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Dear Mr. Reyes,

I hope this message finds you well. I wanted to let you know that I completed these problems on the plane in Latex because I had to travel to Boston to meet my family and didn’t have access to a notebook or WiFi on the plane. I actually had quite a bit of fun learning the latex syntax from the latex cheatsheet I downloaded. I understand if you want me to redo this submission as well on paper now that I have landed. I apologize for the late submission and any inconvenience this may have caused. Thank you for your understanding.

Best regards, Roshan

13

Are the vectors \((4, 5, -2)\) and \((3,-1, 5)\) orthogonal, parallel, or neither?

if dot product is zero, then vectors are perpendicular b/c \(\frac{u \cdot v}{||u||*||v||} = \cos(\theta)\)

$$ \begin{aligned} &(4, 5, -2) \cdot (3,-1, 5) \ \ &= (4 * 3) + (5 * -1) + (-2 * 5) \ &= 12 - 5 - 10 \ &= -3 \end{aligned}$$ Therefore the vectors are Not orthogonal

15

Find the acute angle between the lines \(2𝑥 − 𝑦 = 3\) and \(3𝑥 + 𝑦 = 7\).

Get the slopes of the lines in question

$$\begin{aligned}

2x - y &= 3 \ y &= 2x - 3 \ m_{1} &= 2 \ \

3x + y &= 7 \ y &= 7 - 3x \ m_{2} &= -3 \end{aligned}$$

turn those slopes into vectors

$$\begin{aligned}

m_{1} &= 2 \implies \vec{u} = <1, 2> \ m_{2} &= -3 \implies \vec{v} = <1, -3> \ \ \cos(\theta) &= \frac{u \cdot v}{||u||||v||} \ \cos(\theta) &= \frac{(11) + (2-3)}{\sqrt{1 + 4}\sqrt{1 + 9}}\ \cos(\theta) &= \frac{-5}{5\sqrt{2}} \ \cos(\theta) &= \frac{-1}{\sqrt{2}} \ \cos(\theta) &= \frac{-\sqrt{2}}{2} \ \theta &= \frac{3\pi}{4} \ \text{is not acute } & \text{so we need }180 - \theta \ \theta_{real} &= \pi - \frac{3\pi}{4} \ = \frac{\pi}{4} \end{aligned}$$

18a

Find the distances between the following pairs of points:

a. \((−3,1,2)\) and \((4,−1,1)\)

\[\begin{aligned} \text{if } \vec{u} &= (−3,1,2) \\ \text{and } \vec{v} &= (4, -1, 1) \\ \text{then } \vec{u} - \vec{v} &= (-7, 2, 1) \\ ||\vec{u} - \vec{v}|| &= \sqrt{49 + 4 + 1} \\ &= \sqrt{54} = 3\sqrt{6} \end{aligned}\]

20

Let \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) be vectors in a given Euclidean space and let 𝑐 and 𝑑 be nonzero scalars. Tell whether each of the following expressions is a scalar, a vector, or is not a valid expression.

a. \((\vec{u} \cdot \vec{v})\vec{w}\) is a vector

b. \((\vec{u} \cdot \vec{v}) \cdot \vec{w}\) is a scalar? I think?

c. \(\vec{u} \cdot \vec{v} + c \vec{w}\) is not valid

d. \(\vec{u} \cdot \vec{v} + c\) is a scalar

e. \(c(\vec{u} \cdot \vec{v}) + d\vec{w}\) is not valid

f. \(||\vec{u} \cdot c\vec{v}|| + d\) is not valid

g. \(c\vec{u} \cdot d\vec{v} + ||\vec{w}||\vec{v}\) is not valid

h. \(||\vec{u} \cdot \vec{v}||\) is not valid

22

Show that if \(u + v\) and \(u - v\) are orthogonal, then the vectors \(u\) and \(v\) must have the same length.

$$\begin{aligned}

(\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) =& 0\ =& (u_{1} + v_{1}, u_{2} + v_{2},\dots u_{n} + v_{n}) \cdot(u_{1} - v_{1}, u_{2} -v_{2}, \dots u_{n} - v_{n}) \ =& ((u_{1}+v_{1}) * (u_{1}-v_{1})) + ((u_{2}+v_{2}) * (u_{2}-v_{2})) + \dots + ((u_{n}+v_{n}) * (u_{n}-v_{n})) \ =& (u_{1}^2-v_{1}^2) + (u_{2}^2-v_{2}^2) + \dots + (u_{n}^2-v_{n}^2) = 0 \ =& \vec{u} \cdot \vec{u} - \vec{v} \cdot \vec{v} \ \vec{u} \cdot \vec{u} =& \vec{v} \cdot \vec{v} \ ||\vec{u}||^2 =& ||\vec{v}||^2\ ||\vec{u}|| =& ||\vec{v}||

\end{aligned}$$

23b

If \(𝒖 = 2𝒊 + 3𝒋 + 𝒌\), \(𝒗 = −𝒊 + 2𝒋 + 4𝒌\), and \(𝒘 = 3𝒊 − 7𝒌\), compute the following cross products.

\[ \begin{aligned} & \vec{u} = 2\hat{i} + 3\hat{j} + \hat{k}, \quad \vec{v} = -\hat{i} + 2\hat{j} + 4\hat{k}, \quad \vec{w} = 3\hat{i} - 7\hat{k}. \\ \\ \vec{w} \times \vec{u} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -7 \\ 2 & 3 & 1 \end{vmatrix} \\ &= \hat{i} \begin{vmatrix} 0 & -7 \\ 3 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -7 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 0 \\ 2 & 3 \end{vmatrix} \\ &= \hat{i}((0)(1) - (3)(-7)) - \hat{j}((3)(1) - (2)(-7)) + \hat{k}((3)(3) - (2)(0)) \\ &= \hat{i}(0 + 21) - \hat{j}(3 + 14) + \hat{k}(9 + 0) \\ &= 21\hat{i} - 17\hat{j} + 9\hat{k}. \\ \\ (\vec{w} \times \vec{u}) \times \vec{v} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 21 & -17 & 9 \\ -1 & 2 & 4 \end{vmatrix} \\ &= \hat{i} \begin{vmatrix} -17 & 9 \\ 2 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 21 & 9 \\ -1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 21 & -17 \\ -1 & 2 \end{vmatrix} \\ &= \hat{i}((-17)(4) - (9)(2)) - \hat{j}((21)(4) - (9)(-1)) + \hat{k}((21)(2) - (-17)(-1)) \\ &= \hat{i}(-68 - 18) - \hat{j}(84 + 9) + \hat{k}(42 - 17) \\ &= \hat{i}(-86) - \hat{j}(93) + \hat{k}(25) \\ &= -86\hat{i} - 93\hat{j} + 25\hat{k}. \\ \\ & (\vec{w} \times \vec{u}) \times \vec{v} = -86\hat{i} - 93\hat{j} + 25\hat{k}. \end{aligned} \]

25

Find two unit vectors that are nonzero and orthogonal to both of the vectors (3, 2, 1) and (-1, 1, 0).

\[ \begin{aligned} & \vec{u} = (3, 2, 1), \quad \vec{v} = (-1, 1, 0). \\ \\ & \vec{u} \times \vec{v}\\ \vec{u} \times \vec{v} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ -1 & 1 & 0 \end{vmatrix} \\ &= \hat{i} \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ -1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ -1 & 1 \end{vmatrix} \\ &= \hat{i}((2)(0) - (1)(1)) - \hat{j}((3)(0) - (-1)(1)) + \hat{k}((3)(1) - (-1)(2)) \\ &= \hat{i}(-1) - \hat{j}(1) + \hat{k}(3 + 2) \\ &= -\hat{i} - \hat{j} + 5\hat{k} \\ &= (-1, -1, 5). \\ \\ & \vec{u} \times \vec{v} \text{ to find the unit vector:} \\ ||\vec{u} \times \vec{v}|| &= \sqrt{(-1)^2 + (-1)^2 + (5)^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}. \\ \vec{w}_1 &= \frac{\vec{u} \times \vec{v}}{||\vec{u} \times \vec{v}||} = \frac{(-1, -1, 5)}{3\sqrt{3}} = \left(-\frac{1}{3\sqrt{3}}, -\frac{1}{3\sqrt{3}}, \frac{5}{3\sqrt{3}}\right). \\ \\ & \text{by reversing the direction of } \vec{w}_1: \\ \vec{w}_2 &= \left(\frac{1}{3\sqrt{3}}, \frac{1}{3\sqrt{3}}, -\frac{5}{3\sqrt{3}}\right). \\ \\ & (3, 2, 1) \text{ and } (-1, 1, 0) \text{ are:} \\ \vec{w}_1 &= \left(-\frac{1}{3\sqrt{3}}, -\frac{1}{3\sqrt{3}}, \frac{5}{3\sqrt{3}}\right), \\ \vec{w}_2 &= \left(\frac{1}{3\sqrt{3}}, \frac{1}{3\sqrt{3}}, -\frac{5}{3\sqrt{3}}\right). \end{aligned} \]

28

Find the area of the triangle determined by the points 𝑃1 = (2,2,0), 𝑃2 = (−1,0,2), and 𝑃3 = (0,4,3).

Also find a nonzero vector orthogonal to the plane that contains the three points (and the triangle).

\[ \begin{aligned} \text{Given: } & P_1 = (2, 2, 0), \quad P_2 = (-1, 0, 2), \quad P_3 = (0, 4, 3). \\ \\ \vec{P_1P_2} &= P_2 - P_1 = (-1 - 2, 0 - 2, 2 - 0) = (-3, -2, 2), \\ \vec{P_1P_3} &= P_3 - P_1 = (0 - 2, 4 - 2, 3 - 0) = (-2, 2, 3). \\ \\ \vec{P_1P_2} \times \vec{P_1P_3}: \\ \vec{P_1P_2} \times \vec{P_1P_3} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -2 & 2 \\ -2 & 2 & 3 \end{vmatrix} \\ &= \hat{i} \begin{vmatrix} -2 & 2 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & 2 \\ -2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & -2 \\ -2 & 2 \end{vmatrix} \\ &= \hat{i}((-2)(3) - (2)(2)) - \hat{j}((-3)(3) - (-2)(2)) + \hat{k}((-3)(2) - (-2)(-2)) \\ &= \hat{i}(-6 - 4) - \hat{j}(-9 + 4) + \hat{k}(-6 - 4) \\ &= \hat{i}(-10) - \hat{j}(-5) + \hat{k}(-10) \\ &= -10\hat{i} + 5\hat{j} - 10\hat{k}. \\ \\ ||\vec{P_1P_2} \times \vec{P_1P_3}|| &= \sqrt{(-10)^2 + 5^2 + (-10)^2} = \sqrt{100 + 25 + 100} = \sqrt{225} = 15. \\ \\ \text{Area} &= \frac{1}{2} ||\vec{P_1P_2} \times \vec{P_1P_3}|| = \frac{1}{2}(15) = 7.5. \\ \\ \vec{n} &= -10\hat{i} + 5\hat{j} - 10\hat{k}. \\ \text{Area: } 7.5,& \quad \text{vector orthogonal to plane: } \vec{n} = -10\hat{i} + 5\hat{j} - 10\hat{k}. \end{aligned} \]