Practice Midterm 1

View as PDF ✏️ Solutions PDF ✅ Video Walkthroughs 🎥

This page is meant to give you quick access to problems and their solutions. Refer to the original exam PDF, linked above, for test-taking instructions and formatting. Note that we’ve kept the problem text identical, which is why you may see things like “write your answer in the box below” despite there not being a box on this page.

Problem 1 (9 pts) 🎥 Walkthrough

Consider the points

P=(-7,8,-6),\qquad Q=(8,-12,19)

(2 pts) Let $\vec d$ be the vector pointing from $$P$$ to $$Q$$ . Find $\vec d$ . Give your answer as a vector.

Solution

Subtract the coordinates of the starting point from those of the ending point.

\vec d=\begin{bmatrix}8-(-7)\\-12-8\\19-(-6)\end{bmatrix} =\boxed{\begin{bmatrix}15\\-20\\25\end{bmatrix}}

(3 pts) Find the distance between $$P$$ and $$Q$$ .

Solution

The distance is the length of the vector from $$P$$ to $$Q$$ . Notice that

\vec d=5\begin{bmatrix}3\\-4\\5\end{bmatrix}

The length of a scalar multiple is the absolute value of the scalar times the original length. So,

\Vert \vec d\Vert =5\left\Vert \begin{bmatrix}3\\-4\\5\end{bmatrix}\right\Vert =5\sqrt{3^2+(-4)^2+5^2}=5\sqrt{50}=\boxed{25\sqrt2}

(4 pts) Find the coordinates of the point $$R$$ that is $\frac35$ of the way from $$P$$ to $$Q$$ .

Solution

Start at $$P$$ and add $\frac35$ of the vector pointing from $$P$$ to $$Q$$ .

\begin{bmatrix}-7\\8\\-6\end{bmatrix} +\frac35\begin{bmatrix}15\\-20\\25\end{bmatrix} =\begin{bmatrix}-7\\8\\-6\end{bmatrix} +\begin{bmatrix}9\\-12\\15\end{bmatrix} =\begin{bmatrix}2\\-4\\9\end{bmatrix}

So, $\boxed{R=(2,-4,9)}$ .

Problem 2 (16 pts) 🎥 Walkthrough

Suppose $\vec u,\vec v\in\mathbb R^3$ satisfy

\Vert \vec u\Vert =5,\qquad \Vert \vec v\Vert =4,\qquad \vec u\cdot\vec v=-6

(7 pts)

I)
(5 pts) Find $(5\vec u-6\vec v)\cdot(8\vec u+10\vec v)$ . Show your work.

Solution
Distribute the dot product, using $\vec u\cdot\vec u=\Vert \vec u\Vert ^2$ and $\vec v\cdot\vec v=\Vert \vec v\Vert ^2$ .
$\begin{align*} (5\vec u-6\vec v)\cdot(8\vec u+10\vec v) &=40\Vert \vec u\Vert ^2+(50-48)(\vec u\cdot\vec v)-60\Vert \vec v\Vert ^2\\ &=20\bigl(2\Vert \vec u\Vert ^2-3\Vert \vec v\Vert ^2\bigr)+2(\vec u\cdot\vec v)\\ &=20(50-48)+2(-6)=40-12=\boxed{28} \end{align*}$
II)
(2 pts) What type of angle is formed by $5\vec u-6\vec v$ and $8\vec u+10\vec v$ ? Select one.
Acute Right Obtuse

Solution
Acute Right Obtuse
The dot product is positive, so the angle is $\boxed{\text{acute}}$ . Recall that $\vec a\cdot\vec b=\Vert \vec a\Vert \Vert \vec b\Vert \cos\theta$ ; a positive dot product means $\cos\theta>0$ .

(5 pts) Find $\Vert \vec u-\vec v\Vert$ .

Solution

First find the squared length by taking the dot product of the vector with itself.

\begin{align*} \Vert \vec u-\vec v\Vert ^2 &=(\vec u-\vec v)\cdot(\vec u-\vec v)\\ &=\Vert \vec u\Vert ^2-2(\vec u\cdot\vec v)+\Vert \vec v\Vert ^2\\ &=25-2(-6)+16=53 \end{align*}

Therefore, $\boxed{\Vert \vec u-\vec v\Vert =\sqrt{53}}$ .

(4 pts) Could a unit vector $\vec w\in\mathbb R^3$ satisfy $\vec u\cdot\vec w=9$ ? Explain why or why not.

Solution

$\boxed{\text{No}}$ . By the Cauchy–Schwarz inequality, a unit vector $\vec w$ must satisfy

|\vec u\cdot\vec w|\leq\Vert \vec u\Vert \Vert \vec w\Vert =5(1)=5

Since $$9>5$$ , the proposed dot product is impossible.

Problem 3 (14 pts) 🎥 Walkthrough

(3 pts) Let $\ell$ be the line in $\mathbb R^2$ defined by $$5x+7y=35$$ . Find nonzero vectors $\vec n$ and $\vec d$ such that $\vec n$ is perpendicular to $\ell$ and $\vec d$ is parallel to $\ell$ .

$\vec n=$

$\vec d=$

Solution

The coefficients of $$x$$ and $$y$$ give a normal vector. A direction vector must be orthogonal to it, so one choice is

\boxed{\vec n=\begin{bmatrix}5\\7\end{bmatrix}},\qquad \boxed{\vec d=\begin{bmatrix}7\\-5\end{bmatrix}}

These work because $\vec n\cdot\vec d=5(7)+7(-5)=0$ . Any nonzero scalar multiples also work.

(4 pts) Write the line $\ell$ , defined by $$5x+7y=35$$ , in vector-parametric form.

Solution

The point $$(7,0)$$ is on $\ell$ , and part (a) gives a direction vector. So, one vector-parametric form is

\boxed{\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}7\\0\end{bmatrix} +t\begin{bmatrix}7\\-5\end{bmatrix},\qquad t\in\mathbb R}

Substituting gives $$5(7+7t)+7(-5t)=35$$ for every $$t$$ .

(7 pts) Consider another line $\ell’$ , defined by $$7x-5y=13$$ . Which of the following are valid scalar-parametric forms of $\ell’$ ? Select all that apply. In each option, $t\in\mathbb R$ .

\begin{cases}x=9+5t\\y=10+7t\end{cases}

\begin{cases}x=9+7t\\y=10-5t\end{cases}

\begin{cases}x=-6-10t\\y=-11-14t\end{cases}

\begin{cases}x=14+15t\\y=17+21t\end{cases}

\begin{cases}x=8+5t\\y=9+7t\end{cases}

Solution

Select $\boxed{\text{the first, third, and fourth options}}$ . Substitute each option into $$7x-5y=13$$ :

Option 1: $x=9+5t,\ y=10+7t$ . Substituting gives $$7(9+5t)-5(10+7t)=63+35t-50-35t=13$$ . The equation is satisfied for every $$t$$ , so this option is valid.
Option 2: $x=9+7t,\ y=10-5t$ . Substituting gives $$7(9+7t)-5(10-5t)=63+49t-50+25t=13+74t$$ . The equation is satisfied only when $$t=0$$ , so this option is invalid.
Option 3: $x=-6-10t,\ y=-11-14t$ . Substituting gives $$7(-6-10t)-5(-11-14t)=-42-70t+55+70t=13$$ . The equation is satisfied for every $$t$$ , so this option is valid.
Option 4: $x=14+15t,\ y=17+21t$ . Substituting gives $$7(14+15t)-5(17+21t)=98+105t-85-105t=13$$ . The equation is satisfied for every $$t$$ , so this option is valid.
Option 5: $x=8+5t,\ y=9+7t$ . Substituting gives $7(8+5t)-5(9+7t)=56+35t-45-35t=11\ne13$ . The equation is never satisfied, so this option is invalid.

Problem 4 (14 pts) 🎥 Walkthrough

The vectors

\vec u_1=\frac1{19}\begin{bmatrix}-15\\10\\6\end{bmatrix},\qquad \vec u_2=\frac1{19}\begin{bmatrix}-6\\-15\\10\end{bmatrix},\qquad \vec u_3=\frac1{19}\begin{bmatrix}10\\6\\15\end{bmatrix}

form an orthonormal basis of $\mathbb R^3$ .

(2 pts) Given that these vectors form an orthonormal basis of $\mathbb R^3$ , find the following quantities.

$\vec u_1\cdot\vec u_3=$

$\Vert \vec u_1\Vert =$

Solution

Orthonormal means that distinct basis vectors are orthogonal and each basis vector has length 1. Therefore,

\boxed{\vec u_1\cdot\vec u_3=0},\qquad \boxed{\Vert \vec u_1\Vert =1}

(4 pts) Find $\Vert 5\vec u_1-7\vec u_2\Vert$ .

Solution

Since $\vec u_1$ and $\vec u_2$ are orthogonal unit vectors,

\begin{align*} \Vert 5\vec u_1-7\vec u_2\Vert ^2 &=25\Vert \vec u_1\Vert ^2-70(\vec u_1\cdot\vec u_2)+49\Vert \vec u_2\Vert ^2\\ &=25+49=74 \end{align*}

So, $\boxed{\Vert 5\vec u_1-7\vec u_2\Vert =\sqrt{74}}$ .

(8 pts) Let $\vec w=\begin{bmatrix}-4\\3\\4\end{bmatrix}$ . Write $\vec w$ as a linear combination of $\vec u_1$ , $\vec u_2$ , and $\vec u_3$ .

Solution

For an orthonormal basis, the coefficient of $\vec u_i$ is $\vec w\cdot\vec u_i$ . Taking dot products gives

\begin{align*} \vec w\cdot\vec u_1&=\frac{(-4)(-15)+3(10)+4(6)}{19}=\frac{114}{19}=6\\ \vec w\cdot\vec u_2&=\frac{(-4)(-6)+3(-15)+4(10)}{19}=\frac{19}{19}=1\\ \vec w\cdot\vec u_3&=\frac{(-4)(10)+3(6)+4(15)}{19}=\frac{38}{19}=2 \end{align*}

Therefore,

\boxed{\vec w=6\vec u_1+\vec u_2+2\vec u_3}

Problem 5 (19 pts) 🎥 Walkthrough

Let $$P$$ be the plane through the points

A=(8,-15,6),\qquad B=(13,-5,11),\qquad C=(15,-8,7)

(5 pts) Express $$P$$ in vector-parametric form.

Solution

Use $$A$$ as a starting point. The directions $\vec d_1$ from $$A$$ to $$B$$ and $\vec d_2$ from $$A$$ to $$C$$ are not scalar multiples, so they span the plane.

\boxed{\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}8\\-15\\6\end{bmatrix} +s\underbrace{\begin{bmatrix}5\\10\\5\end{bmatrix}}_{\vec d_1} +t\underbrace{\begin{bmatrix}7\\7\\1\end{bmatrix}}_{\vec d_2},\quad s,t\in\mathbb R}

(8 pts) Write an equation for $$P$$ in the form $$ax+by+cz=d$$ .

Solution

The coefficients $$a,b,c$$ of a nonzero normal vector must satisfy

\begin{cases} 5a+10b+5c=0\\ 7a+7b+c=0 \end{cases}

The second equation gives $$c=-7a-7b$$ . Substituting into the first gives $$-30a-25b=0$$ , or $$6a+5b=0$$ .

Choose $$a=5$$ . Then $$b=-6$$ and $$c=-7(5)-7(-6)=7$$ . Since $$A$$ lies on the plane $$5x-6y+7z=d$$ ,

$$ d=5(8)-6(-15)+7(6)=172 $$

So, the equation is $\boxed{5x-6y+7z=172}$ .

(6 pts) Suppose $$C$$ is replaced by $$C’=(23,15,21)$$ , while $$A$$ and $$B$$ stay the same. Do the three points $$A$$ , $$B$$ , $$C’$$ determine a unique plane? Explain what changes geometrically.

Solution

$\boxed{\text{No}}$ . The new direction vector is

\vec d_3=\begin{bmatrix}23-8\\15-(-15)\\21-6\end{bmatrix} =\begin{bmatrix}15\\30\\15\end{bmatrix} =3\vec d_1

Unlike $\vec d_1$ and $\vec d_2$ , the vectors $\vec d_1$ and $\vec d_3$ are linearly dependent. The three points are now collinear, so infinitely many planes contain their line.

Problem 6 (18 pts) 🎥 Walkthrough

Consider the line $\ell$ in $\mathbb R^3$ given by

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}7\\-8\\11\end{bmatrix} +t\begin{bmatrix}5\\-6\\7\end{bmatrix},\qquad t\in\mathbb R

(10 pts) Find equations for two distinct planes whose intersection is exactly $\ell$ . Your equations should both be of the form $$ax+by+cz=d$$ .

Solution

Each plane must contain $$(7,-8,11)$$ , and its normal must be orthogonal to the line’s direction $\begin{bmatrix}5\\-6\\7\end{bmatrix}$ . Two nonparallel choices are

\vec n_1=\begin{bmatrix}6\\5\\0\end{bmatrix},\qquad \vec n_2=\begin{bmatrix}7\\0\\-5\end{bmatrix}

Their dot products with the direction vector are $$30-30=0$$ and $$35-35=0$$ . Substituting the point into each plane gives

6(7)+5(-8)=2,\qquad 7(7)-5(11)=-6

So, one answer is

\boxed{6x+5y+0z=2},\qquad \boxed{7x+0y-5z=-6}

(8 pts) A student proposes the equations $$11x+15y+5z=12$$ and $$55x+75y+25z=60$$ . Every point on $\ell$ satisfies both. Explain why these equations, unlike those in part (a), do not intersect exactly at the line $\ell$ . Describe their intersection and give one point in it that is not on $\ell$ .

Solution

The second equation is five times the first, so they describe the same plane. Their intersection is that entire plane, which contains points outside $\ell$ .

One such point is $\boxed{(2,1,-5)}$ , since

11(2)+15(1)+5(-5)=12,\qquad 55(2)+75(1)+25(-5)=60

It is not on $\ell$ : its $$x$$ -coordinate would require $$7+5t=2$$ , or $$t=-1$$ , but then the line’s $$y$$ -coordinate would be $$-8-6(-1)=-2$$ , rather than 1.

Problem 7 (10 pts) 🎥 Walkthrough

Let

\vec v=\begin{bmatrix}6\\8\\-8\end{bmatrix},\qquad \ell=\operatorname{span}\left(\begin{bmatrix}5\\-5\\7\end{bmatrix}\right),\qquad P:\ 5x-5y+7z=0

(6 pts) Find the orthogonal projection of $\vec v$ onto $\ell$ .

Solution

Let $\vec w=\begin{bmatrix}5\\-5\\7\end{bmatrix}$ . The projection onto its span is

\operatorname{proj}_{\ell}(\vec v)=\frac{\vec v\cdot\vec w}{\vec w\cdot\vec w}\vec w

Here,

\vec v\cdot\vec w=6(5)+8(-5)+(-8)(7)=-66,\qquad \vec w\cdot\vec w=25+25+49=99

Therefore,

\operatorname{proj}_{\ell}(\vec v) =-\frac23\begin{bmatrix}5\\-5\\7\end{bmatrix} =\boxed{\begin{bmatrix}-10/3\\10/3\\-14/3\end{bmatrix}}

(4 pts) Find the orthogonal projection of $\vec v$ onto $$P$$ .

Solution

The plane passes through the origin and has normal $\vec w$ , so it is perpendicular to $\ell$ . To project onto the plane, subtract the component along the normal.

\operatorname{proj}_{P}(\vec v) =\vec v-\operatorname{proj}_{\ell}(\vec v) =\begin{bmatrix}6\\8\\-8\end{bmatrix} -\begin{bmatrix}-10/3\\10/3\\-14/3\end{bmatrix} =\boxed{\begin{bmatrix}28/3\\14/3\\-10/3\end{bmatrix}}

This vector lies in $$P$$ , since $$5(28/3)-5(14/3)+7(-10/3)=0$$ , and the subtracted component is perpendicular to $$P$$ .