Chapter B
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. $\mathbf{a}=\left(-2.5,3\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathbf{b}=\left(1,2\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\mathbf{c}=\left(2.5,2\right)$
[4pt] $\mathbf{d}=\left(-1,1\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\mathbf{e}=\left(0,0\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\mathbf{f}=\left(2,-0.5\right)$
[4pt] $\mathbf{g}=\left(-0.5,-1.5\right)\phantom{\rule{2em}{0ex}}\mathbf{h}=\left(0,-2\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathbf{i}=\left(-3,-2\right)$
2. $\mathbf{a}=\left(1,2,4\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathbf{b}=\left(-3,-3,-5\right)\phantom{\rule{2em}{0ex}}\mathbf{c}=\left(-3,6,2.5\right)\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{d}=\left(3,0,-1\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1pt}{0ex}}\mathbf{e}=\left(0,0,0\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1pt}{0ex}}\mathbf{f}=\left(0,0,3\right)\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{g}=\left(-3.5,4,0\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathbf{h}=\left(5,-5,-1.5\right)\phantom{\rule{2em}{0ex}}\mathbf{i}=\left(4,1,5\right)$
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 $z$ .
1. (a)Right-handed.
2. (b) ${x}_{\mathrm{u}\mathrm{s}}←{y}_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}$ ,     ${y}_{\mathrm{u}\mathrm{s}}←-{z}_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}$ ,     ${z}_{\mathrm{u}\mathrm{s}}←{x}_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}$
3. (c) ${x}_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}←{z}_{\mathrm{u}\mathrm{s}}$ ,     ${y}_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}←{x}_{\mathrm{u}\mathrm{s}}$ ,     ${z}_{\mathrm{a}\mathrm{e}\mathrm{r}\mathrm{o}}←-{y}_{\mathrm{u}\mathrm{s}}$
5. (a) CW     (b) CCW     (c) CCW     (d) CW
6. (a) 15     (b) 30     (c) 3840     (d) 2016840     (e) 5050
7.  (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) $-6\pi$
8.  (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) $36{}^{\mathrm{o}}$
9. 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.
1. (a) $\left(\mathrm{sin}\left(\alpha \right)/\mathrm{csc}\left(\alpha \right)\right)+\left(\mathrm{cos}\left(\alpha \right)/\mathrm{sec}\left(\alpha \right)\right)={\mathrm{sin}}^{2}\left(\alpha \right)+{\mathrm{cos}}^{2}\left(\alpha \right)=1$
2. (b) $\left({\mathrm{sec}}^{2}\left(\theta \right)-1\right)/{\mathrm{sec}}^{2}\left(\theta \right)=1-\left(1/{\mathrm{sec}}^{2}\left(\theta \right)\right)=1-{\mathrm{cos}}^{2}\left(\theta \right)={\mathrm{sin}}^{2}\left(\theta \right)$
3. (c) $1+{\mathrm{cot}}^{2}\left(t\right)=1+\left({\mathrm{cos}}^{2}\left(t\right)/{\mathrm{sin}}^{2}\left(t\right)\right)=\left({\mathrm{sin}}^{2}\left(t\right)/{\mathrm{sin}}^{2}\left(t\right)\right)+\left({\mathrm{cos}}^{2}\left(t\right)/{\mathrm{sin}}^{2}\left(t\right)\right)=\left({\mathrm{sin}}^{2}\left(t\right)+{\mathrm{cos}}^{2}\left(t\right)\right)/{\mathrm{sin}}^{2}\left(t\right)=1/{\mathrm{sin}}^{2}\left(t\right)={\mathrm{csc}}^{2}\left(t\right)$
4. (d) $\mathrm{cos}\left(\varphi \right)\left(\mathrm{tan}\left(\varphi \right)+\mathrm{cot}\left(\varphi \right)\right)=\mathrm{sin}\left(\varphi \right)+\left({\mathrm{cos}}^{2}\left(\varphi \right)/\mathrm{sin}\left(\varphi \right)\right)=\left({\mathrm{sin}}^{2}\left(\varphi \right)+{\mathrm{cos}}^{2}\left(\varphi \right)\right)/\mathrm{sin}\left(\varphi \right)=1/\mathrm{sin}\left(\varphi \right)=\mathrm{csc}\left(\varphi \right)$

# B.2Chapter 2

1. $\mathbf{a}$ is a 2D row vector. $\mathbf{b}$ is a 3D column vector. $\mathbf{c}$ is a 4D column vector.
2. ${b}_{y}+{c}_{w}+{a}_{x}+{b}_{z}=0+6+\left(-3\right)+5=8$
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.
1. $\mathbf{a}=\left[0,2\right]\phantom{\rule{19pt}{0ex}}\mathbf{b}=\left[0,-2\right]\phantom{\rule{18pt}{0ex}}\mathbf{c}=\left[0.5,2\right]$
[4pt] $\mathbf{d}=\left[0.5,2\right]\phantom{\rule{12pt}{0ex}}\mathbf{e}=\left[0.5,-3\right]\phantom{\rule{12pt}{0ex}}\mathbf{f}=\left[-2,0\right]$
[4pt] $\mathbf{g}=\left[-2,1\right]\phantom{\rule{12pt}{0ex}}\mathbf{h}=\left[2.5,2\right]\phantom{\rule{18pt}{0ex}}\mathbf{i}=\left[6,1\right]$
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 $\left[x,y\right]$ gives the displacement from the point $\left(x,y\right)$ to the origin. False. This is reversed; the vector $\left[x,y\right]$ gives the displacement from the origin to the point $\left(x,y\right)$ .
1. $-\left[\begin{array}{cc}3& 7\end{array}\right]=\left[\begin{array}{cc}-3& -7\end{array}\right]$
2. $\parallel \left[\begin{array}{cc}-12& 5\end{array}\right]\parallel =\sqrt{\left(-12{\right)}^{2}+{5}^{2}}=\sqrt{169}=13$
3. $\parallel \left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]\parallel =\sqrt{{8}^{2}+\left(-3{\right)}^{2}+\left(1/2{\right)}^{2}}=\sqrt{64+9+\left(1/4\right)}\phantom{\rule[-6pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{5.6pc}{0ex}}=\sqrt{293/4}\approx 8.56$
4. $3\left[\begin{array}{ccc}4& -7& 0\end{array}\right]=\left[\begin{array}{ccc}\left(3\right)\left(4\right)& \left(3\right)\left(-7\right)& \left(3\right)\left(0\right)\end{array}\right]=\left[\begin{array}{ccc}12& -21& 0\end{array}\right]$
5. $\left[\begin{array}{cc}4& 5\end{array}\right]/2=\left[\begin{array}{cc}2& 5/2\end{array}\right]$
1. ${\left[\begin{array}{cc}12& 5\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{cc}12& 5\end{array}\right]}{\parallel \left[\begin{array}{cc}12& 5\end{array}\right]\parallel }=\frac{\left[\begin{array}{cc}12& 5\end{array}\right]}{13}=\left[\begin{array}{cc}\frac{12}{13}& \frac{5}{13}\end{array}\right]\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{4.25pc}{0ex}}\approx \left[\begin{array}{cc}0.923& 0.385\end{array}\right]$
2. ${\left[\begin{array}{cc}0& 743.632\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{cc}0& 743.632\end{array}\right]}{\parallel \left[\begin{array}{cc}0& 743.632\end{array}\right]\parallel }=\frac{\left[\begin{array}{cc}0& 743.632\end{array}\right]}{\sqrt{{0}^{2}+{743.632}^{2}}}\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{5.8pc}{0ex}}=\frac{\left[\begin{array}{cc}0& 743.632\end{array}\right]}{743.632}=\left[\begin{array}{cc}0& 1\end{array}\right]$
3. ${\left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]}{\parallel \left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]\parallel }\approx \frac{\left[\begin{array}{ccc}8& -3& 1/2\end{array}\right]}{8.56}\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{6.25pc}{0ex}}\approx \left[\begin{array}{ccc}0.935& -0.350& 0.058\end{array}\right]$
4. ${\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}{\parallel \left[\begin{array}{ccc}-12& 3& -4\end{array}\right]\parallel }=\frac{\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}{\sqrt{\left(-12{\right)}^{2}+{3}^{2}+\left(-4{\right)}^{2}}}\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{6.25pc}{0ex}}=\frac{\left[\begin{array}{ccc}-12& 3& -4\end{array}\right]}{13}=\left[\begin{array}{ccc}\frac{-12}{13}& \frac{3}{13}& \frac{-4}{13}\end{array}\right]$
5. ${\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}_{\text{norm}}=\frac{\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}{\parallel \left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]\parallel }=\frac{\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}{\sqrt{{1}^{2}+{1}^{2}+{1}^{2}+{1}^{2}}}\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{6.25pc}{0ex}}=\frac{\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]}{2}=\left[\begin{array}{cccc}0.5& 0.5& 0.5& 0.5\end{array}\right]$
1. $\left[\begin{array}{ccc}7& -2& -3\end{array}\right]+\left[\begin{array}{ccc}6& 6& -4\end{array}\right]=\left[\begin{array}{ccc}7+6& -2+6& -3+\left(-4\right)\end{array}\right]=\left[\begin{array}{ccc}13& 4& -7\end{array}\right]$
2. $\left[\begin{array}{ccc}2& 9& -1\end{array}\right]+\left[\begin{array}{ccc}-2& -9& 1\end{array}\right]=\left[\begin{array}{ccc}2+\left(-2\right)& 9+\left(-9\right)& -1+1\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
3. $\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right]-\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]=\left[\begin{array}{c}3-8\\ 10-\left(-7\right)\\ 7-4\end{array}\right]=\left[\begin{array}{c}-5\\ 17\\ 3\end{array}\right]$
4. $\left[\begin{array}{c}4\\ 5\\ -11\end{array}\right]-\left[\begin{array}{c}-4\\ -5\\ 11\end{array}\right]=\left[\begin{array}{c}4-\left(-4\right)\\ 5-\left(-5\right)\\ -11-11\end{array}\right]=\left[\begin{array}{c}8\\ 10\\ -22\end{array}\right]$
5. $3\left[\begin{array}{c}a\\ b\\ c\end{array}\right]-4\left[\begin{array}{c}2\\ 10\\ -6\end{array}\right]=\left[\begin{array}{c}3a\\ 3b\\ 3c\end{array}\right]-\left[\begin{array}{c}8\\ 40\\ -24\end{array}\right]=\left[\begin{array}{c}3a-8\\ 3b-40\\ 3c+24\end{array}\right]$
1. $\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\left(\left[\begin{array}{c}10\\ 6\end{array}\right],\left[\begin{array}{c}-14\\ 30\end{array}\right]\right)=\sqrt{\left(10-\left(-14\right){\right)}^{2}+\left(6-30{\right)}^{2}}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{7.8pc}{0ex}}=\sqrt{{24}^{2}+\left(-24{\right)}^{2}}=\sqrt{576+576}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{7.8pc}{0ex}}=\sqrt{1152}\approx 33.94$
2. $\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\left(\left[\begin{array}{c}0\\ 0\end{array}\right],\left[\begin{array}{c}-12\\ 5\end{array}\right]\right)=\sqrt{\left(0-\left(-12\right){\right)}^{2}+\left(0-5{\right)}^{2}}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{7.6pc}{0ex}}=\sqrt{{12}^{2}+\left(-5{\right)}^{2}}=\sqrt{144+25}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{7.6pc}{0ex}}=\sqrt{169}=13$
3. $\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\left(\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right],\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]\right)=\sqrt{\left(3-8{\right)}^{2}+\left(10-\left(-7\right){\right)}^{2}+\left(7-4{\right)}^{2}}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{\left(-5{\right)}^{2}+{17}^{2}+{3}^{2}}=\sqrt{25+289+9}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{323}\approx 17.97$
4. $\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\left(\left[\begin{array}{c}-2\\ -4\\ 9\end{array}\right],\left[\begin{array}{c}6\\ -7\\ 9.5\end{array}\right]\right)=\sqrt{\left(6-\left(-2\right){\right)}^{2}+\left(-7-\left(-4\right){\right)}^{2}+\left(9.5-9{\right)}^{2}}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{{8}^{2}+\left(-3{\right)}^{2}+\left(0.5{\right)}^{2}}=\sqrt{64+9+0.25}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{73.25}\approx 8.56$
5. $\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\left(\left[\begin{array}{c}4\\ -4\\ -4\\ 4\end{array}\right],\left[\begin{array}{c}-6\\ 6\\ 6\\ -6\end{array}\right]\right)=\sqrt{\left(-6-4{\right)}^{2}+\left(6-\left(-4\right){\right)}^{2}+\left(6-\left(-4\right){\right)}^{2}+\left(-6-4{\right)}^{2}}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{\left(-10{\right)}^{2}+\left(10{\right)}^{2}+\left(10{\right)}^{2}+\left(-10{\right)}^{2}}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{100+100+100+100}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{8.4pc}{0ex}}=\sqrt{400}=20$
1. $\left[\begin{array}{c}2\\ 6\end{array}\right]\cdot \left[\begin{array}{c}-3\\ 8\end{array}\right]=\left(2\right)\left(-3\right)+\left(6\right)\left(8\right)=-6+48=42$
2. $-7\left[\begin{array}{cc}1& 2\end{array}\right]\cdot \left[\begin{array}{cc}11& -4\end{array}\right]=\left[\begin{array}{cc}-7& -14\end{array}\right]\cdot \left[\begin{array}{cc}11& -4\end{array}\right]\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{7.4pc}{0ex}}=\left(-7\right)\left(11\right)+\left(-14\right)\left(-4\right)\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{7.4pc}{0ex}}=-21$
3. $10+\left[\begin{array}{c}-5\\ 1\\ 3\end{array}\right]\cdot \left[\begin{array}{c}4\\ -13\\ 9\end{array}\right]=10+\left(\left(-5\right)\left(4\right)+\left(1\right)\left(-13\right)+\left(3\right)\left(9\right)\right)\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{6.7pc}{0ex}}=10+\left(-20+\left(-13\right)+27\right)\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{6.7pc}{0ex}}=10+\left(-6\right)=4$
4. $3\left[\begin{array}{c}-2\\ 0\\ 4\end{array}\right]\cdot \left(\left[\begin{array}{c}8\\ -2\\ 3/2\end{array}\right]+\left[\begin{array}{c}0\\ 9\\ 7\end{array}\right]\right)=\left[\begin{array}{c}-6\\ 0\\ 12\end{array}\right]\cdot \left[\begin{array}{c}8\\ 7\\ 17/2\end{array}\right]\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{9.4pc}{0ex}}=\left(-6\right)\left(8\right)+\left(0\right)\left(7\right)+\left(12\right)\left(17/2\right)=54$
2. ${\mathbf{v}}_{\parallel }=\stackrel{^}{\mathbf{n}}\frac{\mathbf{v}\cdot \stackrel{^}{\mathbf{n}}}{{\parallel \stackrel{^}{\mathbf{n}}\parallel }^{2}}=\stackrel{^}{\mathbf{n}}\frac{\mathbf{v}\cdot \stackrel{^}{\mathbf{n}}}{1}=\stackrel{^}{\mathbf{n}}\left(\mathbf{v}\cdot \stackrel{^}{\mathbf{n}}\right)\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{1pc}{0ex}}=\left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right]\left(\left[\begin{array}{c}4\\ 3\\ -1\end{array}\right]\cdot \left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right]\right)=\left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right]\left(2\sqrt{2}+\frac{3\sqrt{2}}{2}+0\right)\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{1pc}{0ex}}=\left[\begin{array}{c}\sqrt{2}/2\\ \sqrt{2}/2\\ 0\end{array}\right]\frac{7\sqrt{2}}{2}=\left[\begin{array}{c}7/2\\ 7/2\\ 0\end{array}\right]$
${\mathbf{v}}_{\perp }=\mathbf{v}-{\mathbf{v}}_{\parallel }\phantom{\rule[-8pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{1pc}{0ex}}=\left[\begin{array}{c}4\\ 3\\ -1\end{array}\right]-\left[\begin{array}{c}7/2\\ 7/2\\ 0\end{array}\right]=\left[\begin{array}{c}4-7/2\\ 3-7/2\\ -1-0\end{array}\right]=\left[\begin{array}{c}1/2\\ -1/2\\ -1\end{array}\right]$
3. 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:
$\begin{array}{rcl}{\parallel \mathbf{a}-\mathbf{b}\parallel }^{2}& =& \left(\mathbf{a}-\mathbf{b}\right)\cdot \left(\mathbf{a}-\mathbf{b}\right)\\ & =& \mathbf{a}\cdot \mathbf{a}-2\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{b}\\ & =& \mathbf{a}\cdot \mathbf{a}+\mathbf{b}\cdot \mathbf{b}-2\mathbf{a}\cdot \mathbf{b}\\ & =& {\parallel \mathbf{a}\parallel }^{2}+{\parallel \mathbf{b}\parallel }^{2}-2\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \mathrm{cos}\theta \end{array}$
which is the law of cosines.
4. First, let's obtain some information about the vector components.

From the figure, we have
$\begin{array}{rlrl}{a}_{x}& =\parallel \mathbf{a}\parallel \mathrm{cos}{\theta }_{a},& {a}_{y}& =\parallel \mathbf{a}\parallel \mathrm{sin}{\theta }_{a},\\ {b}_{x}& =\parallel \mathbf{b}\parallel \mathrm{cos}{\theta }_{b},& {b}_{y}& =\parallel \mathbf{b}\parallel \mathrm{sin}{\theta }_{b}.\end{array}$
Now we can proceed with the algebraic definition of the dot product and the cosine difference identity:
$\begin{array}{rl}\mathbf{a}\cdot \mathbf{b}& ={a}_{x}{b}_{x}+{a}_{y}{b}_{y}\\ & =\parallel \mathbf{a}\parallel \mathrm{cos}{\theta }_{a}\parallel \mathbf{b}\parallel \mathrm{cos}{\theta }_{b}+\parallel \mathbf{a}\parallel \mathrm{sin}{\theta }_{a}\parallel \mathbf{b}\parallel \mathrm{sin}{\theta }_{b}\\ & =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \left(\mathrm{cos}{\theta }_{a}\mathrm{cos}{\theta }_{b}+\mathrm{sin}{\theta }_{a}\mathrm{sin}{\theta }_{b}\right)\\ & =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \mathrm{cos}\left({\theta }_{b}-{\theta }_{a}\right)\\ & =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \mathrm{cos}\theta .\end{array}$
1. $\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]×\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}\left(-1\right)\left(1\right)-\left(0\right)\left(0\right)\\ \left(0\right)\left(0\right)-\left(0\right)\left(1\right)\\ \left(0\right)\left(0\right)-\left(-1\right)\left(0\right)\end{array}\right]=\left[\begin{array}{c}-1-0\\ 0-0\\ 0-0\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 0\end{array}\right]$
$\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]×\left[\begin{array}{c}0\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}\left(0\right)\left(0\right)-\left(1\right)\left(-1\right)\\ \left(1\right)\left(0\right)-\left(0\right)\left(0\right)\\ \left(0\right)\left(-1\right)-\left(0\right)\left(0\right)\end{array}\right]=\left[\begin{array}{c}0-\left(-1\right)\\ 0-0\\ 0-0\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$
2. $\left[\begin{array}{c}-2\\ 4\\ 1\end{array}\right]×\left[\begin{array}{c}1\\ -2\\ -1\end{array}\right]=\left[\begin{array}{c}\left(4\right)\left(-1\right)-\left(1\right)\left(-2\right)\\ \left(1\right)\left(1\right)-\left(-2\right)\left(-1\right)\\ \left(-2\right)\left(-2\right)-\left(4\right)\left(1\right)\end{array}\right]=\left[\begin{array}{c}-4-\left(-2\right)\\ 1-2\\ 4-4\end{array}\right]=\left[\begin{array}{c}-2\\ -1\\ 0\end{array}\right]$
$\left[\begin{array}{c}1\\ -2\\ -1\end{array}\right]×\left[\begin{array}{c}-2\\ 4\\ 1\end{array}\right]=\left[\begin{array}{c}\left(-2\right)\left(1\right)-\left(-1\right)\left(4\right)\\ \left(-1\right)\left(-2\right)-\left(1\right)\left(1\right)\\ \left(1\right)\left(4\right)-\left(-2\right)\left(-2\right)\end{array}\right]=\left[\begin{array}{c}-2-\left(-4\right)\\ 2-1\\ 4-4\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 0\end{array}\right]$
3. $\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right]×\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]=\left[\begin{array}{c}\left(10\right)\left(4\right)-\left(7\right)\left(-7\right)\\ \left(7\right)\left(8\right)-\left(3\right)\left(4\right)\\ \left(3\right)\left(-7\right)-\left(10\right)\left(8\right)\end{array}\right]=\left[\begin{array}{c}40-\left(-49\right)\\ 56-12\\ -21-80\end{array}\right]=\left[\begin{array}{c}89\\ 44\\ -101\end{array}\right]$
$\left[\begin{array}{c}8\\ -7\\ 4\end{array}\right]×\left[\begin{array}{c}3\\ 10\\ 7\end{array}\right]=\left[\begin{array}{c}\left(-7\right)\left(7\right)-\left(4\right)\left(10\right)\\ \left(4\right)\left(3\right)-\left(8\right)\left(7\right)\\ \left(8\right)\left(10\right)-\left(-7\right)\left(3\right)\end{array}\right]=\left[\begin{array}{c}-49-40\\ 12-56\\ 80-\left(-21\right)\end{array}\right]=\left[\begin{array}{c}-89\\ -44\\ 101\end{array}\right]$
5. 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 $\mathbf{a}\cdot \mathbf{b}=\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \mathrm{cos}\theta$ and $\mathbf{a}×\mathbf{b}=$
$\left[\begin{array}{c}{a}_{y}{b}_{z}-{a}_{z}{b}_{y}\\ {a}_{z}{b}_{x}-{a}_{x}{b}_{z}\\ {a}_{x}{b}_{y}-{a}_{y}{b}_{x}\end{array}\right]$ . From $\parallel \mathbf{a}×\mathbf{b}\parallel$ , we have:
$\begin{array}{rl}& \parallel \mathbf{a}×\mathbf{b}\parallel =\sqrt{\left({a}_{y}{b}_{z}-{a}_{z}{b}_{y}{\right)}^{2}+\left({a}_{z}{b}_{x}-{a}_{x}{b}_{z}{\right)}^{2}+\left({a}_{x}{b}_{y}-{a}_{y}{b}_{x}{\right)}^{2}}\\ & =\sqrt{{a}_{y}^{2}{b}_{z}^{2}-2{a}_{y}{a}_{z}{b}_{y}{b}_{z}+{a}_{z}^{2}{b}_{y}^{2}+{a}_{z}^{2}{b}_{x}^{2}-2{a}_{x}{a}_{z}{b}_{x}{b}_{z}+{a}_{x}^{2}{b}_{z}^{2}+{a}_{x}^{2}{b}_{y}^{2}-2{a}_{x}{a}_{y}{b}_{x}{b}_{y}+{a}_{y}^{2}{b}_{x}^{2}}.\end{array}$
If we now consider $\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \mathrm{sin}\theta$ , we find that:
$\begin{array}{rl}& =\sqrt{\left({a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}\right)\left({b}_{x}^{2}+{b}_{y}^{2}+{b}_{z}^{2}\right)\left(1-\frac{\left({a}_{x}{b}_{x}+{a}_{y}{b}_{y}+{a}_{z}{b}_{z}{\right)}^{2}}{\left({a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}\right)\left({b}_{x}^{2}+{b}_{y}^{2}+{b}_{z}^{2}\right)}\right)}\\ & =\sqrt{\left({a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}\right)\left({b}_{x}^{2}+{b}_{y}^{2}+{b}_{z}^{2}\right)-\left({a}_{x}{b}_{x}+{a}_{y}{b}_{y}+{a}_{z}{b}_{z}{\right)}^{2}}\\ & \phantom{\rule{-4pc}{0ex}}=\sqrt{{a}_{y}^{2}{b}_{z}^{2}-2{a}_{y}{a}_{z}{b}_{y}{b}_{z}+{a}_{z}^{2}{b}_{y}^{2}+{a}_{z}^{2}{b}_{x}^{2}-2{a}_{x}{a}_{z}{b}_{x}{b}_{z}+{a}_{x}^{2}{b}_{z}^{2}+{a}_{x}^{2}{b}_{y}^{2}-2{a}_{x}{a}_{y}{b}_{x}{b}_{y}+{a}_{y}^{2}{b}_{x}^{2}}.\end{array}$
Starting from both ends, we have met in the middle, proving that
$\parallel \mathbf{a}×\mathbf{b}\parallel =\parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel \mathrm{sin}\theta$ .
1. (a)
1. (1) ${\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_{1}=|3|+|4|=7\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_{2}=\sqrt{{|3|}^{2}+{|4|}^{2}}=5\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_{3}=\sqrt[3]{{|3|}^{3}+{|4|}^{3}}=\sqrt[3]{91}\approx 4.498\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cc}3& 4\end{array}\right]\parallel }_{\mathrm{\infty }}=max\left(|3|,|4|\right)=4$
2. (2) ${\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_{1}=|5|+|-12|=17\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_{2}=\sqrt{{|5|}^{2}+{|-12|}^{2}}=13\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_{3}=\sqrt[3]{{|5|}^{3}+{|-12|}^{3}}=\sqrt[3]{1853}\approx 12.283\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cc}5& -12\end{array}\right]\parallel }_{\mathrm{\infty }}=max\left(|5|,|-12|\right)=12$
3. (3) ${\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_{1}=|-2|+|10|+|-7|=19\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_{2}=\sqrt{{|-2|}^{2}+{|10|}^{2}+{|-7|}^{2}}=\sqrt{153}\approx 12.369\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_{3}=\sqrt[3]{{|-2|}^{3}+{|10|}^{3}+{|-7|}^{3}}=\sqrt[3]{1351}\approx 11.055\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{ccc}-2& 10& -7\end{array}\right]\parallel }_{\mathrm{\infty }}=max\left(|-2|,|10|,|-7|\right)=10$
4. (4) ${\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_{1}=|6|+|1|+|-9|=16\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_{2}=\sqrt{{|6|}^{2}+{|1|}^{2}+{|-9|}^{2}}=\sqrt{118}\approx 10.863\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_{3}=\sqrt[3]{{|6|}^{3}+{|1|}^{3}+{|-9|}^{3}}=\sqrt[3]{946}\approx 9.817\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{ccc}6& 1& -9\end{array}\right]\parallel }_{\mathrm{\infty }}=max\left(|6|,|1|,|-9|\right)=9$
5. (5) ${\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_{1}=|-2|+|-2|+|-2|+|-2|=8\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_{2}=\sqrt{{|-2|}^{2}+{|-2|}^{2}+{|-2|}^{2}+{|-2|}^{2}}=4\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_{3}=\sqrt[3]{{|-2|}^{3}+{|-2|}^{3}+{|-2|}^{3}+{|-2|}^{3}}=\sqrt[3]{32}\approx 3.175\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\parallel \left[\begin{array}{cccc}-2& -2& -2& -2\end{array}\right]\parallel }_{\mathrm{\infty }}=max\left(|-2|,|-2|,|-2|,|-2|\right)=2$
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 $\left[1,0\right]$ , $\left[0,1\right]$ , $\left[-1,0\right]$ , $\left[0,-1\right]$ .
6. 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.
7. 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:
$\begin{array}{rcl}\mathbf{s}& =& \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}+\mathbf{e}+\mathbf{f}\\ & =& \left[\begin{array}{c}-1\\ 3\end{array}\right]+\left[\begin{array}{c}1\\ 3\end{array}\right]+\left[\begin{array}{c}3\\ -2\end{array}\right]+\left[\begin{array}{c}-1\\ -2\end{array}\right]+\left[\begin{array}{c}-6\\ 4\end{array}\right]+\left[\begin{array}{c}-1\\ -3\end{array}\right]\\ & =& \left[\begin{array}{c}\left(-1\right)+1+3+\left(-1\right)+\left(-6\right)+\left(-1\right)\\ 3+3+\left(-2\right)+\left(-2\right)+4+\left(-3\right)\end{array}\right]\\ & =& \left[\begin{array}{c}-5\\ 3\end{array}\right]\end{array}$
8. Left-handed.
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
$\begin{array}{rl}{\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}-{r}_{x}\\ {c}_{y}+{r}_{y}\\ {c}_{z}+{r}_{z}\end{array}\right],& \phantom{\rule{2em}{0ex}}{\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}+{r}_{x}\\ {c}_{y}+{r}_{y}\\ {c}_{z}+{r}_{z}\end{array}\right],\\ {\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}-{r}_{x}\\ {c}_{y}-{r}_{y}\\ {c}_{z}+{r}_{z}\end{array}\right],& \phantom{\rule{2em}{0ex}}{\mathbf{p}}_{\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}+{r}_{x}\\ {c}_{y}-{r}_{y}\\ {c}_{z}+{r}_{z}\end{array}\right],\end{array}$
$\begin{array}{rl}{\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}-{r}_{x}\\ {c}_{y}+{r}_{y}\\ {c}_{z}-{r}_{z}\end{array}\right],& \phantom{\rule{2em}{0ex}}{\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{U}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}+{r}_{x}\\ {c}_{y}+{r}_{y}\\ {c}_{z}-{r}_{z}\end{array}\right],\\ {\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{L}\mathrm{e}\mathrm{f}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}-{r}_{x}\\ {c}_{y}-{r}_{y}\\ {c}_{z}-{r}_{z}\end{array}\right],& \phantom{\rule{2em}{0ex}}{\mathbf{p}}_{\mathrm{B}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{L}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{R}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}=\left[\begin{array}{c}{c}_{x}+{r}_{x}\\ {c}_{y}-{r}_{y}\\ {c}_{z}-{r}_{z}\end{array}\right].\end{array}$
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,
$\begin{array}{r}\mathbf{v}\cdot \left(\mathbf{x}-\mathbf{p}\right)=\parallel \mathbf{v}\parallel \parallel \mathbf{x}-\mathbf{p}\parallel \mathrm{cos}\theta ,\end{array}$
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 $\mathrm{cos}\theta$ . If $\mathrm{cos}\theta >0$ then $\theta$ is less than ${90}^{\circ }$ and $\mathbf{x}$ is in front of the NPC. Similarly, if $\mathrm{cos}\theta <0$ then $\theta$ is greater than ${90}^{\circ }$ and $\mathbf{x}$ is behind the NPC.
The special case of $\mathbf{v}\cdot \left(\mathbf{x}-\mathbf{p}\right)=0$ 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.
1. (1) $\mathbf{x}$ is in front of the NPC.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}0\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}3\\ -4\end{array}\right]=\left(5\right)\left(3\right)+\left(-2\right)\left(-4\right)=23\end{array}$
2. (2) $\mathbf{x}$ is in front of the NPC.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}1\\ 6\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}4\\ 2\end{array}\right]=\left(5\right)\left(4\right)+\left(-2\right)\left(2\right)=16\end{array}$
3. (3) $\mathbf{x}$ is behind the NPC.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-6\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-3\\ -4\end{array}\right]=\left(5\right)\left(-3\right)+\left(-2\right)\left(-4\right)\\ & =-7\end{array}$
4. (4) $\mathbf{x}$ is behind the NPC.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-4\\ 7\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 3\end{array}\right]=\left(5\right)\left(-1\right)+\left(-2\right)\left(3\right)=-11\end{array}$
5. (5) $\mathbf{x}$ is in front of the NPC.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}5\\ 5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}8\\ 1\end{array}\right]=\left(5\right)\left(8\right)+\left(-2\right)\left(1\right)=38\end{array}$
6. (6) $\mathbf{x}$ is in front of the NPC.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-3\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}0\\ -4\end{array}\right]=\left(5\right)\left(0\right)+\left(-2\right)\left(-4\right)=8\end{array}$
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.
$\begin{array}{rl}\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-6\\ -3.5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)& =\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-3\\ -7.5\end{array}\right]=\left(5\right)\left(-3\right)+\left(-2\right)\left(-7.5\right)=0\end{array}$
1. To determine whether the point $\mathbf{x}$ is visible to the NPC, compare $\mathrm{cos}\theta$ to $\mathrm{cos}\left(\varphi /2\right)$ . If $\mathrm{cos}\theta \ge \mathrm{cos}\left(\varphi /2\right)$ , then $\mathbf{x}$ is visible to the NPC.
The value of $\mathrm{cos}\left(\varphi /2\right)$ can be obtained from the FOV angle. To get $\mathrm{cos}\theta$ use the dot product
$\mathrm{cos}\theta =\frac{\mathbf{v}\cdot \left(\mathbf{x}-\mathbf{p}\right)}{\parallel \mathbf{v}\parallel \parallel \mathbf{x}-\mathbf{p}\parallel }.$
2. The NPC's FOV is ${90}^{\circ }$ , so the value we are interested in is $\mathrm{cos}\left({45}^{\circ }\right)\approx 0.707$ .
1. (1) $\mathbf{x}$ is visible to the NPC.
$\begin{array}{r}\mathrm{cos}\theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}0\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}0\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{23}{\left(\sqrt{29}\right)\left(\sqrt{\left(}25\right)}\approx 0.854\ge 0.707\end{array}$
2. (2) $\mathbf{x}$ is not visible to the NPC.
$\begin{array}{r}\mathrm{cos}\theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}1\\ 6\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}1\\ 6\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{16}{\left(\sqrt{29}\right)\left(\sqrt{20}\right)}\approx 0.664<0.707\end{array}$
3. (3) $\mathbf{x}$ is not visible to the NPC.
$\begin{array}{rl}\mathrm{cos}\theta & =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-6\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}-6\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{-7}{\left(\sqrt{29}\right)\left(\sqrt{25}\right)}\\ & \approx -0.260<0.707\end{array}$
4. (4) $\mathbf{x}$ is not visible to the NPC.
$\begin{array}{r}\phantom{\rule{-1pc}{0ex}}\mathrm{cos}\theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-4\\ 7\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}-4\\ 7\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{-11}{\left(\sqrt{29}\right)\left(\sqrt{10}\right)}\approx -0.646<0.707\end{array}$
5. (5) $\mathbf{x}$ is visible to the NPC.
$\begin{array}{r}\phantom{\rule{-1pc}{0ex}}\mathrm{cos}\theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}5\\ 5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)}{\parallel \left[\begin{array}{c}5\\ -2\end{array}\right]\parallel \parallel \left[\begin{array}{c}5\\ 5\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\parallel }=\frac{38}{\left(\sqrt{29}\right)\left(\sqrt{65}\right)}\approx 0.875\ge 0.707\end{array}$
6. (6) $\mathbf{x}$ is not visible to the NPC.
$\begin{array}{r}\phantom{\rule{-1pc}{0ex}}\mathrm{cos}\theta =\frac{\left[\begin{array}{c}5\\ -2\end{array}\right]\cdot \left(\left[\begin{array}{c}-3\\ 0\end{array}\right]-\left[\begin{array}{c}-3\\ 4\end{array}\right]\right)}{\parallel \left[}\end{array}$