose two forces are applied to some object. Each of these would be represented by a force vector and the two forces acting together would yield an overall force acting on the object which would also be a force vector known as the resultant. Suppose the two vectors are $\vec{u}=\sum_{k=1}^{n}u_{i}\vec{e}_{i}$ and $\vec{v}=\sum_{k=1}^{n}v_{i}\vec{e}_{i}$. Then the vector $\vec{u}$ involves a component in the $i^{th}$ direction given by $u_{i}\vec{e}_{i}$, while the component in the $i^{th}$ direction of $\vec{v}$ is $v_{i}\vec{e}_{i}.$ Then the vector $\vec{u} + \vec{v}$ should have a component in the $i^{th}$ direction equal to $\left( u_{i}+v_{i}\right) \vec{e}_{i}.$ This is exactly what is obtained when the vectors, $\vec{u}$ and $\vec{v}$ are added. $\begin{aligned} \vec{u}+\vec{v}& =\left[ u_{1}+v_{1} \cdots u_{n}+v_{n}\right]^T \\ & =\sum_{i=1}^{n}\left( u_{i}+v_{i}\right) \vec{e}_{i}\end{aligned}$

Thus the addition of vectors according to the rules of addition in $\mathbb{R}^{n}$ which were presented earlier, yields the appropriate vector which duplicates the cumulative effect of all the vectors in the sum.

Consider now some examples of vector addition.

示例 1：三个力的合力 The Resultant of Three Forces

There are three ropes attached to a car and three people pull on these ropes. The first exerts a force of $\vec{F}_{1} = 2\vec{i} + 3\vec{j} -2 \vec{k}$ Newtons, the second exerts a force of $\vec{F}_{2} = 3\vec{i}+5\vec{j}+\vec{k}$ Newtons and the third exerts a force of $5\vec{i}-\vec{j}+2\vec{k}$ Newtons. Find the total force in the direction of $\vec{i}$.

解 Solution

To find the total force, we add the vectors as described above. This is given by $\begin{aligned} &(2\vec{i}+3\vec{j}-2\vec{k}) + (3\vec{i}+5\vec{j}+\vec{k}) + (5\vec{i}-\vec{j}+2\vec{k}) \\ &= (2 + 3 + 5) \vec{i} + (3 + 5 - 1) \vec{j} + (-2+1+2) \vec{k} \\ &= 10 \vec{i} + 7 \vec{j} + \vec{k}\end{aligned}$ Hence, the total force is $10\vec{i}+7\vec{j}+\vec{k}$ Newtons. Therefore, the force in the $\vec{i}$ direction is $10$ Newtons.

Consider another example.

示例 2：由几何描述求向量 Finding a Vector from Geometric Description

An airplane flies North East at $100$ miles per hour. Write this as a vector.

解 Solution

A picture of this situation follows.

Therefore, we need to find the vector $\vec{u}$ which has length $100$ and direction as shown in this diagram. We can consider the vector $\vec{u}$ as the hypotenuse of a right triangle having equal sides, since the direction of $\vec{u}$ corresponds with the $45 ^{\circ}$ line. The sides, corresponding to the $\vec{i}$ and $\vec{j}$ directions, should be each of length $100/\sqrt{2}.$ Therefore, the vector is given by $\vec{u} = \frac{100}{\sqrt{2}} \vec{i}+ \frac{100}{\sqrt{2 }}\vec{j} = \left[ \begin{array}{rr} \frac{100}{\sqrt{2}} & \frac{100}{\sqrt{2}} \end{array} \right]^T$

This example also motivates the concept of velocity, defined below.

定义 2：速率与速度 Speed and Velocity

The speed of an object is a measure of how fast it is going. It is measured in units of length per unit time. For example, miles per hour, kilometers per minute, feet per second. The velocity is a vector having the speed as the magnitude but also specifying the direction.

Thus the velocity vector in the above example is $\frac{100}{\sqrt{2}}\vec{i}+ \frac{100}{\sqrt{2}}\vec{j}$, while the speed is $100$ miles per hour.

Consider the following example.

The velocity of an airplane is $100\vec{i}+\vec{j}+\vec{k}$ measured in kilometers per hour and at a certain instant of time its position is $\left( 1,2,1\right) .$ Find the position of this airplane one minute later.

解 Solution

Here imagine a Cartesian coordinate system in which the third component is altitude and the first and second components are measured on a line from West to East and a line from South to North.

Consider the vector $\left[ \begin{array}{rrr} 1 & 2 & 1 \end{array} \right]^T ,$ which is the initial position vector of the airplane. As the plane moves, the position vector changes according to the velocity vector. After one minute (considered as $\frac{1}{60}$ of an hour) the airplane has moved in the $\vec{i}$ direction a distance of $100\times \frac{1}{60}= \frac{5}{3}$ kilometer. In the $\vec{j}$ direction it has moved $\frac{1}{60}$ kilometer during this same time, while it moves $\frac{1}{60}$ kilometer in the $\vec{k}$ direction. Therefore, the new displacement vector for the airplane is $\left[ \begin{array}{rrr} 1 & 2 & 1 \end{array} \right]^T + \left[ \begin{array}{rrr} \frac{5}{3} & \frac{1}{60} & \frac{1}{60} \end{array} \right]^T =\left[ \begin{array}{rrr} \frac{8}{3} & \frac{121}{60} & \frac{121}{60} \end{array} \right]^T$

Now consider an example which involves combining two velocities.

示例 4：两个速度之和 Sum of Two Velocities

A certain river is one half kilometer wide with a current flowing at $4$ kilometers per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of $3$ kilometers per hour. How far down the river does he find himself when he has swam across? How far does he end up swimming?

解 Solution

Consider the following picture which demonstrates the above scenario.

First we want to know the total time of the swim across the river. The velocity in the direction across the river is $3$ kilometers per hour, and the river is $\frac{1}{2}$ kilometer wide. It follows the trip takes $1/6$ hour or $10$ minutes.

Now, we can compute how far downstream he will end up. Since the river runs at a rate of $4$ kilometers per hour, and the trip takes $1/6$ hour, the distance traveled downstream is given by $4 \left(\frac{1}{6}\right) = \frac{2}{3}$ kilometers.

The distance traveled by the swimmer is given by the hypotenuse of a right triangle. The two arms of the triangle are given by the distance across the river, $\frac{1}{2}$ km, and the distance traveled downstream, $\frac{2}{3}$ km. Then, using the Pythagorean Theorem, we can calculate the total distance $d$ traveled. $d = \sqrt{ \left(\frac{2}{3} \right)^2 + \left( \frac{1}{2} \right) ^2 } = \frac{5}{6} \mbox{ km}$

Therefore, the swimmer travels a total distance of $\frac{5}{6}$ kilometers.

功 Work

The mathematical concept of work is an application of vectors in $\mathbb{R}^n$. The physical concept of work differs from the notion of work employed in ordinary conversation. For example, suppose you were to slide a $150$ pound weight off a table which is three feet high and shuffle along the floor for $50$ yards, keeping the height always three feet and then deposit this weight on another three foot high table. The physical concept of work would indicate that the force exerted by your arms did no work during this project. The reason for this definition is that even though your arms exerted considerable force on the weight, the direction of motion was at right angles to the force they exerted. The only part of a force which does work in the sense of physics is the component of the force in the direction of motion.

Work is defined to be the magnitude of the component of this force times the distance over which it acts, when the component of force points in the direction of motion. In the case where the force points in exactly the opposite direction of motion work is given by $\left( -1\right)$ times the magnitude of this component times the distance. Thus the work done by a force on an object as the object moves from one point to another is a measure of the extent to which the force contributes to the motion. This is illustrated in the following picture in the case where the given force contributes to the motion.

Recall that for any vector $\vec{u}$ in $\mathbb{R}^n$, we can write $\vec{u}$ as a sum of two vectors, as in $\vec{u} = \vec{u}_{\|} + \vec{u}_{\perp}$ For any force $\vec{F}$, we can write this force as the sum of a vector in the direction of the motion and a vector perpendicular to the motion. In other words, $\vec{F} = \vec{F}_{\|} + \vec{F}_{\bot}$

In the above picture the force, $\vec{F}$ is applied to an object which moves on the straight line from $P$ to $Q.$ There are two vectors shown, $\vec{F}_{\|}$ and $\vec{F}_{\bot }$ and the picture is intended to indicate that when you add these two vectors you get $\vec{F}$. In other words, $\vec{F} = \vec{F}_{\|} + \vec{F}_{\bot}$. Notice that $\vec{F}_{\|}$ acts in the direction of motion and $\vec{F}_{\bot }$ acts perpendicular to the direction of motion. Only $\vec{F}_{\|}$ contributes to the work done by $\vec{F}$ on the object as it moves from $P$ to $Q$. $\vec{F}_{\|}$ is called the component of the force in the direction of motion. From trigonometry, you see the magnitude of $\vec{F}_{\|}$ should equal $\|\vec{F}\| \left\| \cos \theta \right\| .$ Thus, since $\vec{F}_{\|}$ points in the direction of the vector from $P$ to $Q,$ the total work done should equal $\|\vec{F}\| \|\vec{PQ}\| \cos \theta = \|\vec{F}\| \|\vec{q}-\vec{p}\| \cos \theta$

Now, suppose the included angle had been obtuse. Then the work done by the force $\vec{F}$ on the object would have been negative because $\vec{F}_{\|}$ would point in $-1$ times the direction of the motion. In this case, $\cos \theta$ would also be negative and so it is still the case that the work done would be given by the above formula. Thus from the geometric description of the dot product given above, the work equals $\|\vec{F}\| \|\vec{q}-\vec{p}\| \cos \theta =\vec{F}\bullet \left( \vec{q}-\vec{p}\right)$ This explains the following definition.

定义 3：力对物体所做的功 Work Done on an Object by a Force

Let $\vec{F}$ be a force acting on an object which moves from the point $P$ to the point $Q$, which have position vectors given by $\vec{p}$ and $\vec{q}$ respectively. Then the work done on the object by the given force equals $\vec{F}\bullet \left( \vec{q}-\vec{p}\right) .$

Consider the following example.

示例 4：求功 Finding Work

Let $\vec{F}= \left[ \begin{array}{rrr} 2 & 7 & -3 \end{array} \right]^T$ Newtons. Find the work done by this force in moving from the point $\left( 1,2,3\right)$ to the point $\left( -9,-3,4\right)$ along the straight line segment joining these points where distances are measured in meters.

解 Solution

First, compute the vector $\vec{q} - \vec{p}$, given by $\left[ \begin{array}{rrr} -9 & -3 & 4 \end{array} \right]^T - \left[ \begin{array}{rrr} 1 & 2 & 3 \end{array} \right]^T = \left[ \begin{array}{rrr} -10 & -5 & 1 \end{array} \right]^T$

According to Definition 3 the work done is $\begin{aligned} \left[ \begin{array}{rrr} 2 & 7 & 3 \end{array} \right]^T \bullet \left[ \begin{array}{rrr} -10 & -5 & 1 \end{array} \right]^T & =-20+\left( -35\right) +\left( -3\right) \\ & =-58 \text{ Newton meters}\end{aligned}$

Note that if the force had been given in pounds and the distance had been given in feet, the units on the work would have been foot pounds. In general, work has units equal to units of a force times units of a length. Recall that $1$ Newton meter is equal to $1$ Joule. Also notice that the work done by the force can be negative as in the above example.

习题 Exercises

习题 1 Exercise 1

Find $-3\left[\begin{array}{c}5\\-1\\2\\-3\end{array}\right]+5\left[\begin{array}{c}-8\\2\\-3\\6\end{array}\right]$.

答案 Answer：$\left[\begin{array}{c}-55\\13\\-21\\39\end{array}\right]$

习题 2 Exercise 2

Find $-7\left[\begin{array}{c}6\\0\\4\\-1\end{array}\right]+6\left[\begin{array}{c}-13\\-1\\1\\6\end{array}\right]$.

习题 3 Exercise 3

Decide whether $\vec{v}=\left[\begin{array}{c}4\\4\\-3\end{array}\right]$ is a linear combination of the vectors $\vec{u}_{1}=\left[\begin{array}{c}3\\1\\-1\end{array}\right]\quad\text{and}\quad\vec{u}_{2}=\left[\begin{array}{c}2\\-2\\1\end{array}\right].$

答案 Answer：$\left[\begin{array}{c}4\\4\\-3\end{array}\right]=2\left[\begin{array}{c}3\\1\\-1\end{array}\right]-\left[\begin{array}{c}2\\-2\\1\end{array}\right]$

习题 4 Exercise 4

Decide whether $\vec{v}=\left[\begin{array}{c}4\\4\\4\end{array}\right]$ is a linear combination of the vectors $\vec{u}_{1}=\left[\begin{array}{c}3\\1\\-1\end{array}\right]\quad\text{and}\quad\vec{u}_{2}=\left[\begin{array}{c}2\\-2\\1\end{array}\right].$

答案 Answer：The system $\left[\begin{array}{c}4\\4\\4\end{array}\right]=a_1\left[\begin{array}{c}3\\1\\-1\end{array}\right]+a_2\left[\begin{array}{c}2\\-2\\1\end{array}\right]$ has no solution.

习题 5 Exercise 5

Find the vector equation for the line through $(-7, 6, 0)$ and $(-1, 1, 4)$. Then, find the parametric equations for this line.

习题 6 Exercise 6

Find parametric equations for the line through the point $(7, 7, 1)$ with a direction vector $\vec{d}=\left[\begin{array}{c}1\\6\\2\end{array}\right]$.

习题 7 Exercise 7

Parametric equations of the line are $\begin{aligned}x&=t+2 \\ y&=6-3t \\ x&=-t=6\end{aligned}$ Find a direction vector for the line and a point on the line.

习题 8 Exercise 8

Find the vector equation for the line through the two points $(-5, 5, 1),$ $(2, 2, 4)$. Then, find the parametric equations.

习题 9 Exercise 9

The equation of a line in two dimensions is written as $y = x-5$. Find parametric equations for this line.

习题 10 Exercise 10

Find parametric equations for the line through $(6, 5,-2)$ and $(5, 1, 2)$.

习题 11 Exercise 11

Find the vector equation and parametric equations for the line through the point $(-7, 10,-6)$ with a direction vector $\vec{d}=\left[\begin{array}{c}1\\1\\3\end{array}\right]$.

习题 12 Exercise 12

Parametric equations of the line are $\begin{aligned}x&=2t+2 \\ y&=5-4t \\ z&=-t-3\end{aligned}$ Find a direction vector for the line and a point on the line, and write the vector equation of the line.

习题 13 Exercise 13

Find the vector equation and parametric equations for the line through the two points $(4, 10, 0),$ $(1,-5,-6)$.

习题 14 Exercise 14

Find the point on the line segment from $P = (-4, 7, 5)$ to $Q = (2,-2,-3)$ which is $\frac{1}{7}$ of the way from $P$ to $Q$.

习题 15 Exercise 15

Suppose a triangle in $\mathbb{R}^n$ has vertices at $P_1,$ $P_2,$ and $P_3$. Consider the lines which are drawn from a vertex to the mid point of the opposite side. Show these three lines intersect in a point and find the coordinates of this point.

习题 16 Exercise 16

Find $\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]\bullet\left[\begin{array}{c}2\\0\\1\\3\end{array}\right]$.

答案 Answer：$\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]\bullet\left[\begin{array}{c}2\\0\\1\\3\end{array}\right]=17$

习题 17 Exercise 17

Use the formula given in Proposition 4.7.2 to verify the Cauchy Schwarz inequality and to show that equality occurs if and only if one of the vectors is a scalar multiple of the other.

答案 Answer：This formula says that $|\vec{u}\bullet\vec{v}| = \|\vec{u}\|\|\vec{v}\|\cos\theta$ where $\theta$ is the included angle between the two vectors. Thus $|\vec{u}\bullet\vec{v}|=\|\vec{u}\|\|\vec{v}\||\cos\theta|\leq \|\vec{u}\|\|\vec{v}\|$ and equality holds if and only if $\theta = 0$ or $\pi$. This means that the two vectors either point in the same direction or opposite directions. Hence one is a multiple of the other.

习题 18 Exercise 18

For $\vec{u}$, $\vec{v}$ vectors in $\mathbb{R}^3$, define the product, $\vec{u}\ast\vec{v} = u_1v_1 +2u_2v_2 +3u_3v_3$. Show the axioms for a dot product all hold for this product. Prove $|\vec{u}\ast\vec{v}|\leq (\vec{u}\ast\vec{u})^{1/2}(\vec{v}\ast\vec{v})^{1/2}$

答案 Answer：This follows from the Cauchy Schwarz inequality and the proof of Theorem 4.7.1 which only used the properties of the dot product. Since this new product has the same properties the Cauchy Schwarz inequality holds for it as well.

习题 19 Exercise 19

Let $\vec{a}$, $\vec{b}$ be vectors. Show that $\left(\vec{a}\bullet\vec{b}\right)=\frac{1}{4}\left(\|\vec{a}+\vec{b}\|^2-\|\vec{a}-\vec{b}\|^2\right).$

习题 20 Exercise 20

Using the axioms of the dot product, prove the parallelogram identity: $\|\vec{a}+\vec{b}\|^2+\|\vec{a}-\vec{b}\|^2=2\|\vec{a}\|^2+2\|\vec{b}\|^2$

习题 21 Exercise 21

Let $A$ be a real $m\times n$ matrix and let $\vec{u} \in \mathbb{R}^n$ and $\vec{v} \in \mathbb{R}^m$. Show $A\vec{u}\bullet\vec{v} =\vec{u}\bullet A^T\vec{v}$. Hint: Use the definition of matrix multiplication to do this.

答案 Answer：$A\vec{x}\bullet\vec{y}=\sum_k(A\vec{x})_ky_k=\sum_k\sum_iA_{ki}x_iy_k=\sum_i\sum_kA^T_{ik}x_iy_k=\vec{x}\bullet A^T\vec{y}$

习题 22 Exercise 22

Use the result of Problem 21 to verify directly that $(AB)^T = B^TA^T$ without making any reference to subscripts.

答案 Answer：$\begin{aligned}AB\vec{x}\bullet\vec{y}&=B\vec{x}\bullet A^T\vec{y} \\ &=\vec{x}\bullet B^TA^T\vec{y} \\ &=\vec{x}\bullet (AB)^T\vec{y}\end{aligned}$ Since this is true for all $\vec{x}$, it follows that, in particular, it holds for $\vec{x}=B^TA^T\vec{y}-(AB)^T\vec{y}$ and so from the axioms of the dot product, $\left(B^TA^T\vec{y}-(AB)^T\vec{y}\right)\bullet\left(B^TA^T\vec{y}-(AB)^T\vec{y}\right)=0$ and so $B^TA^T\vec{y}-(AB)^T\vec{y}=\vec{0}$. However, this is true for all $\vec{y}$ and so $B^TA^T-(AB)^T=0$.

习题 23 Exercise 23

Find the angle between the vectors $\vec{u}=\left[\begin{array}{r}3\\-1\\-1\end{array}\right],$ $\vec{v}=\left[\begin{array}{c}1\\4\\2\end{array}\right]$

答案 Answer：$\frac{\left[\begin{array}{ccc}3&-1&-1\end{array}\right]^T\bullet\left[\begin{array}{ccc}1&4&2\end{array}\right]^T}{\sqrt{9+1+1}\sqrt{1+16+4}}=\frac{-3}{\sqrt{11}\sqrt{21}}=-0.19739=\cos\theta$ Therefore we need to solve $-0.19739=\cos\theta$ Thus $\theta=1.7695$ radians.

习题 24 Exercise 24

Find the angle between the vectors $\vec{u}=\left[\begin{array}{r}1\\-2\\1\end{array}\right],$ $\vec{v}=\left[\begin{array}{r}1\\2\\-7\end{array}\right]$

答案 Answer：$\frac{-10}{\sqrt{1+4+1}\sqrt{1+4+49}}=-0.55555=\cos\theta$ Therefore we need to solve $-0.55555 = \cos \theta$, which gives $\theta = 2.0313$ radians.

习题 25 Exercise 25

Find $\text{proj}_{\vec{v}}(\vec{w})$ where $\vec{w}=\left[\begin{array}{r}1\\0\\-2\end{array}\right]$ and $\vec{v}=\left[\begin{array}{c}1\\2\\3\end{array}\right]$.

答案 Answer：$\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{-5}{14}\left[\begin{array}{c}1\\2\\3\end{array}\right]=\left[\begin{array}{r}-\frac{5}{14}\\-\frac{5}{7}\\-\frac{15}{14}\end{array}\right]$

习题 26 Exercise 26

Find $\text{proj}_{\vec{v}}(\vec{w})$ where $\vec{w}=\left[\begin{array}{r}1\\2\\-2\end{array}\right]$ and $\vec{v}=\left[\begin{array}{c}1\\0\\3\end{array}\right]$.

答案 Answer：$\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{-5}{10}\left[\begin{array}{c}1\\0\\3\end{array}\right]=\left[\begin{array}{r}-\frac{1}{2}\\0\\-\frac{3}{2}\end{array}\right]$

习题 27 Exercise 27

Find $\text{proj}_{\vec{v}}(\vec{w})$ where $\vec{w}=\left[\begin{array}{r}1\\2\\-2\\1\end{array}\right]$ and $\vec{v}=\left[\begin{array}{c}1\\2\\3\\0\end{array}\right]$.

答案 Answer：$\frac{\vec{u}\bullet\vec{v}}{\vec{u}\bullet\vec{u}}\vec{u}=\frac{\left[\begin{array}{cccc}1&2&-2&1\end{array}\right]^T\bullet\left[\begin{array}{cccc}1&2&3&0\end{array}\right]^T}{1+4+9}\left[\begin{array}{c}1\\2\\3\\0\end{array}\right]=\left[\begin{array}{r}-\frac{1}{14}\\-\frac{1}{7}\\-\frac{3}{14}\\0\end{array}\right]$

习题 28 Exercise 28

Let $P = (1, 2, 3)$ be a point in $\mathbb{R}^3$. Let $L$ be the line through the point $P_0 = (1, 4, 5)$ with direction vector $\vec{d} =\left[\begin{array}{r}1\\-1\\1\end{array}\right]$. Find the shortest distance from $P$ to $L$, and find the point $Q$ on $L$ that is closest to $P$.

习题 29 Exercise 29

Let $P = (0, 2, 1)$ be a point in $\mathbb{R}^3$. Let $L$ be the line through the point $P_0 = (1, 1, 1)$ with direction vector $\vec{d} =\left[\begin{array}{c}3\\0\\1\end{array}\right]$. Find the shortest distance from $P$ to $L$, and find the point $Q$ on $L$ that is closest to $P$.

习题 30 Exercise 30

Does it make sense to speak of $\text{proj}_{\vec{0}} (\vec{w})$?

答案 Answer：No, it does not. The $0$ vector has no direction. The formula for $\text{proj}_{\vec{0}} (\vec{w})$ doesn't make sense either.

习题 31 Exercise 31

Prove the Cauchy Schwarz inequality in $\mathbb{R}^n$ as follows. For $\vec{u},$ $\vec{v}$ vectors, consider $(\vec{w}-\text{proj}_{\vec{v}}\vec{w})\bullet (\vec{w}-\text{proj}_{\vec{v}}\vec{w})\geq 0$ Simplify using the axioms of the dot product and then put in the formula for the projection. Notice that this expression equals $0$ and you get equality in the Cauchy Schwarz inequality if and only if $\vec{w} = \text{proj}_{\vec{v}}\vec{w}$. What is the geometric meaning of $\vec{w}= \text{proj}_{\vec{v}}\vec{w}$?

答案 Answer：$\left(\vec{u}-\frac{\vec{u}\bullet\vec{v}}{\|\vec{v}\|^2}\vec{v}\right)\bullet\left(\vec{u}-\frac{\vec{u}\bullet\vec{v}}{\|\vec{v}\|^2}\vec{v}\right)=\|\vec{u}\|^2-2(\vec{u}\bullet\vec{v})^2\frac{1}{\|\vec{v}\|^2}+(\vec{u}\bullet\vec{v})^2\frac{1}{\|\vec{v}\|^2}\geq 0$ And so $\|\vec{u}\|^2\|\vec{v}\|^2\geq (\vec{u}\bullet\vec{v})^2$ You get equality exactly when $\vec{u}=\text{proj}_{\vec{v}}\vec{u}=\frac{\vec{u}\bullet\vec{v}}{\|\vec{v}\|^2}\vec{v}$ in other words, when $\vec{u}$ is a multiple of $\vec{v}$.

习题 32 Exercise 32

Let $\vec{v},$ $\vec{w},$ $\vec{u}$ be vectors. Show that $(\vec{w}+\vec{u})_{\perp}=\vec{w}_{\perp} +\vec{u}_{\perp}$ where $\vec{w}_{\perp} =\vec{w}-\text{proj}_{\vec{v}}(\vec{w})$.

答案 Answer：$\begin{aligned}\vec{w}-\text{proj}_{\vec{v}}(\vec{w})+\vec{u}-\text{proj}_{\vec{v}}(\vec{u})&=\vec{w}+\vec{u}-(\text{proj}_{\vec{v}}(\vec{w})+\text{proj}_{\vec{v}}(\vec{u})) \\ &=\vec{w}+\vec{u}-\text{proj}_{\vec{v}}(\vec{w}+\vec{u})\end{aligned}$ This follows because $\begin{aligned}\text{proj}_{\vec{v}}(\vec{w})+\text{proj}_{\vec{v}}(\vec{u})&=\frac{\vec{u}\bullet\vec{v}}{\|\vec{v}\|^2}\vec{v}+\frac{\vec{w}\bullet\vec{v}}{\|\vec{v}\|^2}\vec{v} \\ &=\frac{(\vec{u}+\vec{w})\bullet\vec{v}}{\|\vec{v}\|^2}\vec{v} \\ &=\text{proj}_{\vec{v}}(\vec{w}+\vec{u})\end{aligned}$

习题 33 Exercise 33

Show that $(\vec{v}-\text{proj}_{\vec{u}}(\vec{v}),\vec{u})=(\vec{v}-\text{proj}_{\vec{u}}(\vec{v}))\bullet\vec{u}=0$ and conclude every vector in $\mathbb{R}^n$ can be written as the sum of two vectors, one which is perpendicular and one which is parallel to the given vector.

答案 Answer：$(\vec{v}-\text{proj}_{\vec{u}}(\vec{v}))\bullet\vec{u}=\vec{v}\bullet\vec{u}-\left(\frac{(\vec{v}\cdot\vec{u}}{\|\vec{u}\|^2}\vec{u}\right)\bullet\vec{u}=\vec{v}\bullet\vec{u}-\vec{v}\bullet\vec{u}=0$. Therefore, $\vec{v}=\vec{v}-\text{proj}_{\vec{u}}(\vec{v})+\text{proj}_{\vec{u}}(\vec{v})$. The first is perpendicular to $\vec{u}$ and the second is a multiple of $\vec{u}$ so it is parallel to $\vec{u}$.

习题 34 Exercise 34

Show that if $\vec{a}\times\vec{u}=\vec{0}$ for any unit vector $\vec{u}$, then $\vec{a}=\vec{0}$.

答案 Answer：If $\vec{a}\neq\vec{0}$, then the condition says that $\|\vec{a}\times\vec{u}\|=\|\vec{a}\|\sin\theta =0$ for all angles $\theta$. Hence $\vec{a}=\vec{0}$ after all.

习题 35 Exercise 35

Find the area of the triangle determined by the three points $(1, 2, 3),$ $(4, 2, 0)$ and $(-3, 2, 1)$.

答案 Answer：$\left[\begin{array}{r}3\\0\\-3\end{array}\right]\times\left[\begin{array}{r}-4\\0\\-2\end{array}\right]=\left[\begin{array}{r}0\\18\\0\end{array}\right]$. So the area is $9$.

习题 36 Exercise 36

Find the area of the triangle determined by the three points $(1, 0, 3),$ $(4, 1, 0)$ and $(-3, 1, 1)$.

答案 Answer：$\left[\begin{array}{r}3\\1\\-3\end{array}\right]\times\left[\begin{array}{r}-4\\1\\-2\end{array}\right]=\left[\begin{array}{c}1\\18\\7\end{array}\right]$. The area is given by $\frac{1}{2}\sqrt{1+(18)^2+49}=\frac{1}{2}\sqrt{374}$

习题 37 Exercise 37

Find the area of the triangle determined by the three points, $(1, 2, 3),$ $(2, 3, 4)$ and $(3, 4, 5)$. Did something interesting happen here? What does it mean geometrically?

答案 Answer：$\left[\begin{array}{ccc}1&1&1\end{array}\right]\times\left[\begin{array}{ccc}2&2&2\end{array}\right]=\left[\begin{array}{ccc}0&0&0\end{array}\right]$. The area is $0$. It means the three points are on the same line.

习题 38 Exercise 38

Find the area of the parallelogram determined by the vectors $\left[\begin{array}{c}1\\2\\3\end{array}\right]$, $\left[\begin{array}{r}3\\-2\\1\end{array}\right]$.

答案 Answer：$\left[\begin{array}{c}1\\2\\3\end{array}\right]\times\left[\begin{array}{r}3\\-2\\1\end{array}\right]=\left[\begin{array}{r}8\\8\\-8\end{array}\right]$. The area is $8\sqrt{3}$.

习题 39 Exercise 39

Find the area of the parallelogram determined by the vectors $\left[\begin{array}{c}1\\0\\3\end{array}\right]$, $\left[\begin{array}{r}4\\-2\\1\end{array}\right]$.

答案 Answer：$\left[\begin{array}{c}1\\0\\3\end{array}\right]\times\left[\begin{array}{r}4\\-2\\1\end{array}\right]=\left[\begin{array}{r}6\\11\\-2\end{array}\right]$. The area is $\sqrt{36+121+4}=\sqrt{161}$.

习题 40 Exercise 40

Is $\vec{u}\times (\vec{v}\times\vec{w})=(\vec{u}\times\vec{v})\times\vec{w}$? What is the meaning of $\vec{u}\times\vec{v}\times\vec{w}$? Explain. Hint: Try $\left(\vec{i}\times\vec{j}\right)\times\vec{k}$.

答案 Answer：$\left(\vec{i}\times\vec{j}\right)\times\vec{j}=\vec{k}\times\vec{j}=i\vec{i}$. However, $\vec{i}\times\left(\vec{j}\times\vec{j}\right)=\vec{0}$ and so the cross product is not associative.

习题 41 Exercise 41

Verify directly that the coordinate description of the cross product, $\vec{u}\times\vec{v}$ has the property that it is perpendicular to both $\vec{u}$ and $\vec{v}$. Then show by direct computation that this coordinate description satisfies $\begin{aligned} \|\vec{u}\times\vec{v}\|^2&=\|\vec{u}\|^2\|\vec{v}\|^2-(\vec{u}\bullet\vec{v})^2 \\ &=\|\vec{u}\|^2\|\vec{v}\|^2(1-\cos^2(\theta ))\end{aligned}$ where $\theta$ is the angle included between the two vectors. Explain why $\|\vec{u}\times\vec{v}\|$ has the correct magnitude.

答案 Answer：Verify directly from the coordinate description of the cross product that the right hand rule applies to the vectors $\vec{i},\vec{j},\vec{k}$. Next verify that the distributive law holds for the coordinate description of the cross product. This gives another way to approach the cross product. First define it in terms of coordinates and then get the geometric properties from this. However, this approach does not yield the right hand rule property very easily. From the coordinate description, $\vec{a}\times\vec{b}\cdot\vec{a}=\epsilon_{ijk}a_jb_ka_i=-\epsilon_{jik}a_kb_ka_i=-\epsilon_{jik}b_ka_ia_j=-\vec{a}\times\vec{b}\cdot\vec{a}$ and so $\vec{a}\times\vec{b}$ is perpendicular to $\vec{a}$. Similarly, $\vec{a}\times\vec{b}$ is perpendicular to $\vec{b}$. Now we need that $\|\vec{a}\times\vec{b}\|^2=\|\vec{a}\|^2\|\vec{b}\|^2(1-\cos^2\theta )=\|\vec{a}\|^2\|\vec{b}\|^2\sin^2\theta$ and so $\|\vec{a}\times\vec{b}\|=\|\vec{a}\|\|\vec{b}\|\sin\theta$, the area of the parallelogram determined by $\vec{a}$, $\vec{b}$. Only the right hand rule is a little problematic. However, you can see right away from the component definition that the right hand rule holds for each of the standard unit vectors. Thus $\vec{i}\times\vec{j}=\vec{k}$ etc. $\left\|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\1&0&0\\0&1&0\end{array}\right\|=\vec{k}$

习题 42 Exercise 42

Suppose $A$ is a $3\times 3$ skew symmetric matrix such that $A^T = -A$. Show there exists a vector $\vec{\Omega}$ such that for all $\vec{u} \in \mathbb{R}^3$ $A\vec{u}=\vec{\Omega}\times\vec{u}$ Hint: Explain why since $A$ is skew symmetric it is of the form $A=\left[\begin{array}{ccc}0&-\omega_3&\omega_2 \\ \omega_3&0&-\omega_1 \\ -\omega_2&\omega_1&0\end{array}\right]$ where the $\omega_i$ are numbers. Then consider $\omega_1\vec{i}+\omega_2\vec{j}+\omega_3\vec{k}$.

习题 43 Exercise 43

Find the volume of the parallelepiped determined by the vectors $\left[\begin{array}{r}1\\-7\\-5\end{array}\right]$, $\left[\begin{array}{r}1\\-2\\-6\end{array}\right]$, and $\left[\begin{array}{c}3\\2\\3\end{array}\right]$.

答案 Answer：$\left\|\begin{array}{ccc}1&-7&-5 \\ 1&-2&-6 \\ 3&2&3\end{array}\right\|=113$

习题 44 Exercise 44

Suppose $\vec{u}$, $\vec{v}$, and $\vec{w}$ are three vectors whose components are all integers. Can you conclude the volume of the parallelepiped determined from these three vectors will always be an integer?

答案 Answer：Yes. It will involve the sum of product of integers and so it will be an integer.

习题 45 Exercise 45

What does it mean geometrically if the box product of three vectors gives zero?

答案 Answer：It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.

习题 46 Exercise 46

Using Problem 45, find an equation of a plane containing the two position vectors, $\vec{p}$ and $\vec{q}$ and the point $0$. Hint: If $(x, y,z)$ is a point on this plane, the volume of the parallelepiped determined by $(x, y,z)$ and the vectors $\vec{p}$, $\vec{q}$ equals $0$.

答案 Answer：$\vec{x}\bullet\left(\vec{a}\times\vec{b}\right)=0$

习题 47 Exercise 47

Using the notion of the box product yielding either plus or minus the volume of the parallelepiped determined by the given three vectors, show that $(\vec{u}\times\vec{v})\bullet\vec{w}=\vec{u}\bullet (\vec{v}\times\vec{w})$ In other words, the dot and the cross can be switched as long as the order of the vectors remains the same. Hint: There are two ways to do this, by the coordinate description of the dot and cross product and by geometric reasoning.

习题 48 Exercise 48

Simplify $(\vec{u}\times\vec{v})\bullet [(\vec{v}\times\vec{w})\times (\vec{w}\times\vec{z})]$.

答案 Answer：Here $[\vec{v},\vec{w},\vec{z}]$ denotes the box product. Consider the cross product term. From the above, $\begin{aligned}(\vec{v}\times\vec{w})\times(\vec{w}\times\vec{z})&=[\vec{v},\vec{w},\vec{z}]\vec{w}-[\vec{w},\vec{w},\vec{z}]\vec{v} \\ &=[\vec{v},\vec{w},\vec{z}]\vec{w}\end{aligned}$ Thus it reduces to $(\vec{u}\times\vec{v})\bullet [\vec{v},\vec{w},\vec{z}]\vec{w}=[\vec{v},\vec{w},\vec{z}][\vec{u},\vec{v},\vec{w}]$

习题 49 Exercise 49

Simplify $\|\vec{u}\times\vec{v}\|^2+(\vec{u}\bullet\vec{v})^2-\|\vec{u}\|^2\|\vec{v}\|^2$.

答案 Answer：$\begin{aligned}\|\vec{u}\times\vec{v}\|^2&=\epsilon_{ijk}u_jv_k\epsilon_{irs}u_rv_s=(\delta_{jr}\delta_{ks}-\delta_{kr}\delta_{js})u_rv_su_jv_k \\ &=u_jv_ku_jv_k-u_kv_ju_jv_k=\|\vec{u}\|^2\|\vec{v}\|^2-(\vec{u}\bullet\vec{v})^2\end{aligned}$ It follows that the expression reduces to $0$. You can also do the following. $\begin{aligned}\|\vec{u}\times\vec{v}\|^2&=\|\vec{u}\|^2\|\vec{v}\|^2\sin^2\theta \\ &=\|\vec{u}\|^2\|\vec{v}\|^2(1-\cos^2\theta ) \\ &=\|\vec{u}\|^2\|\vec{v}\|^2-\|\vec{u}\|^2\|\vec{v}\|^2\cos^2\theta \\ &=\|\vec{u}\|^2\|\vec{v}\|^2-(\vec{u}\bullet\vec{v})^2\end{aligned}$ which implies the expression equals $0$.

习题 50 Exercise 50

For $\vec{u},$ $\vec{v},$ $\vec{w}$ functions of $t$, prove the following product rules: $\begin{aligned}(\vec{u}\times\vec{v})'&=\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}' \\ (\vec{u}\bullet\vec{v})'&=\vec{u}'\bullet\vec{v}+\vec{u}\bullet\vec{v}'\end{aligned}$

答案 Answer：We will show it using the summation convention and permutation symbol $\begin{aligned}((\vec{u}\times\vec{v})')_i&=((\vec{u}\times\vec{v})_i)'=(\epsilon_{ijk}u_jv_k)' \\ &=\epsilon_{ijk}u_j'v_k+\epsilon_{ijk}u_kv_k'=(\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}')_i\end{aligned}$ and so $(\vec{u}\times\vec{v})'=\vec{u}'\times\vec{v}+\vec{u}\times\vec{v}'$.

习题 51 Exercise 51

Here are some vectors. $\left[\begin{array}{r}1\\1\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\2\\-2\end{array}\right],$ $\left[\begin{array}{r}2\\7\\-4\end{array}\right],$ $\left[\begin{array}{r}5\\7\\-10\end{array}\right],$ $\left[\begin{array}{r}12\\17\\-24\end{array}\right]$ Describe the span of these vectors as the span of as few vectors as possible.

习题 52 Exercise 52

Here are some vectors. $\left[\begin{array}{r}1\\2\\-2\end{array}\right],$ $\left[\begin{array}{r}12\\29\\-24\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-2\end{array}\right],$ $\left[\begin{array}{r}2\\9\\-4\end{array}\right],$ $\left[\begin{array}{r}5\\12\\-10\end{array}\right].$ Describe the span of these vectors as the span of as few vectors as possible.

习题 53 Exercise 53

Here are some vectors. $\left[\begin{array}{r}1\\2\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\-2\\-2\end{array}\right],$ $\left[\begin{array}{r}-1\\0\\2\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-1\end{array}\right]$ Describe the span of these vectors as the span of as few vectors as possible.

习题 54 Exercise 54

Here are some vectors. $\left[\begin{array}{r}1\\1\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\2\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\-3\\-2\end{array}\right],$ $\left[\begin{array}{r}-1\\1\\2\end{array}\right]$ Now here is another vector: $\left[\begin{array}{r}1\\2\\-1\end{array}\right]$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

习题 55 Exercise 55

Here are some vectors. $\left[\begin{array}{r}1\\1\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\2\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\-3\\-2\end{array}\right],$ $\left[\begin{array}{r}-1\\1\\2\end{array}\right]$ Now here is another vector: $\left[\begin{array}{r}2\\-3\\-4\end{array}\right]$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

习题 56 Exercise 56

Here are some vectors. $\left[\begin{array}{r}1\\1\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\2\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\-3\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\2\\-1\end{array}\right]$ Now here is another vector: $\left[\begin{array}{r}1\\9\\1\end{array}\right]$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

习题 57 Exercise 57

Here are some vectors, $\left[\begin{array}{r}1\\-1\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\0\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\-5\\-2\end{array}\right],$ $\left[\begin{array}{r}-1\\5\\2\end{array}\right]$ Now here is another vector: $\left[\begin{array}{r}1\\1\\-1\end{array}\right]$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

习题 58 Exercise 58

Here are some vectors. $\left[\begin{array}{r}1\\-1\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\0\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\-5\\-2\end{array}\right],$ $\left[\begin{array}{r}-1\\5\\2\end{array}\right]$ Now here is another vector: $\left[\begin{array}{r}1\\1\\-1\end{array}\right]$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

习题 59 Exercise 59

Here are some vectors. $\left[\begin{array}{r}1\\0\\-2\end{array}\right],$ $\left[\begin{array}{r}1\\1\\-2\end{array}\right],$ $\left[\begin{array}{r}2\\-2\\-3\end{array}\right],$ $\left[\begin{array}{r}-1\\4\\2\end{array}\right]$ Now here is another vector: $\left[\begin{array}{r}-1\\-4\\2\end{array}\right]$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

习题 60 Exercise 60

Suppose $\{\vec{x}_{1},\cdots ,\vec{x}_{k}\}$ is a set of vectors from $\mathbb{R}^n$. Show that $\vec{0}$ is in $\mathrm{span}\{\vec{x}_{1},\cdots ,\vec{x}_{k}\}$.

答案 Answer：$\sum\limits_{i=1}^k 0\vec{x}_{k}=\vec{0}$

习题 61 Exercise 61

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\4\\0\\1\end{array}\right],$ $\left[\begin{array}{r}1\\10\\2\\1\end{array}\right]$

习题 62 Exercise 62

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}-1\\-2\\2\\3\end{array}\right],$ $\left[\begin{array}{r}-3\\-4\\3\\3\end{array}\right],$ $\left[\begin{array}{r}0\\-1\\4\\3\end{array}\right],$ $\left[\begin{array}{r}0\\-1\\6\\4\end{array}\right]$

习题 63 Exercise 63

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}1\\6\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}-1\\-4\\1\\-1\end{array}\right],$ $\left[\begin{array}{r}1\\6\\-2\\1\end{array}\right]$

习题 64 Exercise 64

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}1\\-1\\3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\6\\34\\1\end{array}\right],$ $\left[\begin{array}{r}1\\0\\7\\1\end{array}\right],$ $\left[\begin{array}{r}1\\0\\8\\1\end{array}\right]$

习题 65 Exercise 65

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. $\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}-3\\-10\\3\\-3\end{array}\right],$ $\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]$

习题 66 Exercise 66

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\4\\-5\\1\end{array}\right],$ $\left[\begin{array}{r}1\\4\\-4\\1\end{array}\right],$ $\left[\begin{array}{r}1\\10\\-14\\1\end{array}\right]$

习题 67 Exercise 67

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}1\\0\\3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\1\\8\\1\end{array}\right],$ $\left[\begin{array}{r}1\\7\\34\\1\end{array}\right],$ $\left[\begin{array}{r}1\\1\\7\\1\end{array}\right]$

习题 68 Exercise 68

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\7\\-5\\1\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right]$

习题 69 Exercise 69

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. $\left[\begin{array}{r}1\\2\\2\\-4\end{array}\right],$ $\left[\begin{array}{r}3\\4\\1\\-4\end{array}\right],$ $\left[\begin{array}{r}0\\-1\\0\\4\end{array}\right],$ $\left[\begin{array}{r}0\\-1\\-2\\5\end{array}\right]$

习题 70 Exercise 70

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $\left[\begin{array}{r}2\\3\\1\\-3\end{array}\right],$ $\left[\begin{array}{r}-5\\-6\\0\\3\end{array}\right],$ $\left[\begin{array}{r}-1\\-2\\1\\3\end{array}\right],$ $\left[\begin{array}{r}-1\\-2\\0\\4\end{array}\right]$

习题 71 Exercise 71

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\1\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\2\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\-2\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\2\\0\\1\end{array}\right],$ $\left[\begin{array}{r}1\\-1\\-1\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 72 Exercise 72

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\2\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}4\\3\\-1\\4\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-2\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 73 Exercise 73

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\1\\0\\1\end{array}\right],$ $\left[\begin{array}{r}1\\2\\1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\-2\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}2\\-5\\-7\\2\end{array}\right],$ $\left[\begin{array}{r}1\\2\\2\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 74 Exercise 74

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\2\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\-1\\1\\1\end{array}\right],$ $\left[\begin{array}{r}2\\-3\\3\\2\end{array}\right],$ $\left[\begin{array}{r}1\\3\\-2\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 75 Exercise 75

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}4\\11\\-1\\4\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 76 Exercise 76

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}-\frac{3}{2}\\-\frac{9}{2}\\ \frac{3}{2}\\ -\frac{3}{2}\end{array}\right],$ $\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}2\\-1\\-2\\2\end{array}\right],$ $\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 77 Exercise 77

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\3\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\4\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}1\\0\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}2\\-1\\-2\\2\end{array}\right],$ $\left[\begin{array}{r}1\\4\\0\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 78 Exercise 78

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\4\\-2\\1\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\1\\1\\1\end{array}\right],$ $\left[\begin{array}{r}2\\1\\3\\2\end{array}\right],$ $\left[\begin{array}{r}1\\5\\-2\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 79 Exercise 79

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\-1\\3\\1\end{array}\right],$ $\left[\begin{array}{r}1\\0\\7\\1\end{array}\right],$ $\left[\begin{array}{r}1\\0\\8\\1\end{array}\right],$ $\left[\begin{array}{r}4\\-9\\-6\\4\end{array}\right],$ $\left[\begin{array}{r}1\\0\\8\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 80 Exercise 80

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\-1\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}-3\\3\\3\\-3\end{array}\right],$ $\left[\begin{array}{r}1\\0\\-1\\1\end{array}\right],$ $\left[\begin{array}{r}2\\-9\\-2\\2\end{array}\right],$ $\left[\begin{array}{r}1\\0\\0\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 81 Exercise 81

Here are some vectors in $\mathbb{R}^4$. $\left[\begin{array}{r}1\\b+1\\a\\1\end{array}\right],$ $\left[\begin{array}{r}3\\3b+3\\3a\\3\end{array}\right],$ $\left[\begin{array}{r}1\\b+2\\2a+1\\1\end{array}\right],$ $\left[\begin{array}{r}2\\2b-5\\-5a-7\\2\end{array}\right],$ $\left[\begin{array}{r}1\\b+2\\2a+2\\1\end{array}\right]$ These vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

习题 82 Exercise 82

Let $H=\mathrm{span}\left\{\left[\begin{array}{r}2\\1\\1\\1\end{array}\right],$ $\left[\begin{array}{r}-1\\0\\-1\\-1\end{array}\right],$ $\left[\begin{array}{r}5\\2\\3\\3\end{array}\right],$ $\left[\begin{array}{r}-1\\1\\-2\\-2\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 83 Exercise 83

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}0\\1\\1\\-1\end{array}\right],$ $\left[\begin{array}{r}-1\\-1\\-2\\2\end{array}\right],$ $\left[\begin{array}{r}2\\3\\5\\-5\end{array}\right],$ $\left[\begin{array}{r}0\\1\\2\\-2\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 84 Exercise 84

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}-2\\1\\1\\-3\end{array}\right],\ \left[\begin{array}{r}-9\\4\\3\\-9\end{array}\right],\ \left[\begin{array}{r}-33\\15\\12\\-36\end{array}\right],\ \left[\begin{array}{r}-22\\10\\8\\-24\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 85 Exercise 85

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}-1\\1\\-1\\-2\end{array}\right],\ \left[\begin{array}{r}-4\\3\\-2\\-4\end{array}\right],\ \left[\begin{array}{r}-3\\2\\-1\\-2\end{array}\right],\ \left[\begin{array}{r}-1\\1\\-2\\-4\end{array}\right],\ \left[\begin{array}{r}-7\\5\\-3\\-6\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 86 Exercise 86

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}2\\3\\2\\1\end{array}\right],\ \left[\begin{array}{r}8\\15\\6\\3\end{array}\right],\ \left[\begin{array}{r}3\\6\\2\\1\end{array}\right],\ \left[\begin{array}{r}4\\6\\6\\3\end{array}\right],\ \left[\begin{array}{r}8\\15\\6\\3\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 87 Exercise 87

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}0\\2\\0\\-1\end{array}\right],\ \left[\begin{array}{r}-1\\6\\0\\-2\end{array}\right],\ \left[\begin{array}{r}-2\\16\\0\\-6\end{array}\right],\ \left[\begin{array}{r}-3\\22\\0\\-8\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 88 Exercise 88

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}5\\1\\1\\4\end{array}\right],\ \left[\begin{array}{r}14\\3\\2\\8\end{array}\right],\ \left[\begin{array}{r}38\\8\\6\\24\end{array}\right],\ \left[\begin{array}{r}47\\10\\7\\28\end{array}\right],\ \left[\begin{array}{r}10\\2\\3\\12\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 89 Exercise 89

Let $H$ denote $\mathrm{span}\left\{\left[\begin{array}{r}6\\1\\1\\5\end{array}\right],\ \left[\begin{array}{r}17\\3\\2\\10\end{array}\right],\ \left[\begin{array}{r}52\\9\\7\\35\end{array}\right],\ \left[\begin{array}{r}18\\3\\4\\20\end{array}\right]\right\}.$ Find the dimension of $H$ and determine a basis.

习题 90 Exercise 90

Let $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:\sin(u_{1})=1\right\}.$ Is $M$ a subspace? Explain.

答案 Answer：No. Let $\vec{u}=\left[\begin{array}{c}\frac{\pi}{2}\\0\\0\\0\end{array}\right]$. Then $2\vec{u}\notin M$ although $\vec{u}\in M$.

习题 91 Exercise 91

Let $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:\left|u_{1}\right|\leq 4\right\}.$ Is $M$ a subspace? Explain.

答案 Answer：No. $\left[\begin{array}{c}1\\0\\0\\0\end{array}\right]\in M$ but $10\left[\begin{array}{c}1\\0\\0\\0\end{array}\right]\notin M$.

习题 92 Exercise 92

Let $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:u_{i}\geq 0\text{ for each }i=1,2,3,4 \right\}.$ Is $M$ a subspace? Explain.

答案 Answer：This is not a subspace. $\left[\begin{array}{c}1\\1\\1\\1\end{array}\right]$ is in it. However, $(-1)\left[\begin{array}{c}1\\1\\1\\1\end{array}\right]$ is not.

习题 93 Exercise 93

Let $\vec{w}$, $\vec{w}_{1}$ be given vectors in $\mathbb{R}^{4}$ and define $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:\vec{w}\bullet\vec{u}=0\text{ and }\vec{w}_{1}\bullet\vec{u}=0\right\}.$ Is $M$ a subspace? Explain.

答案 Answer：This is a subspace because it is closed with respect to vector addition and scalar multiplication.

习题 94 Exercise 94

Let $\vec{w}\in\mathbb{R}^{4}$ and let $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:\vec{w}\bullet\vec{u}=0\right\}.$ Is $M$ a subspace? Explain.

答案 Answer：Yes, this is a subspace because it is closed with respect to vector addition and scalar multiplication.

习题 95 Exercise 95

Let $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:u_{3}\geq u_{1}\right\}.$ Is $M$ a subspace? Explain.

答案 Answer：This is not a subspace. $\left[\begin{array}{c}0\\0\\1\\0\end{array}\right]$ is in it. However $(-1)\left[\begin{array}{c}0\\0\\1\\0\end{array}\right]=\left[\begin{array}{r}0\\0\\-1\\0\end{array}\right]$ is not.

习题 96 Exercise 96

Let $M=\left\{\vec{u}=\left[\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u_{4}\end{array}\right]\in\mathbb{R}^{4}:u_{3}=u_{1}=0\right\}.$ Is $M$ a subspace? Explain.

答案 Answer：This is a subspace. It is closed with respect to vector addition and scalar multiplication.

习题 97 Exercise 97

Consider the set of vectors $S$ given by $S=\left\{\left[\begin{array}{c}4u+v-5w\\12u+6v-6w\\4u+4v+4w\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is $S$ a subspace of $\mathbb{R}^{3}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 98 Exercise 98

Consider the set of vectors $S$ given by $S=\left\{\left[\begin{array}{c}2u+6v+7w\\-3u-9v-12w\\2u+6v+6w\\u+3v+3w\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is $S$ a subspace of $\mathbb{R}^{4}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 99 Exercise 99

Consider the set of vectors $S$ given by $S=\left\{\left[\begin{array}{c}2u+v\\6v-3u+3w\\3v-6u+3w\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is this set of vectors a subspace of $\mathbb{R}^{3}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 100 Exercise 100

Consider the vectors of the form $\left\{\left[\begin{array}{c}2u+v+7w\\u-2v+w\\-6v-6w\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is this set of vectors a subspace of $\mathbb{R}^{3}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 101 Exercise 101

Consider the vectors of the form $\left\{\left[\begin{array}{c}3u+v+11w\\18u+6v+66w\\28u+8v+100w\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is this set of vectors a subspace of $\mathbb{R}^{3}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 102 Exercise 102

Consider the vectors of the form $\left\{\left[\begin{array}{c}3u+v\\2w-4u\\2w-2v-8u\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is this set of vectors a subspace of $\mathbb{R}^{3}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 103 Exercise 103

Consider the set of vectors $S$ given by $\left\{\left[\begin{array}{c}u+v+w\\2u+2v+4w\\u+v+w\\0\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is $S$ a subspace of $\mathbb{R}^{4}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 104 Exercise 104

Consider the set of vectors $S$ given by $\left\{\left[\begin{array}{c}v\\-3u-3w\\8u-4v+4w\end{array}\right]:u,v,w\in\mathbb{R}\right\}.$ Is $S$ a subspace of $\mathbb{R}^{4}$? If so, explain why, give a basis for the subspace and find its dimension.

习题 105 Exercise 105

If you have $5$ vectors in $\mathbb{R}^{5}$ and the vectors are linearly independent, can it always be concluded they span $\mathbb{R}^{5}$? Explain.

答案 Answer：Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.

习题 106 Exercise 106

If you have $6$ vectors in $\mathbb{R}^{5}$, is it possible they are linearly independent? Explain.

答案 Answer：They can't be.

习题 107 Exercise 107

Suppose $A$ is an $m\times n$ matrix and $\left\{\vec{w}_{1},\cdots,\vec{w}_{k}\right\}$ is a linearly independent set of vectors in $A(\mathbb{R}^{n})\subseteq\mathbb{R}^{m}$. Now suppose $A\vec{z}_{i}=\vec{w}_{i}$. Show $\left\{\vec{z}_{1},\cdots,\vec{z}_{k}\right\}$ is also independent.

答案 Answer：Say $\sum\limits_{i=1}^{k}c_{i}\vec{z}_{i}=\vec{0}$. Then apply $A$ to it as follows. $\sum\limits_{i=1}^{k}c_{i}A\vec{z}_{i}=\sum\limits_{i=1}^{k}c_{i}\vec{w}_{i}=\vec{0}$ and so, by linear independence of the $\vec{w}_{i}$, it follows that each $c_{i}=0$.

习题 108 Exercise 108

Suppose $V$, $W$ are subspaces of $\mathbb{R}^{n}$. Let $V\cap W$ be all vectors which are in both $V$ and $W$. Show that $V\cap W$ is a subspace also.

答案 Answer：If $\vec{x},\vec{y}\in V\cap W$, then for scalars $\alpha,\beta$, the linear combination $\alpha\vec{x}+\beta\vec{y}$ must be in both $V$ and $W$ since they are both subspaces.

习题 109 Exercise 109

Suppose $V$ and $W$ both have dimension equal to $7$ and they are subspaces of $\mathbb{R}^{10}$. What are the possibilities for the dimension of $V\cap W$? Hint: Remember that a linear independent set can be extended to form a basis.

习题 110 Exercise 110

Suppose $V$ has dimension $p$ and $W$ has dimension $q$ and they are each contained in a subspace, $U$ which has dimension equal to $n$ where $n>\max(p,q)$. What are the possibilities for the dimension of $V\cap W$? Hint: Remember that a linearly independent set can be extended to form a basis.

答案 Answer：Let $\left\{\vec{x}_{1},\cdots,\vec{x}_{k}\right\}$ be a basis for $V\cap W$. Then there is a basis for $V$ and $W$ which are respectively $\left\{\vec{x}_{1},\cdots,\vec{x}_{k},\vec{y}_{k+1},\cdots,\vec{y}_{p}\right\}$, $\left\{\vec{x}_{1},\cdots,\vec{x}_{k},\vec{z}_{k+1},\cdots,\vec{z}_{q}\right\}$. It follows that you must have $k+p-k+q-k\leq n$ and so you must have $p+q-n\leq k$.

习题 111 Exercise 111

Suppose $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix. Show that $\text{dim}(\text{ker}(AB))\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B))$. Consider the subspace, $B(\mathbb{R}^{p})\cap\text{ker}(A)$ and suppose a basis for this subspace is $\left\{\vec{w}_{1},\cdots,\vec{w}_{k}\right\}$. Now suppose $\left\{\vec{u}_{1},\cdots,\vec{u}_{r}\right\}$ is a basis for $\text{ker}(B)$. Let $\left\{\vec{z}_{1},\cdots,\vec{z}_{k}\right\}$ be such that $B\vec{z}_{i}=\vec{w}_{i}$ and argue that $\text{ker}(AB)\subseteq\text{span}\left\{\vec{u}_{1},\cdots,\vec{u}_{r},\vec{z}_{1},\cdots,\vec{z}_{k}\right\}$.

答案 Answer：Here is how you do this. Suppose $AB\vec{x}=\vec{0}$. Then $B\vec{x}\in\text{ker}(A)\cap B(\mathbb{R}^{p})$ and so $B\vec{x}=\sum\limits_{i=1}^{k}B\vec{z}_{i}$ showing that $\vec{x}-\sum\limits_{i=1}^{k}\vec{z}_{i}\in\text{ker}(B)$. Consider $B(\mathbb{R}^{p})\cap\text{ker}(A)$ and let a basis be $\left\{\vec{w}_{1},\cdots,\vec{w}_{k}\right\}$. Then each $\vec{w}_{i}$ is of the form $B\vec{z}_{i}=\vec{w}_{i}$. Therefore, $\left\{\vec{z}_{1},\cdots,\vec{z}_{k}\right\}$ is linearly independent and $AB\vec{z}_{i}=0$. Now let $\left\{\vec{u}_{1},\cdots,\vec{u}_{r}\right\}$ be a basis for $\text{ker}(B)$. If $AB\vec{x}=\vec{0}$, then $B\vec{x}\in\text{ker}(A)\cap B(\mathbb{R}^{p})$ and so $B\vec{x}=\sum\limits_{i=1}^{k}c_{i}B\vec{z}_{i}$ which implies $\vec{x}-\sum\limits_{i=1}^{k}c_{i}\vec{z}_{i}\in\text{ker}(B)$ and so it is of the form $\vec{x}-\sum\limits_{i=1}^{k}c_{i}\vec{z}_{i}=\sum\limits_{j=1}^{r}d_{j}\vec{u}_{j}$. It follows that if $AB\vec{x}=\vec{0}$ so that $\vec{x}\in\text{ker}(AB)$, then $\vec{x}\in\text{span}\left(\vec{z}_{1},\cdots,\vec{z}_{k},\vec{u}_{1},\cdots,\vec{u}_{r}\right)$. Therefore, $\begin{aligned}\text{dim}(\text{ker}(AB))&\leq k+r=\text{dim}(B(\mathbb{R}^{p})\cap\text{ker}(A))+\text{dim}(\text{ker}(B))\\ &\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B))\end{aligned}$

习题 112 Exercise 112

Show that if $A$ is an $m\times n$ matrix, then $\text{ker}(A)$ is a subspace of $\mathbb{R}^{n}$.

答案 Answer：If $\vec{x}$, $\vec{y}\in\text{ker}(A)$ then $A(a\vec{x}+b\vec{y})=aA\vec{x}+bA\vec{y}=a\vec{0}+b\vec{0}=\vec{0}$ and so $\text{ker}(A)$ is closed under linear combinations. Hence it is a subspace.

习题 113 Exercise 113

Find the rank of the following matrix. Also find a basis for the row and column spaces. $\left[\begin{array}{rrrrrr}1&3&0&-2&0&3\\3&9&1&-7&0&8\\1&3&1&-3&1&-1\\1&3&-1&-1&-2&10\end{array}\right]$

习题 114 Exercise 114

Find the rank of the following matrix. Also find a basis for the row and column spaces. $\left[\begin{array}{rrrrrr}1&3&0&-2&7&3\\3&9&1&-7&23&8\\1&3&1&-3&9&2\\1&3&-1&-1&5&4\end{array}\right]$

习题 115 Exercise 115

Find the rank of the following matrix. Also find a basis for the row and column spaces. $\left[\begin{array}{rrrrrr}1&0&3&0&7&0\\3&1&10&0&23&0\\1&1&4&1&7&0\\1&-1&2&-2&9&1\end{array}\right]$

习题 116 Exercise 116

Find the rank of the following matrix. Also find a basis for the row and column spaces. $\left[\begin{array}{rrr}1&0&3\\3&1&10\\1&1&4\\1&-1&2\end{array}\right]$

习题 117 Exercise 117

Find the rank of the following matrix. Also find a basis for the row and column spaces. $\left[\begin{array}{rrrrr}0&0&-1&0&1\\1&2&3&-2&-18\\1&2&2&-1&-11\\-1&-2&-2&1&11\end{array}\right]$

习题 118 Exercise 118

Find the rank of the following matrix. Also find a basis for the row and column spaces. $\left[\begin{array}{rrrr}1&0&3&0\\3&1&10&0\\-1&1&-2&1\\1&-1&2&-2\end{array}\right]$

习题 119 Exercise 119

Find $\text{ker}(A)$ for the following matrices.

1. $A=\left[\begin{array}{rr}2&3\\4&6\end{array}\right]$

2. $A=\left[\begin{array}{rrr}1&0&-1\\-1&1&3\\3&2&1\end{array}\right]$

3. $A=\left[\begin{array}{rrr}2&4&0\\3&6&-2\\1&2&-2\end{array}\right]$

4. $A=\left[\begin{array}{rrrr}2&-1&3&5\\2&0&1&2\\6&4&-5&-6\\0&2&-4&-6\end{array}\right]$

习题 120 Exercise 120

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $\left[\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3}\\\frac{1}{3}\sqrt{2}\sqrt{3}\\-\frac{1}{6}\sqrt{2}\sqrt{3}\end{array}\right],\ \left[\begin{array}{c}\frac{1}{2}\sqrt{2}\\0\\\frac{1}{2}\sqrt{2}\end{array}\right],\ \left[\begin{array}{c}-\frac{1}{3}\sqrt{3}\\\frac{1}{3}\sqrt{3}\\\frac{1}{3}\sqrt{3}\end{array}\right]$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

习题 121 Exercise 121

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $\left[\begin{array}{r}1\\2\\-1\end{array}\right],\ \left[\begin{array}{r}1\\0\\1\end{array}\right],\ \left[\begin{array}{r}-1\\1\\1\end{array}\right]$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

习题 122 Exercise 122

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $\left[\begin{array}{r}1\\-1\\1\end{array}\right],\ \left[\begin{array}{r}2\\1\\-1\end{array}\right],\ \left[\begin{array}{r}0\\1\\1\end{array}\right]$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

习题 123 Exercise 123

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $\left[\begin{array}{r}1\\-1\\1\end{array}\right],\ \left[\begin{array}{r}2\\1\\-1\end{array}\right],\ \left[\begin{array}{r}1\\2\\1\end{array}\right]$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

习题 124 Exercise 124

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $\left[\begin{array}{r}1\\0\\0\\0\end{array}\right],\ \left[\begin{array}{r}0\\1\\-1\\0\end{array}\right],\ \left[\begin{array}{r}0\\0\\0\\1\end{array}\right]$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

习题 125 Exercise 125

Here are some matrices. Label according to whether they are symmetric, skew symmetric, or orthogonal.

1. $\left[\begin{array}{ccc}1&0&0\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{array}\right]$

2. $\left[\begin{array}{ccc}1&2&-3\\2&1&4\\-3&4&7\end{array}\right]$

3. $\left[\begin{array}{ccc}0&-2&-3\\2&0&-4\\3&4&0\end{array}\right]$

答案 Answer：1. Orthogonal; 2. Symmetric; 3. Skew Symmetric

习题 126 Exercise 126

For $U$ an orthogonal matrix, explain why $\left\|U\vec{x}\right\|=\left\|\vec{x}\right\|$ for any vector $\vec{x}$. Next explain why if $U$ is an $n\times n$ matrix with the property that $\left\|U\vec{x}\right\|=\left\|\vec{x}\right\|$ for all vectors, $\vec{x}$, then $U$ must be orthogonal. Thus the orthogonal matrices are exactly those which preserve length.

答案 Answer：$\left\|U\vec{x}\right\|^{2}=U\vec{x}\bullet U\vec{x}=U^{T}U\vec{x}\bullet\vec{x}=I\vec{x}\bullet\vec{x}=\left\|\vec{x}\right\|^{2}$. Next suppose distance is preserved by $U$. Then $\begin{aligned} (U(\vec{x}+\vec{y}))\bullet (U(\vec{x}+\vec{y}))&=\left\|Ux\right\|^{2}+\left\|Uy\right\|^{2}+2(Ux\bullet Uy)\\ &=\left\|\vec{x}\right\|^{2}+\left\|\vec{y}\right\|^{2}+2(U^{T}U\vec{x}\bullet\vec{y})\end{aligned}$ But since $U$ preserves distances, it is also the case that $U(\vec{x}+\vec{y})\bullet U(\vec{x}+\vec{y}))=\left\|\vec{x}\right\|^{2}+\left\|\vec{y}\right\|^{2}+2(\vec{x}\bullet\vec{y})$. Hence $\vec{x}\bullet\vec{y}=U^{T}U\vec{x}\bullet\vec{y}$ and so $(U^{T}U-I)\vec{x})\bullet\vec{y}=0$. Since $y$ is arbitrary, it follows that $U^{T}U-I=0$. Thus $U$ is orthogonal.

习题 127 Exercise 127

Suppose $U$ is an orthogonal $n\times n$ matrix. Explain why $\text{rank}(U)=n$.

答案 Answer：You could observe that $\det(UU^{T})=(\det(U))^{2}-1$ so $\det(U)\neq 0$.

习题 128 Exercise 128

Fill in the missing entries to make the matrix orthogonal. $\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\underline{\hspace{1.2em}}&\underline{\hspace{1.2em}}\\\underline{\hspace{1.2em}}&\frac{\sqrt{6}}{3}&\underline{\hspace{1.2em}}\end{array}\right].$

答案 Answer：$$\begin{aligned} &\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a\\0&\frac{\sqrt{6}}{3}&b\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&a\\0&\frac{\sqrt{6}}{3}&b\end{array}\right]^{T}\\ =&\left[\begin{array}{ccc}1&\frac{1}{3}\sqrt{3}a-\frac{1}{3}&\frac{1}{3}\sqrt{3}b-\frac{1}{3}\\ \frac{1}{3}\sqrt{3}a-\frac{1}{3}&a^{2}+\frac{2}{3}&ab-\frac{1}{3}\\ \frac{1}{3}\sqrt{3}b-\frac{1}{3}&ab-\frac{1}{3}&b^{2}+\frac{2}{3}\end{array}\right]\end{aligned}$$ This requires $$a=1/\sqrt{3},\quad b=1/\sqrt{3}.$$ $$\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3}\\0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}&1/\sqrt{3}\\0&\frac{\sqrt{6}}{3}&1/\sqrt{3}\end{array}\right]^{T}=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$$

习题 129 Exercise 129

Fill in the missing entries to make the matrix orthogonal. $\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\\frac{2}{3}&\underline{\hspace{1.2em}}&\underline{\hspace{1.2em}}\\\underline{\hspace{1.2em}}&0&\underline{\hspace{1.2em}}\end{array}\right]$

答案 Answer：$$\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\\frac{2}{3}&\frac{-\sqrt{2}}{2}&a\\-\frac{1}{3}&0&b\end{array}\right]\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\\frac{2}{3}&\frac{-\sqrt{2}}{2}&a\\-\frac{1}{3}&0&b\end{array}\right]^{T}=\left[\begin{array}{ccc}1&\frac{1}{6}\sqrt{2}a-\frac{1}{18}&\frac{1}{6}\sqrt{2}b-\frac{2}{9}\\\frac{1}{6}\sqrt{2}a-\frac{1}{18}&a^{2}+\frac{17}{18}&ab-\frac{2}{9}\\\frac{1}{6}\sqrt{2}b-\frac{2}{9}&ab-\frac{2}{9}&b^{2}+\frac{1}{9}\end{array}\right]$$ This requires $$a=\frac{1}{3\sqrt{2}},\quad b=\frac{4}{3\sqrt{2}}.$$ $$\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\\frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}}\\-\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}\frac{2}{3}&\frac{\sqrt{2}}{2}&\frac{1}{6}\sqrt{2}\\\frac{2}{3}&\frac{-\sqrt{2}}{2}&\frac{1}{3\sqrt{2}}\\-\frac{1}{3}&0&\frac{4}{3\sqrt{2}}\end{array}\right]^{T}=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$$

习题 130 Exercise 130

Fill in the missing entries to make the matrix orthogonal. $\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\underline{\hspace{1.2em}}\\\frac{2}{3}&0&\underline{\hspace{1.2em}}\\\underline{\hspace{1.2em}}&\underline{\hspace{1.2em}}&\frac{4}{15}\sqrt{5}\end{array}\right]$

答案 Answer：Try $$\begin{aligned}&\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c\\\frac{2}{3}&0&d\\\frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&c\\\frac{2}{3}&0&d\\\frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]^{T}\\ =&\left[\begin{array}{ccc}c^{2}+\frac{41}{45}&cd+\frac{2}{9}&\frac{4}{15}\sqrt{5}c-\frac{8}{45}\\cd+\frac{2}{9}&d^{2}+\frac{4}{9}&\frac{4}{15}\sqrt{5}d+\frac{4}{9}\\\frac{4}{15}\sqrt{5}c-\frac{8}{45}&\frac{4}{15}\sqrt{5}d+\frac{4}{9}&1\end{array}\right]\end{aligned}$$ This requires that $c=\frac{2}{3\sqrt{5}}$, $d=\frac{-5}{3\sqrt{5}}$. $$\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}}\\\frac{2}{3}&0&\frac{-5}{3\sqrt{5}}\\\frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}&-\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{5}}\\\frac{2}{3}&0&\frac{-5}{3\sqrt{5}}\\\frac{2}{3}&\frac{1}{\sqrt{5}}&\frac{4}{15}\sqrt{5}\end{array}\right]^{T}=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$$

习题 131 Exercise 131

Find an orthonormal basis for the span of each of the following sets of vectors.

1. $\left[\begin{array}{r}3\\-4\\0\end{array}\right],\ \left[\begin{array}{r}7\\-1\\0\end{array}\right],\ \left[\begin{array}{r}1\\7\\1\end{array}\right]$

2. $\left[\begin{array}{r}3\\0\\-4\end{array}\right],\ \left[\begin{array}{r}11\\0\\2\end{array}\right],\ \left[\begin{array}{r}1\\1\\7\end{array}\right]$

3. $\left[\begin{array}{r}3\\0\\-4\end{array}\right],\ \left[\begin{array}{r}5\\0\\10\end{array}\right],\ \left[\begin{array}{r}-7\\1\\1\end{array}\right]$

答案 Answer：1. $\left[\begin{array}{c}\frac{3}{5}\\-\frac{4}{5}\\0\end{array}\right],\ \left[\begin{array}{c}\frac{4}{5}\\\frac{3}{5}\\0\end{array}\right],\ \left[\begin{array}{c}0\\0\\1\end{array}\right]$. 2. $\left[\begin{array}{c}\frac{3}{5}\\0\\-\frac{4}{5}\end{array}\right],\ \left[\begin{array}{c}\frac{4}{5}\\0\\\frac{3}{5}\end{array}\right],\ \left[\begin{array}{c}0\\1\\0\end{array}\right]$. 3. $\left[\begin{array}{c}\frac{3}{5}\\0\\-\frac{4}{5}\end{array}\right],\ \left[\begin{array}{c}\frac{4}{5}\\0\\\frac{3}{5}\end{array}\right],\ \left[\begin{array}{c}0\\1\\0\end{array}\right]$

习题 132 Exercise 132

Using the Gram Schmidt process find an orthonormal basis for the following span: $\mathrm{span}\left\{\left[\begin{array}{r}1\\2\\1\end{array}\right],\ \left[\begin{array}{r}2\\-1\\3\end{array}\right],\ \left[\begin{array}{r}1\\0\\0\end{array}\right]\right\}$

答案 Answer：A solution is $\left[\begin{array}{c}\frac{1}{6}\sqrt{6}\\\frac{1}{3}\sqrt{6}\\\frac{1}{6}\sqrt{6}\end{array}\right],\ \left[\begin{array}{c}\frac{3}{10}\sqrt{2}\\-\frac{2}{5}\sqrt{2}\\\frac{1}{2}\sqrt{2}\end{array}\right],\ \left[\begin{array}{c}\frac{7}{15}\sqrt{3}\\-\frac{1}{15}\sqrt{3}\\-\frac{1}{3}\sqrt{3}\end{array}\right]$

习题 133 Exercise 133

Using the Gram Schmidt process find an orthonormal basis for the following span: $\mathrm{span}\left\{\left[\begin{array}{r}1\\2\\1\\0\end{array}\right],\ \left[\begin{array}{r}2\\-1\\3\\1\end{array}\right],\ \left[\begin{array}{r}1\\0\\0\\1\end{array}\right]\right\}$

答案 Answer：Then a solution is $\left[\begin{array}{c}\frac{1}{6}\sqrt{6}\\\frac{1}{3}\sqrt{6}\\\frac{1}{6}\sqrt{6}\\0\end{array}\right],\ \left[\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3}\\-\frac{2}{9}\sqrt{2}\sqrt{3}\\\frac{5}{18}\sqrt{2}\sqrt{3}\\\frac{1}{9}\sqrt{2}\sqrt{3}\end{array}\right],\ \left[\begin{array}{c}\frac{5}{111}\sqrt{3}\sqrt{37}\\\frac{1}{133}\sqrt{3}\sqrt{37}\\-\frac{17}{333}\sqrt{3}\sqrt{37}\\\frac{22}{333}\sqrt{3}\sqrt{37}\end{array}\right]$

习题 134 Exercise 134

The set $V=\left\{\left[\begin{array}{c}x\\y\\z\end{array}\right]:2x+3y-z=0\right\}$ is a subspace of $\mathbb{R}^{3}$. Find an orthonormal basis for this subspace.

答案 Answer：The subspace is of the form $\left[\begin{array}{c}x\\y\\2x+3y\end{array}\right]$ and a basis is $\left[\begin{array}{c}1\\0\\2\end{array}\right],\ \left[\begin{array}{c}0\\1\\3\end{array}\right]$. Therefore, an orthonormal basis is $\left[\begin{array}{c}\frac{1}{5}\sqrt{5}\\0\\\frac{2}{5}\sqrt{5}\end{array}\right],\ \left[\begin{array}{c}-\frac{3}{35}\sqrt{5}\sqrt{14}\\\frac{1}{14}\sqrt{5}\sqrt{14}\\\frac{3}{70}\sqrt{5}\sqrt{14}\end{array}\right]$

习题 135 Exercise 135

Consider the following scalar equation of a plane. $2x-3y+z=0$ Find the orthogonal complement of the vector $\vec{v}=\left[\begin{array}{c}3\\4\\1\end{array}\right]$. Also find the point on the plane which is closest to $(3,4,1)$.

习题 136 Exercise 136

Consider the following scalar equation of a plane. $x+3y+z=0$ Find the orthogonal complement of the vector $\vec{v}=\left[\begin{array}{c}1\\2\\1\end{array}\right]$. Also find the point on the plane which is closest to $(3,4,1)$.

习题 137 Exercise 137

Let $\vec{v}$ be a vector and let $\vec{n}$ be a normal vector for a plane through the origin. Find the equation of the line through the point determined by $\vec{v}$ which has direction vector $\vec{n}$. Show that it intersects the plane at the point determined by $\vec{v}-\text{proj}_{\vec{n}}\vec{v}$. Hint: The line:$\vec{v}+t\vec{n}$. It is in the plane if $\vec{n}\bullet(\vec{v}+t\vec{n})=0$. Determine $t$. Then substitute in to the equation of the line.

习题 138 Exercise 138

As shown in the above problem, one can find the closest point to $\vec{v}$ in a plane through the origin by finding the intersection of the line through $\vec{v}$ having direction vector equal to the normal vector to the plane with the plane. If the plane does not pass through the origin, this will still work to find the point on the plane closest to the point determined by $\vec{v}$. Here is a relation which defines a plane $2x+y+z=11$ and here is a point: $(1,1,2)$. Find the point on the plane which is closest to this point. Then determine the distance from the point to the plane by taking the distance between these two points. Hint: Line: $(x,y,z)=(1,1,2)+t(2,1,1)$. Now require that it intersect the plane.

习题 139 Exercise 139

In general, you have a point $(x_{0},y_{0},z_{0})$ and a scalar equation for a plane $ax+by+cz=d$ where $a^{2}+b^{2}+c^{2}>0$. Determine a formula for the closest point on the plane to the given point. Then use this point to get a formula for the distance from the given point to the plane. Hint: Find the line perpendicular to the plane which goes through the given point: $(x,y,z)=(x_{0},y_{0},z_{0})+t(a,b,c)$. Now require that this point satisfy the equation for the plane to determine $t$.

习题 140 Exercise 140

Find the least squares solution to the following system. $\begin{aligned}x+2y&=1\\2x+3y&=2\\3x+5y&=4\end{aligned}$

答案 Answer：$$\begin{aligned}\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]^{T}\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]&=\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\\ &=\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]^{T}\left[\begin{array}{c}1\\2\\4\end{array}\right]=\left[\begin{array}{c}17\\28\end{array}\right]\end{aligned}$$ $$\begin{aligned}\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]&=\left[\begin{array}{c}17\\28\end{array}\right]\\ \left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]&=\left[\begin{array}{c}17\\28\end{array}\right],\end{aligned}$$ Solution is: $$\left[\begin{array}{c}\frac{2}{3}\\\frac{1}{3}\end{array}\right]$$

习题 141 Exercise 141

You are doing experiments and have obtained the ordered pairs, $(0,1),(1,2),(2,3.5),(3,4)$ Find $m$ and $b$ such that $\vec{y}=m\vec{x}+b$ approximates these four points as well as possible.

习题 142 Exercise 142

Suppose you have several ordered triples, $(x_{i},y_{i},z_{i})$. Describe how to find a polynomial such as $z=a+bx+cy+dxy+ex^{2}+fy^{2}$ giving the best fit to the given ordered triples.

习题 143 Exercise 143

The wind blows from the South at $20$ kilometers per hour and an airplane which flies at $600$ kilometers per hour in still air is heading East. Find the velocity of the airplane and its location after two hours.

习题 144 Exercise 144

The wind blows from the West at $30$ kilometers per hour and an airplane which flies at $400$ kilometers per hour in still air is heading North East. Find the velocity of the airplane and its position after two hours.

习题 145 Exercise 145

The wind blows from the North at $10$ kilometers per hour. An airplane which flies at $300$ kilometers per hour in still air is supposed to go to the point whose coordinates are at $\left(100,100\right).$ In what direction should the airplane fly?

习题 146 Exercise 146

Three forces act on an object. Two are $\left[\begin{array}{r}3\\-1\\-1\end{array}\right]$ and $\left[\begin{array}{r}1\\-3\\4\end{array}\right]$ Newtons. Find the third force if the object is not to move.

习题 147 Exercise 147

Three forces act on an object. Two are $\left[\begin{array}{r}6\\-3\\3\end{array}\right]$ and $\left[\begin{array}{r}2\\1\\3\end{array}\right]$ Newtons. Find the third force if the total force on the object is to be $\left[\begin{array}{r}7\\1\\3\end{array}\right].$

习题 148 Exercise 148

A river flows West at the rate of $b$ miles per hour. A boat can move at the rate of $8$ miles per hour. Find the smallest value of $b$ such that it is not possible for the boat to proceed directly across the river.

习题 149 Exercise 149

The wind blows from West to East at a speed of $50$ miles per hour and an airplane which travels at $400$ miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?

答案 Answer：The velocity is the sum of two vectors. $50\vec{i}+\frac{300}{\sqrt{2}}\left(\vec{i}+\vec{j}\right)=\left(50+\frac{300}{\sqrt{2}}\right)\vec{i}+\frac{300}{\sqrt{2}}\vec{j}.$ The component in the direction of North is then $\frac{300}{\sqrt{2}}=150\sqrt{2}$ and the velocity relative to the ground is $\left(50+\frac{300}{\sqrt{2}}\right)\vec{i}+\frac{300}{\sqrt{2}}\vec{j}$

习题 150 Exercise 150

The wind blows from West to East at a speed of $60$ miles per hour and an airplane can travel travels at $100$ miles per hour in still air. How many degrees West of North should the airplane head in order to travel exactly North?

习题 151 Exercise 151

The wind blows from West to East at a speed of $50$ miles per hour and an airplane which travels at $400$ miles per hour in still air heading somewhat West of North so that, with the wind, it is flying due North. It uses $30.0$ gallons of gas every hour. If it has to travel $600.0$ miles due North, how much gas will it use in flying to its destination?

习题 152 Exercise 152

An airplane is flying due north at $150.0$ miles per hour but it is not actually going due North because there is a wind which is pushing the airplane due east at $40.0$ miles per hour. After one hour, the plane starts flying $30^{\circ}$ East of North. Assuming the plane starts at $\left(0,0\right),$ where is it after $2$ hours? Let North be the direction of the positive $y$ axis and let East be the direction of the positive $x$ axis.

答案 Answer：Velocity of plane for the first hour: $\left[\begin{array}{cc}0&150\end{array}\right]+\left[\begin{array}{cc}40&0\end{array}\right]=\left[\begin{array}{cc}40&150\end{array}\right].$ After one hour it is at $\left(40,150\right).$ Next the velocity of the plane is $150\left[\begin{array}{cc}\frac{1}{2}&\frac{\sqrt{3}}{2}\end{array}\right]+\left[\begin{array}{cc}40&0\end{array}\right]$ in miles per hour. After two hours it is then at $\left(40,150\right)+150\left[\begin{array}{cc}\frac{1}{2}&\frac{\sqrt{3}}{2}\end{array}\right]+\left[\begin{array}{cc}40&0\end{array}\right]=\left[\begin{array}{cc}155&75\sqrt{3}+150\end{array}\right]=\left[\begin{array}{cc}155.0&279.9\end{array}\right]$

习题 153 Exercise 153

City A is located at the origin $\left(0,0\right)$ while city B is located at $\left(300,500\right)$ where distances are in miles. An airplane flies at $250$ miles per hour in still air. This airplane wants to fly from city A to city B but the wind is blowing in the direction of the positive $y$ axis at a speed of $50$ miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city B having flown the shortest possible distance. How long will it take to get there?

答案 Answer：Wind: $\left[\begin{array}{cc}0&50\end{array}\right].$ Direction it needs to travel: $\left(3,5\right)\frac{1}{\sqrt{34}}.$ Then you need $250\left[\begin{array}{cc}a&b\end{array}\right]+\left[\begin{array}{cc}0&50\end{array}\right]$ to have this direction where $\left[\begin{array}{cc}a&b\end{array}\right]$ is an appropriate unit vector. Thus you need $\begin{aligned}a^{2}+b^{2}&=1\\ \frac{250b+50}{250a}&=\frac{5}{3}\end{aligned}$ Thus $a=\frac{3}{5}$, $b=\frac{4}{5}.$ The velocity of the plane relative to the ground is $\left[\begin{array}{cc}150&250\end{array}\right].$ The speed of the plane relative to the ground is given by $\sqrt{\left(150\right)^{2}+\left(250\right)^{2}}=291.55$ miles per hour. It has to go a distance of $\sqrt{\left(300\right)^{2}+\left(500\right)^{2}}=583.10$ miles. Therefore, it takes $\frac{583.1}{291.55}=2$ hours.

习题 154 Exercise 154

A certain river is one half mile wide with a current flowing at $2$ miles per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of $3$ miles per hour. How far down the river does he find himself when he has swam across? How far does he end up traveling?

答案 Answer：Water:$\left[\begin{array}{rr}-2&0\end{array}\right]$ Swimmer:$\left[\begin{array}{rr}0&3\end{array}\right]$ Speed relative to earth: $\left[\begin{array}{rr}-2&3\end{array}\right].$ It takes him $1/6$ of an hour to get across. Therefore, he ends up traveling $\frac{1}{6}\sqrt{4+9}=\frac{1}{6}\sqrt{13}$ miles. He ends up $1/3$ mile down stream.

习题 155 Exercise 155

A certain river is one half mile wide with a current flowing at 2 miles per hour from East to West. A man can swim at $3$ miles per hour in still water. In what direction should he swim in order to travel directly across the river? What would the answer to this problem be if the river flowed at 3 miles per hour and the man could swim only at the rate of 2 miles per hour?

答案 Answer：Man: $3\left[\begin{array}{rr}a&b\end{array}\right]$ Water: $\left[\begin{array}{rr}-2&0\end{array}\right]$ Then you need $3a=2$ and so $a=2/3$ and hence $b=\sqrt{5}/3$. The vector is then $\left[\begin{array}{cc}\frac{2}{3}&\frac{\sqrt{5}}{3}\end{array}\right].$

In the second case, he could not do it. You would need to have a unit vector $\left[\begin{array}{rr}a&b\end{array}\right]$ such that $2a=3$ which is not possible.

习题 156 Exercise 156

Three forces are applied to a point which does not move. Two of the forces are $2\vec{i}+2\vec{j}-6\vec{k}$ Newtons and $8\vec{i}+8\vec{j}+3\vec{k}$ Newtons. Find the third force.

习题 157 Exercise 157

The total force acting on an object is to be $4\vec{i}+2\vec{j}-3\vec{k}$ Newtons. A force of $-3\vec{i}-1\vec{j}+8\vec{k}$ Newtons is being applied. What other force should be applied to achieve the desired total force?

习题 158 Exercise 158

A bird flies from its nest $8$ km in the direction $\frac{5}{6}\pi$ north of east where it stops to rest on a tree. It then flies $1$ km in the direction due southeast and lands atop a telephone pole. Place an $xy$ coordinate system so that the origin is the bird's nest, and the positive $x$ axis points east and the positive $y$ axis points north. Find the displacement vector from the nest to the telephone pole.

习题 159 Exercise 159

If $\vec{F}$ is a force and $\vec{D}$ is a vector, show $\text{proj}_{\vec{D}}\left(\vec{F}\right)=\left(\left\|\vec{F}\right\|\cos\theta\right)\vec{u}$ where $\vec{u}$ is the unit vector in the direction of $\vec{D}$, where $\vec{u}=\vec{D}/\|\vec{D}\|$ and $\theta$ is the included angle between the two vectors, $\vec{F}$ and $\vec{D}$. $\|\vec{F}\|\cos\theta$ is sometimes called the component of the force, $\vec{F}$ in the direction, $\vec{D}$.

答案 Answer：$$\text{proj}_{\vec{D}}\left(\vec{F}\right)=\frac{\vec{F}\bullet\vec{D}}{\|\vec{D}\|}\frac{\vec{D}}{\|\vec{D}\|}=\left(\|\vec{F}\|\cos\theta\right)\frac{\vec{D}}{\|\vec{D}\|}=\left(\|\vec{F}\|\cos\theta\right)\vec{u}$$

习题 160 Exercise 160

A boy drags a sled for $100$ feet along the ground by pulling on a rope which is $20$ degrees from the horizontal with a force of $40$ pounds. How much work does this force do?

答案 Answer：$$40\cos\left(\frac{20}{180}\pi\right)100=3758.8$$

习题 161 Exercise 161

A girl drags a sled for $200$ feet along the ground by pulling on a rope which is $30$ degrees from the horizontal with a force of $20$ pounds. How much work does this force do?

答案 Answer：$$20\cos\left(\frac{\pi}{6}\right)200=3464.1$$

习题 162 Exercise 162

A large dog drags a sled for $300$ feet along the ground by pulling on a rope which is $45$ degrees from the horizontal with a force of $20$ pounds. How much work does this force do?

答案 Answer：$$20\left(\cos\frac{\pi}{4}\right)300=4242.6$$

习题 163 Exercise 163

How much work does it take to slide a crate $20$ meters along a loading dock by pulling on it with a $200$ Newton force at an angle of $30^{\circ}$ from the horizontal? Express your answer in Newton meters.

答案 Answer：$$200\left(\cos\left(\frac{\pi}{6}\right)\right)20=3464.1$$

习题 164 Exercise 164

An object moves $10$ meters in the direction of $\vec{j}$. There are two forces acting on this object, $\vec{F}_{1}=\vec{i}+\vec{j}+2\vec{k}$, and $\vec{F}_{2}=-5\vec{i}+2\vec{j}-6\vec{k}$. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

答案 Answer：$\left[\begin{array}{r}-4\\3\\-4\end{array}\right]\bullet\left[\begin{array}{r}0\\1\\0\end{array}\right]\times 10=30$ You can consider the resultant of the two forces because of the properties of the dot product.

习题 165 Exercise 165

An object moves $10$ meters in the direction of $\vec{j}+\vec{i}$. There are two forces acting on this object, $\vec{F}_{1}=\vec{i}+2\vec{j}+2\vec{k}$, and $\vec{F}_{2}=5\vec{i}+2\vec{j}-6\vec{k}$. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

答案 Answer：$$\begin{aligned}\vec{F}_{1}\bullet\left[\begin{array}{r}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{array}\right]10+\vec{F}_{2}\bullet\left[\begin{array}{r}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{array}\right]10&=\left(\vec{F}_{1}+\vec{F}_{2}\right)\bullet\left[\begin{array}{r}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{array}\right]10\\ &=\left[\begin{array}{r}6\\4\\-4\end{array}\right]\bullet\left[\begin{array}{r}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{array}\right]10\\ &=50\sqrt{2}\end{aligned}$$

习题 166 Exercise 166

An object moves $20$ meters in the direction of $\vec{k}+\vec{j}$. There are two forces acting on this object, $\vec{F}_{1}=\vec{i}+\vec{j}+2\vec{k}$, and $\vec{F}_{2}=\vec{i}+2\vec{j}-6\vec{k}$. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force.

答案 Answer：$$\left[\begin{array}{r}2\\3\\-4\end{array}\right]\bullet\left[\begin{array}{r}0\\\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right]20=-10\sqrt{2}$$