Chapter B

Answers to the Exercises

I believe that every human has a finite number of heart-beats.
I don't intend to waste any of mine running around doing exercises.

— Buzz Aldrin (1930–)

B.1Chapter 1

1. $\mathbf{a}=(-2.5,3)\mathbf{b}=(1,2)\mathbf{c}=(2.5,2)$
   [4pt] $\mathbf{d}=(-1,1)\mathbf{e}=(0,0)\mathbf{f}=(2,-0.5)$
   [4pt] $\mathbf{g}=(-0.5,-1.5)\mathbf{h}=(0,-2)\mathbf{i}=(-3,-2)$
2. $$\begin{gathered} \mathbf{a}=(1,2,4)\mathbf{b}=(-3,-3,-5)\mathbf{c}=(-3,6,2.5) \\ \mathbf{d}=(3,0,-1)\mathbf{e}=(0,0,0)\mathbf{f}=(0,0,3) \\ \mathbf{g}=(-3.5,4,0)\mathbf{h}=(5,-5,-1.5)\mathbf{i}=(4,1,5) \end{gathered}$$
3. See the table below.
   

   Left-handed						Right-handed
   East	Up	North	East	Up	North	East	Up	North	East	Up	North
   $+x$	$+y$	$+z$	$-x$	$-y$	$+z$	$-x$	$-y$	$-z$	$+x$	$+y$	$-z$
   $+x$	$-y$	$-z$	$-x$	$+y$	$-z$	$-x$	$+y$	$+z$	$+x$	$-y$	$+z$
   $+x$	$+z$	$-y$	$-x$	$-z$	$-y$	$-x$	$-z$	$+y$	$+x$	$+z$	$+y$
   $+x$	$-z$	$+y$	$-x$	$+z$	$+y$	$-x$	$+z$	$-y$	$+x$	$-z$	$-y$
   $+y$	$+z$	$+x$	$-y$	$-z$	$+x$	$-y$	$-z$	$-x$	$+y$	$+z$	$-x$
   $+y$	$-z$	$-x$	$-y$	$+z$	$-x$	$-y$	$+z$	$+x$	$+y$	$-z$	$+x$
   $+y$	$+x$	$-z$	$-y$	$-x$	$-z$	$-y$	$-x$	$+z$	$+y$	$+x$	$+z$
   $+y$	$-x$	$+z$	$-y$	$+x$	$+z$	$-y$	$+x$	$-z$	$+y$	$-x$	$-z$
   $+z$	$+x$	$+y$	$-z$	$-x$	$+y$	$-z$	$-x$	$-y$	$+z$	$+x$	$-y$
   $+z$	$-x$	$-y$	$-z$	$+x$	$-y$	$-z$	$+x$	$+y$	$+z$	$-x$	$+y$
   $+z$	$+y$	$-x$	$-z$	$-y$	$-x$	$-z$	$-y$	$+x$	$+z$	$+y$	$+x$
   $+z$	$-y$	$+x$	$-z$	$+y$	$+x$	$-z$	$+y$	$-x$	$+z$	$-y$	$-x$

4. (a) Right-handed.     (b) Swap $y$ and $z$ .     (c) Swap $y$ and $z$ .
5. 1. (a)Right-handed.
   2. (b) $x_{\mathrm{u}\mathrm{s}}\leftarrow y_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}$ ,     $y_{\mathrm{u}\mathrm{s}}\leftarrow -z_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}$ ,     $z_{\mathrm{u}\mathrm{s}}\leftarrow x_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}$
   3. (c) $x_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}\leftarrow z_{\mathrm{u}\mathrm{s}}$ ,     $y_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}\leftarrow x_{\mathrm{u}\mathrm{s}}$ ,     $z_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}\leftarrow -y_{\mathrm{u}\mathrm{s}}$
6. (a) CW     (b) CCW     (c) CCW     (d) CW
7. (a) 15     (b) 30     (c) 3840     (d) 2016840     (e) 5050

8. (a) $\pi /6$	(b) $-\pi /4$	(c) $\pi /3$	(d) $\pi /2$	(e) $-\pi$
   (f) $5\pi /4$	(g) $-3\pi /2$	(h) $2.923$	(i) $9.198$	(j) $-6\pi$

9. (a) $-30{}^{\mathrm{o}}$	(b) $120{}^{\mathrm{o}}$	(c) $270{}^{\mathrm{o}}$	(d) $-240{}^{\mathrm{o}}$	(e) $360{}^{\mathrm{o}}$
   (f) $1{}^{\mathrm{o}}$	(g) $10{}^{\mathrm{o}}$	(h) $-900{}^{\mathrm{o}}$	(i) $1800{}^{\mathrm{o}}$	(j) $36{}^{\mathrm{o}}$

10. The scarecrow should have said:
    The sum of the squares of the legs of a right triangle is equal to the square of the remaining side.
    since the Pythagorean theorem is $c^2=a^2+b^2$ , where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse.
11. 1. (a) $(\sin (\alpha )/\csc (\alpha ))+(\cos (\alpha )/\sec (\alpha ))={\sin }^2(\alpha )+{\cos }^2(\alpha )=1$
    2. (b) $({\sec }^2(\theta )-1)/{\sec }^2(\theta )=1-(1/{\sec }^2(\theta ))=1-{\cos }^2(\theta )={\sin }^2(\theta )$
    3. (c) $1+{\cot }^2(t)=1+({\cos }^2(t)/{\sin }^2(t))=({\sin }^2(t)/{\sin }^2(t))+({\cos }^2(t)/{\sin }^2(t))=({\sin }^2(t)+{\cos }^2(t))/{\sin }^2(t)=1/{\sin }^2(t)={\csc }^2(t)$
    4. (d) $\cos (\varphi )(\tan (\varphi )+\cot (\varphi ))=\sin (\varphi )+({\cos }^2(\varphi )/\sin (\varphi ))=({\sin }^2(\varphi )+{\cos }^2(\varphi ))/\sin (\varphi )=1/\sin (\varphi )=\csc (\varphi )$

B.2Chapter 2

1. 1. $\mathbf{a}$ is a 2D row vector. $\mathbf{b}$ is a 3D column vector. $\mathbf{c}$ is a 4D column vector.
   2. $b_y+c_w+a_x+b_z=0+6+(-3)+5=8$
2. 1. “How much do you weigh?” Your weight is a scalar quantity. But the force of gravity, which pulls you downwards, is a vector, and so if you said that weight was a vector for that reason, you are also correct. (“My weight is 150 lbs of force in the downward direction.”)
   2. “Do you have any idea how fast you were going?” The officer is probably referring to the speed of your vehicle, which is a scalar quantity.
   3. “It's two blocks north of here.” Vector quantity.
   4. “We're cruising from Los Angeles to New York at 600 mph, at an altitude of 33,000 ft.” The speed “600 mph” is a scalar quantity. Since New York is east of Los Angeles, you could reasonably infer an eastward direction, so “600 mph eastward” is a velocity, which is a vector quantity. Likewise, “33,000 ft” is a scalar quantity, although if you're a stickler, you might say that a direction of “up” is implied, in which case “33,000 ft up” is a vector quantity.
3. $\mathbf{a}=[0,2]\mathbf{b}=[0,-2]\mathbf{c}=[0.5,2]$
   [4pt] $\mathbf{d}=[0.5,2]\mathbf{e}=[0.5,-3]\mathbf{f}=[-2,0]$
   [4pt] $\mathbf{g}=[-2,1]\mathbf{h}=[2.5,2]\mathbf{i}=[6,1]$
4. 1. The size of a vector in a diagram doesn't matter; we just need to draw it in the right place. False. This is reversed; for vectors, size matters (meaning the length of the vector), position doesn't.
   2. The displacement expressed by a vector can be visualized as a sequence of axially aligned displacements. True.
   3. These axially aligned displacements from the previous question must occur in order. False. We can apply them in any order and get the same end result.
   4. The vector $[x,y]$ gives the displacement from the point $(x,y)$ to the origin. False. This is reversed; the vector $[x,y]$ gives the displacement from the origin to the point $(x,y)$ .
5. 1. $-\left[\begin{array}{cc}3& 7\end{array}\right]=\left[\begin{array}{cc}-3& -7\end{array}\right]$
   2. $\parallel \left[\begin{array}{cc}-12& 5\end{array}\right]\parallel =\sqrt{(-12{)}^2+5^2}=\sqrt{169}=13$
   3. $$\begin{gathered} \parallel \left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]\parallel =\sqrt{8^2+(-3{)}^2+(1/2{)}^2}=\sqrt{64+9+(1/4)} \\ =\sqrt{293/4}\approx 8.56 \end{gathered}$$
   4. $3\left[\begin{array}{ccc}4& -7& 0\end{array}\right]=\left[\begin{array}{ccc}(3)(4)& (3)(-7)& (3)(0)\end{array}\right]=\left[\begin{array}{ccc}12& -21& 0\end{array}\right]$
   5. $\left[\begin{array}{cc}4& 5\end{array}\right]/2=\left[\begin{array}{cc}2& 5/2\end{array}\right]$
6. 1. $$\begin{gathered} {\left[\begin{array}{cc}12& 5\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{cc}12& 5\end{array}\right]}{\parallel \left[\begin{array}{cc}12& 5\end{array}\right]\parallel }=\frac{\left[\begin{array}{cc}12& 5\end{array}\right]}{13}=\left[\begin{array}{cc}{\displaystyle \frac{12}{13}}& {\displaystyle \frac{5}{13}}\end{array}\right] \\ \approx \left[\begin{array}{cc}0.923& 0.385\end{array}\right] \end{gathered}$$
   2. $$\begin{gathered} {\left[\begin{array}{cc}0& 743.632\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{cc}0& 743.632\end{array}\right]}{\parallel \left[\begin{array}{cc}0& 743.632\end{array}\right]\parallel }=\frac{\left[\begin{array}{cc}0& 743.632\end{array}\right]}{\sqrt{0^2+{743.632}^2}} \\ =\frac{\left[\begin{array}{cc}0& 743.632\end{array}\right]}{743.632}=\left[\begin{array}{cc}0& 1\end{array}\right] \end{gathered}$$
   3. $$\begin{gathered} {\left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]}{\parallel \left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]\parallel }\approx \frac{\left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]}{8.56} \\ \approx \left[\begin{array}{ccc}0.935& -0.350& 0.058\end{array}\right] \end{gathered}$$
   4. $$\begin{gathered} {\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}{\parallel \left[\begin{array}{ccc}-12& 3& -4\end{array}\right]\parallel }=\frac{\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}{\sqrt{(-12{)}^2+3^2+(-4{)}^2}} \\ =\frac{\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}{13}=\left[\begin{array}{ccc}{\displaystyle \frac{-12}{13}}& {\displaystyle \frac{3}{13}}& {\displaystyle \frac{-4}{13}}\end{array}\right] \end{gathered}$$
   5. $$\begin{gathered} {\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}{\parallel \left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]\parallel }=\frac{\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}{\sqrt{1^2+1^2+1^2+1^2}} \\ =\frac{\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}{2}=\left[\begin{array}{cccc}0.5& 0.5& 0.5& 0.5\end{array}\right] \end{gathered}$$
7. 1. $\left[\begin{array}{ccc}7& -2& -3\end{array}\right]+\left[\begin{array}{ccc}6& 6& -4\end{array}\right]=\left[\begin{array}{ccc}7+6& -2+6& -3+(-4)\end{array}\right]=\left[\begin{array}{ccc}13& 4& -7\end{array}\right]$
   2. $\left[\begin{array}{ccc}2& 9& -1\end{array}\right]+\left[\begin{array}{ccc}-2& -9& 1\end{array}\right]=\left[\begin{array}{ccc}2+(-2)& 9+(-9)& -1+1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
   3. $\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right]-\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]=\left[\begin{array}{c}3-8\\ 10-(-7)\\ 7-4\end{array}\right]=\left[\begin{array}{c}-5\\ 17\\ 3\end{array}\right]$
   4. $\left[\begin{array}{c}4\\ 5\\ -11\end{array}\right]-\left[\begin{array}{c}-4\\ -5\\ 11\end{array}\right]=\left[\begin{array}{c}4-(-4)\\ 5-(-5)\\ -11-11\end{array}\right]=\left[\begin{array}{c}8\\ 10\\ -22\end{array}\right]$
   5. $3\left[\begin{array}{c}a\\ b\\ c\end{array}\right]-4\left[\begin{array}{c}2\\ 10\\ -6\end{array}\right]=\left[\begin{array}{c}3a\\ 3b\\ 3c\end{array}\right]-\left[\begin{array}{c}8\\ 40\\ -24\end{array}\right]=\left[\begin{array}{c}3a-8\\ 3b-40\\ 3c+24\end{array}\right]$
8. 1. $$\begin{gathered} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}(\left[\begin{array}{c}10\\ 6\end{array}\right],\left[\begin{array}{c}-14\\ 30\end{array}\right])=\sqrt{(10-(-14){)}^2+(6-30{)}^2} \\ =\sqrt{{24}^2+(-24{)}^2}=\sqrt{576+576} \\ =\sqrt{1152}\approx 33.94 \end{gathered}$$
   2. $$\begin{gathered} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{c}-12\\ 5\end{array}\right])=\sqrt{(0-(-12){)}^2+(0-5{)}^2} \\ =\sqrt{{12}^2+(-5{)}^2}=\sqrt{144+25} \\ =\sqrt{169}=13 \end{gathered}$$
   3. $$\begin{gathered} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}(\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right],\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right])=\sqrt{(3-8{)}^2+(10-(-7){)}^2+(7-4{)}^2} \\ =\sqrt{(-5{)}^2+{17}^2+3^2}=\sqrt{25+289+9} \\ =\sqrt{323}\approx 17.97 \end{gathered}$$
   4. $$\begin{gathered} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}(\left[\begin{array}{c}-2\\ -4\\ 9\end{array}\right],\left[\begin{array}{c}6\\ -7\\ 9.5\end{array}\right])=\sqrt{(6-(-2){)}^2+(-7-(-4){)}^2+(9.5-9{)}^2} \\ =\sqrt{8^2+(-3{)}^2+(0.5{)}^2}=\sqrt{64+9+0.25} \\ =\sqrt{73.25}\approx 8.56 \end{gathered}$$
   5. $$\begin{gathered} \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}(\left[\begin{array}{c}4\\ -4\\ -4\\ 4\end{array}\right],\left[\begin{array}{c}-6\\ 6\\ 6\\ -6\end{array}\right])=\sqrt{(-6-4{)}^2+(6-(-4){)}^2+(6-(-4){)}^2+(-6-4{)}^2} \\ =\sqrt{(-10{)}^2+(10{)}^2+(10{)}^2+(-10{)}^2} \\ =\sqrt{100+100+100+100} \\ =\sqrt{400}=20 \end{gathered}$$
9. 1. $\left[\begin{array}{c}2\\ 6\end{array}\right]\cdot \left[\begin{array}{c}-3\\ 8\end{array}\right]=(2)(-3)+(6)(8)=-6+48=42$
   2. $$\begin{gathered} -7\left[\begin{array}{cc}1& 2\end{array}\right]\cdot \left[\begin{array}{cc}11& -4\end{array}\right]=\left[\begin{array}{cc}-7& -14\end{array}\right]\cdot \left[\begin{array}{cc}11& -4\end{array}\right] \\ =(-7)(11)+(-14)(-4) \\ =-21 \end{gathered}$$
   3. $$\begin{gathered} 10+\left[\begin{array}{c}-5\\ 1\\ 3\end{array}\right]\cdot \left[\begin{array}{c}4\\ -13\\ 9\end{array}\right]=10+((-5)(4)+(1)(-13)+(3)(9)) \\ =10+(-20+(-13)+27) \\ =10+(-6)=4 \end{gathered}$$
   4. $$\begin{gathered} 3\left[\begin{array}{c}-2\\ 0\\ 4\end{array}\right]\cdot (\left[\begin{array}{c}8\\ -2\\ 3/2\end{array}\right]+\left[\begin{array}{c}0\\ 9\\ 7\end{array}\right])=\left[\begin{array}{c}-6\\ 0\\ 12\end{array}\right]\cdot \left[\begin{array}{c}8\\ 7\\ 17/2\end{array}\right] \\ =(-6)(8)+(0)(7)+(12)(17/2)=54 \end{gathered}$$
10. $$\begin{gathered} {\mathbf{v}}_{\parallel }=\hat{\mathbf{n}}\frac{\mathbf{v}\cdot \hat{\mathbf{n}}}{{\parallel \hat{\mathbf{n}}\parallel }^2}=\hat{\mathbf{n}}\frac{\mathbf{v}\cdot \hat{\mathbf{n}}}{1}=\hat{\mathbf{n}}(\mathbf{v}\cdot \hat{\mathbf{n}}) \\ =\left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right](\left[\begin{array}{c}4\\ 3\\ -1\end{array}\right]\cdot \left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right])=\left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right](2\sqrt{2}+\frac{3\sqrt{2}}{2}+0) \\ =\left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right]\frac{7\sqrt{2}}{2}=\left[\begin{array}{c}7/2\\ 7/2\\ 0\end{array}\right] \end{gathered}$$
    $$\begin{gathered} {\mathbf{v}}_{\perp }=\mathbf{v}-{\mathbf{v}}_{\parallel } \\ =\left[\begin{array}{c}4\\ 3\\ -1\end{array}\right]-\left[\begin{array}{c}7/2\\ 7/2\\ 0\end{array}\right]=\left[\begin{array}{c}4-7/2\\ 3-7/2\\ -1-0\end{array}\right]=\left[\begin{array}{c}1/2\\ -1/2\\ -1\end{array}\right] \end{gathered}$$
11. Define a triangle using the vectors $\mathbf{a}$ , $\mathbf{b}$ , and $\mathbf{a}-\mathbf{b}$ , and let $\theta$ be the angle between $\mathbf{a}$ and $\mathbf{b}$ . Then the squared length of the edge $\mathbf{a}-\mathbf{b}$ is:
    $$\begin{array}{rcl}{\parallel \mathbf{a}-\mathbf{b}\parallel }^2& =& (\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})\\ & =& \mathbf{a}\cdot \mathbf{a}-2\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{b}\\ & =& \mathbf{a}\cdot \mathbf{a}+\mathbf{b}\cdot \mathbf{b}-2\mathbf{a}\cdot \mathbf{b}\\ & =& {\parallel \mathbf{a}\parallel }^2+{\parallel \mathbf{b}\parallel }^2-2\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \cos \theta \end{array}$$
    which is the law of cosines.
12. First, let's obtain some information about the vector components.
    
    From the figure, we have
    $$\begin{array}{rlrl}{a}_x& =\parallel \mathbf{a}\parallel \cos {\theta }_a,& a_y& =\parallel \mathbf{a}\parallel \sin {\theta }_a,\\ b_x& =\parallel \mathbf{b}\parallel \cos {\theta }_b,& b_y& =\parallel \mathbf{b}\parallel \sin {\theta }_b.\end{array}$$
    Now we can proceed with the algebraic definition of the dot product and the cosine difference identity:
    $$\begin{array}{rl}\mathbf{a}\cdot \mathbf{b}& =a_x{b}_x+a_y{b}_y\\ & =\parallel \mathbf{a}\parallel \cos {\theta }_a\parallel \mathbf{b}\parallel \cos {\theta }_b+\parallel \mathbf{a}\parallel \sin {\theta }_a\parallel \mathbf{b}\parallel \sin {\theta }_b\\ & =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel (\cos {\theta }_a\cos {\theta }_b+\sin {\theta }_a\sin {\theta }_b)\\ & =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \cos ({\theta }_b-{\theta }_a)\\ & =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \cos \theta .\end{array}$$
13. 1. $\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]\times \left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}(-1)(1)-(0)(0)\\ (0)(0)-(0)(1)\\ (0)(0)-(-1)(0)\end{array}\right]=\left[\begin{array}{c}-1-0\\ 0-0\\ 0-0\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 0\end{array}\right]$
       $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\times \left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}(0)(0)-(1)(-1)\\ (1)(0)-(0)(0)\\ (0)(-1)-(0)(0)\end{array}\right]=\left[\begin{array}{c}0-(-1)\\ 0-0\\ 0-0\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$
    2. $\left[\begin{array}{c}-2\\ 4\\ 1\end{array}\right]\times \left[\begin{array}{c}1\\ -2\\ -1\end{array}\right]=\left[\begin{array}{c}(4)(-1)-(1)(-2)\\ (1)(1)-(-2)(-1)\\ (-2)(-2)-(4)(1)\end{array}\right]=\left[\begin{array}{c}-4-(-2)\\ 1-2\\ 4-4\end{array}\right]=\left[\begin{array}{c}-2\\ -1\\ 0\end{array}\right]$
       $\left[\begin{array}{c}1\\ -2\\ -1\end{array}\right]\times \left[\begin{array}{c}-2\\ 4\\ 1\end{array}\right]=\left[\begin{array}{c}(-2)(1)-(-1)(4)\\ (-1)(-2)-(1)(1)\\ (1)(4)-(-2)(-2)\end{array}\right]=\left[\begin{array}{c}-2-(-4)\\ 2-1\\ 4-4\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]$
    3. $\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right]\times \left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]=\left[\begin{array}{c}(10)(4)-(7)(-7)\\ (7)(8)-(3)(4)\\ (3)(-7)-(10)(8)\end{array}\right]=\left[\begin{array}{c}40-(-49)\\ 56-12\\ -21-80\end{array}\right]=\left[\begin{array}{c}89\\ 44\\ -101\end{array}\right]$
       $\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]\times \left[\begin{array}{c}3\\ 10\\ 7\end{array}\right]=\left[\begin{array}{c}(-7)(7)-(4)(10)\\ (4)(3)-(8)(7)\\ (8)(10)-(-7)(3)\end{array}\right]=\left[\begin{array}{c}-49-40\\ 12-56\\ 80-(-21)\end{array}\right]=\left[\begin{array}{c}-89\\ -44\\ 101\end{array}\right]$
14. Let $\mathbf{a}=\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]$ and $\mathbf{b}=\left[\begin{array}{c}{b}_x\\ b_y\\ b_z\end{array}\right]$ . Then $\mathbf{a}\cdot \mathbf{b}=\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \cos \theta$ and $\mathbf{a}\times \mathbf{b}=$
    $\left[\begin{array}{c}{a}_y{b}_z-a_z{b}_y\\ a_z{b}_x-a_x{b}_z\\ a_x{b}_y-a_y{b}_x\end{array}\right]$ . From $\parallel \mathbf{a}\times \mathbf{b}\parallel$ , we have:
    $$\begin{array}{rl}& \parallel \mathbf{a}\times \mathbf{b}\parallel =\sqrt{(a_y{b}_z-a_z{b}_y{)}^2+(a_z{b}_x-a_x{b}_z{)}^2+(a_x{b}_y-a_y{b}_x{)}^2}\\ & =\sqrt{a_y^2{b}_z^2-2a_y{a}_z{b}_y{b}_z+a_z^2{b}_y^2+a_z^2{b}_x^2-2a_x{a}_z{b}_x{b}_z+a_x^2{b}_z^2+a_x^2{b}_y^2-2a_x{a}_y{b}_x{b}_y+a_y^2{b}_x^2}.\end{array}$$
    If we now consider $\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \sin \theta$ , we find that:
    $$\begin{array}{rl}& \parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \sin \theta =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \sqrt{1-{\cos }^2\theta }\\ & =\sqrt{a_x^2+a_y^2+a_z^2}\sqrt{b_x^2+b_y^2+b_z^2}\sqrt{1-{\left(\frac{a_x{b}_x+a_y{b}_y+a_z{b}_z}{\sqrt{a_x^2+a_y^2+a_z^2}\sqrt{b_x^2+b_y^2+b_z^2}}\right)}^2}\end{array}$$
    $$\begin{array}{rl}& =\sqrt{(a_x^2+a_y^2+a_z^2)(b_x^2+b_y^2+b_z^2)(1-\frac{(a_x{b}_x+a_y{b}_y+a_z{b}_z{)}^2}{(a_x^2+a_y^2+a_z^2)(b_x^2+b_y^2+b_z^2)})}\\ & =\sqrt{(a_x^2+a_y^2+a_z^2)(b_x^2+b_y^2+b_z^2)-(a_x{b}_x+a_y{b}_y+a_z{b}_z{)}^2}\\ & =\sqrt{a_y^2{b}_z^2-2a_y{a}_z{b}_y{b}_z+a_z^2{b}_y^2+a_z^2{b}_x^2-2a_x{a}_z{b}_x{b}_z+a_x^2{b}_z^2+a_x^2{b}_y^2-2a_x{a}_y{b}_x{b}_y+a_y^2{b}_x^2}.\end{array}$$
    Starting from both ends, we have met in the middle, proving that
    $\parallel \mathbf{a}\times \mathbf{b}\parallel =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \sin \theta$ .
15. 1. (a)
       1. (1) $$\begin{gathered} {\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_1=\left|3\right|+\left|4\right|=7 \\ {\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_2=\sqrt{{\left|3\right|}^2+{\left|4\right|}^2}=5 \\ {\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_3=\sqrt[3]{{\left|3\right|}^3+{\left|4\right|}^3}=\sqrt[3]{91}\approx 4.498 \\ {\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_{\mathrm{\infty }}=max(\left|3\right|,\left|4\right|)=4 \end{gathered}$$
       2. (2) $$\begin{gathered} {\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_1=\left|5\right|+\left|-12\right|=17 \\ {\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_2=\sqrt{{\left|5\right|}^2+{\left|-12\right|}^2}=13 \\ {\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_3=\sqrt[3]{{\left|5\right|}^3+{\left|-12\right|}^3}=\sqrt[3]{1853}\approx 12.283 \\ {\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_{\mathrm{\infty }}=max(\left|5\right|,\left|-12\right|)=12 \end{gathered}$$
       3. (3) $$\begin{gathered} {\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_1=\left|-2\right|+\left|10\right|+\left|-7\right|=19 \\ {\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_2=\sqrt{{\left|-2\right|}^2+{\left|10\right|}^2+{\left|-7\right|}^2}=\sqrt{153}\approx 12.369 \\ {\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_3=\sqrt[3]{{\left|-2\right|}^3+{\left|10\right|}^3+{\left|-7\right|}^3}=\sqrt[3]{1351}\approx 11.055 \\ {\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_{\mathrm{\infty }}=max(\left|-2\right|,\left|10\right|,\left|-7\right|)=10 \end{gathered}$$
       4. (4) $$\begin{gathered} {\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_1=\left|6\right|+\left|1\right|+\left|-9\right|=16 \\ {\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_2=\sqrt{{\left|6\right|}^2+{\left|1\right|}^2+{\left|-9\right|}^2}=\sqrt{118}\approx 10.863 \\ {\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_3=\sqrt[3]{{\left|6\right|}^3+{\left|1\right|}^3+{\left|-9\right|}^3}=\sqrt[3]{946}\approx 9.817 \\ {\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_{\mathrm{\infty }}=max(\left|6\right|,\left|1\right|,\left|-9\right|)=9 \end{gathered}$$
       5. (5) $$\begin{gathered} {\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_1=\left|-2\right|+\left|-2\right|+\left|-2\right|+\left|-2\right|=8 \\ {\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_2=\sqrt{{\left|-2\right|}^2+{\left|-2\right|}^2+{\left|-2\right|}^2+{\left|-2\right|}^2}=4 \\ {\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_3=\sqrt[3]{{\left|-2\right|}^3+{\left|-2\right|}^3+{\left|-2\right|}^3+{\left|-2\right|}^3}=\sqrt[3]{32}\approx 3.175 \\ {\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_{\mathrm{\infty }}=max(\left|-2\right|,\left|-2\right|,\left|-2\right|,\left|-2\right|)=2 \end{gathered}$$
    2. (b)
       1. (1)The unit circle for the $L^1$ norm is a square with sides of length $\sqrt{2}$ rotated by ${45}^{\circ }$ .
       2. (2)The unit circle for the $L^2$ norm is the well-known unit circle we all know and love.
       3. (3)The unit circle for the infinity norm is a square with sides of length $2$ .
       Note that all three unit circles include the vectors $[1,0]$ , $[0,1]$ , $[-1,0]$ , $[0,-1]$ .
16. The man buys a box or has a piece of luggage that is 2 feet long, 2 feet wide, and 2 feet tall. If the object is very thin, such as a sword, then he can put the object diagonally in the box or luggage. The longest such object he could carry on is $\sqrt{2^2+2^2+2^2}\approx 3.46$ feet.
17. Let $\mathbf{s}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}+\mathbf{e}+\mathbf{f}$ . From inspection of Figure 2.11 we see that
    $$\mathbf{s}=\left[\begin{array}{c}-5\\ 3\end{array}\right].$$
    We confirm this numerically using the above equation and the values of the other vectors, also obtained from inspection of Figure 2.11:
    $$\begin{array}{rcl}\mathbf{s}& =& \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}+\mathbf{e}+\mathbf{f}\\ & =& \left[\begin{array}{c}-1\\ 3\end{array}\right]+\left[\begin{array}{c}1\\ 3\end{array}\right]+\left[\begin{array}{c}3\\ -2\end{array}\right]+\left[\begin{array}{c}-1\\ -2\end{array}\right]+\left[\begin{array}{c}-6\\ 4\end{array}\right]+\left[\begin{array}{c}-1\\ -3\end{array}\right]\\ & =& \left[\begin{array}{c}(-1)+1+3+(-1)+(-6)+(-1)\\ 3+3+(-2)+(-2)+4+(-3)\end{array}\right]\\ & =& \left[\begin{array}{c}-5\\ 3\end{array}\right]\end{array}$$
18. Left-handed.
19. 1. Let $\mathbf{c}=\left[\begin{array}{c}{c}_x\\ c_y\end{array}\right]$ and $\mathbf{r}=\left[\begin{array}{c}{r}_x\\ r_y\end{array}\right]$ . Then
       $$\begin{array}{rlrl}{\mathbf{p}}_{\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}& =\left[\begin{array}{c}{c}_x-r_x\\ c_y+r_y\end{array}\right],& {\mathbf{p}}_{\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}& =\left[\begin{array}{c}{c}_x+r_x\\ c_y+r_y\end{array}\right],\\ {\mathbf{p}}_{\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}& =\left[\begin{array}{c}{c}_x-r_x\\ c_y-r_y\end{array}\right],& {\mathbf{p}}_{\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}& =\left[\begin{array}{c}{c}_x+r_x\\ c_y-r_y\end{array}\right].\end{array}$$
    2. Let $\mathbf{c}=\left[\begin{array}{c}{c}_x\\ c_y\\ c_z\end{array}\right]$ and $\mathbf{r}=\left[\begin{array}{c}{r}_x\\ r_y\\ r_z\end{array}\right]$ . Then
       $$\begin{array}{rl}{\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_x-r_x\\ c_y+r_y\\ c_z+r_z\end{array}\right],& {\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_x+r_x\\ c_y+r_y\\ c_z+r_z\end{array}\right],\\ {\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_x-r_x\\ c_y-r_y\\ c_z+r_z\end{array}\right],& {\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_x+r_x\\ c_y-r_y\\ c_z+r_z\end{array}\right],\end{array}$$
       $$\begin{array}{rl}{\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_x-r_x\\ c_y+r_y\\ c_z-r_z\end{array}\right],& {\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_x+r_x\\ c_y+r_y\\ c_z-r_z\end{array}\right],\\ {\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_x-r_x\\ c_y-r_y\\ c_z-r_z\end{array}\right],& {\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_x+r_x\\ c_y-r_y\\ c_z-r_z\end{array}\right].\end{array}$$
20. 1. Use the sign of the dot product between $\mathbf{v}$ and $\mathbf{x}-\mathbf{p}$ to determine whether the point $\mathbf{x}$ is in front of or behind the NPC. This follows from the geometric interpretation of the dot product,
       $$\begin{array}{r}\mathbf{v}\cdot (\mathbf{x}-\mathbf{p})=\parallel \mathbf{v}\parallel \parallel \mathbf{x}-\mathbf{p}\parallel \cos \theta ,\end{array}$$
       where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{x}-\mathbf{p}$ .
       Both $\parallel \mathbf{v}\parallel$ and $\parallel \mathbf{x}-\mathbf{p}\parallel$ are always positive, leaving the sign of the dot product entirely up to the value of $\cos \theta$ . If $\cos \theta >0$ then $\theta$ is less than ${90}^{\circ }$ and $\mathbf{x}$ is in front of the NPC. Similarly, if $\cos \theta <0$ then $\theta$ is greater than ${90}^{\circ }$ and $\mathbf{x}$ is behind the NPC.
       The special case of $\mathbf{v}\cdot (\mathbf{x}-\mathbf{p})=0$ means that $\mathbf{x}$ lies either directly to the left or right of the NPC. If this case does not need to be handled explicitly, it can arbitrarily be assigned to mean either in front of or behind.
    2. 1. (1) $\mathbf{x}$ is in front of the NPC.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}0\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}3\\ -4\end{array}\right]=(5)(3)+(-2)(-4)=23\end{array}$$
       2. (2) $\mathbf{x}$ is in front of the NPC.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}1\\ 6\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}4\\ 2\end{array}\right]=(5)(4)+(-2)(2)=16\end{array}$$
       3. (3) $\mathbf{x}$ is behind the NPC.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-6\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-3\\ -4\end{array}\right]=(5)(-3)+(-2)(-4)\\ & =-7\end{array}$$
       4. (4) $\mathbf{x}$ is behind the NPC.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-4\\ 7\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 3\end{array}\right]=(5)(-1)+(-2)(3)=-11\end{array}$$
       5. (5) $\mathbf{x}$ is in front of the NPC.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}5\\ 5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}8\\ 1\end{array}\right]=(5)(8)+(-2)(1)=38\end{array}$$
       6. (6) $\mathbf{x}$ is in front of the NPC.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-3\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}0\\ -4\end{array}\right]=(5)(0)+(-2)(-4)=8\end{array}$$
       7. (7) $\mathbf{x}$ can be either in front of or behind the NPC, depending on how we've decided to handle this special case.
          $$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-6\\ -3.5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-3\\ -7.5\end{array}\right]=(5)(-3)+(-2)(-7.5)=0\end{array}$$
21. 1. To determine whether the point $\mathbf{x}$ is visible to the NPC, compare $\cos \theta$ to $\cos (\varphi /2)$ . If $\cos \theta \ge \cos (\varphi /2)$ , then $\mathbf{x}$ is visible to the NPC.
       The value of $\cos (\varphi /2)$ can be obtained from the FOV angle. To get $\cos \theta$ use the dot product
       $$\cos \theta =\frac{\mathbf{v}\cdot (\mathbf{x}-\mathbf{p})}{\parallel \mathbf{v}\parallel \parallel \mathbf{x}-\mathbf{p}\parallel }.$$
    2. The NPC's FOV is ${90}^{\circ }$ , so the value we are interested in is $\cos ({45}^{\circ })\approx 0.707$ .
       1. (1) $\mathbf{x}$ is visible to the NPC.
          $$\begin{array}{r}\cos \theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}0\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}0\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{23}{(\sqrt{29})(\sqrt{(}25)}\approx 0.854\ge 0.707\end{array}$$
       2. (2) $\mathbf{x}$ is not visible to the NPC.
          $$\begin{array}{r}\cos \theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}1\\ 6\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}1\\ 6\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{16}{(\sqrt{29})(\sqrt{20})}\approx 0.664<0.707\end{array}$$
       3. (3) $\mathbf{x}$ is not visible to the NPC.
          $$\begin{array}{rl}\cos \theta & =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-6\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}-6\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{-7}{(\sqrt{29})(\sqrt{25})}\\ & \approx -0.260<0.707\end{array}$$
       4. (4) $\mathbf{x}$ is not visible to the NPC.
          $$\begin{array}{r}\cos \theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-4\\ 7\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}-4\\ 7\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{-11}{(\sqrt{29})(\sqrt{10})}\approx -0.646<0.707\end{array}$$
       5. (5) $\mathbf{x}$ is visible to the NPC.
          $$\begin{array}{r}\cos \theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}5\\ 5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}5\\ 5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{38}{(\sqrt{29})(\sqrt{65})}\approx 0.875\ge 0.707\end{array}$$
       6. (6) $\mathbf{x}$ is not visible to the NPC.
          $$\begin{array}{r}\cos \theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot (\left[\begin{array}{c}-3\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right])}{\parallel [}\end{array}$$