Midterm 1

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Problem 1 (10 pts)

Let

\vec d=\begin{bmatrix}6\\-9\\18\end{bmatrix}

You may leave numbers as unsimplified fractions.

(4 pts) Find a unit vector pointing in the same direction as $\vec d$ .

Solution

First factor out $$3$$ :

\vec d=\begin{bmatrix}6\\-9\\18\end{bmatrix} =3\begin{bmatrix}2\\-3\\6\end{bmatrix}

The length of the vector that remains is $\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7$ . So $|\vec d|=3(7)=21$ . Dividing by the length gives the unit vector:

\frac{\vec d}{\|\vec d\|} =\frac3{21}\begin{bmatrix}2\\-3\\6\end{bmatrix} =\frac17\begin{bmatrix}2\\-3\\6\end{bmatrix} =\boxed{\begin{bmatrix}2/7\\-3/7\\6/7\end{bmatrix}}

(3 pts) Find a vector of length $$6$$ pointing in the direction opposite to $\vec d$ .

Solution

Multiply the unit vector from part (a) by $$-6$$ . The factor $$6$$ sets the length, and the negative sign reverses the direction:

-6\frac{\vec d}{\|\vec d\|} =-\frac6{21}\begin{bmatrix}6\\-9\\18\end{bmatrix} =\boxed{\begin{bmatrix}-12/7\\18/7\\-36/7\end{bmatrix}}

(3 pts) A robot starts at $$P=(5,-6,8)$$ and moves $$x$$ units in the direction of $\vec d$ , ending at $$(9,-12,20)$$ . What is the value of $$x$$ ?

Solution

The displacement from the starting point to the ending point is

\begin{bmatrix}9-5\\-12-(-6)\\20-8\end{bmatrix} =\begin{bmatrix}4\\-6\\12\end{bmatrix} =\frac23\vec d

The positive scalar confirms that the robot moves in the direction of $\vec d$ . The distance traveled is the length of this displacement:

x=\left\|\frac23\vec d\right\|=\frac23(21)=\boxed{14}

Problem 2 (14 pts)

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

\|\vec u\|=3,\qquad \|\vec v\|=4,\qquad \vec u\cdot\vec v=-10

(6 pts) Find the value of $$c$$ for which $\vec u+c\vec v$ is orthogonal to $2\vec u-\vec v$ .

Solution

Orthogonal vectors have dot product $$0$$ . Distribute the dot product and use $\vec u\cdot\vec u=|\vec u|^2$ :

\begin{align*} 0&=(\vec u+c\vec v)\cdot(2\vec u-\vec v)\\ &=2\|\vec u\|^2-\vec u\cdot\vec v+2c(\vec u\cdot\vec v)-c\|\vec v\|^2\\ &=2(9)-(-10)+2c(-10)-16c\\ &=28-36c \end{align*}

Therefore, $c=\dfrac{28}{36}=\boxed{\dfrac79}$ .

(5 pts) Find $|3\vec u+\vec v|$ .

Solution

First find the squared length by taking a dot product:

\begin{align*} \|3\vec u+\vec v\|^2 &=(3\vec u+\vec v)\cdot(3\vec u+\vec v)\\ &=9\|\vec u\|^2+6(\vec u\cdot\vec v)+\|\vec v\|^2\\ &=9(9)+6(-10)+16=37 \end{align*}

Taking the nonnegative square root gives $\boxed{|3\vec u+\vec v|=\sqrt{37}}$ .

(3 pts) Find the angle between $\vec u$ and $\vec v$ . Then select the angle type.

Acute Right Obtuse

Solution

Acute Right Obtuse

Using the formula for cosine similarity (which comes from the geometric definition of the dot product):

\cos\theta=\frac{\vec u\cdot\vec v}{\|\vec u\|\|\vec v\|} =\frac{-10}{3(4)}=-\frac56

Thus $\boxed{\theta=\cos^{-1}(-5/6)}$ . The dot product is negative, so the angle is obtuse (greater than $90^\circ$ ).

Problem 3 (15 pts)

Consider the line $\ell$ in $\mathbb R^2$ given by

\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}5\\3\end{bmatrix} +t\begin{bmatrix}6\\5\end{bmatrix},\qquad t\in\mathbb R

(4 pts) Write an equation for $\ell$ in the form $$ax+by=c$$ .

Solution

A normal $\begin{bmatrix}a\\b\end{bmatrix}$ must be perpendicular to the direction $\begin{bmatrix}6\\5\end{bmatrix}$ , so $$6a+5b=0$$ . An easy choice is $$a=5$$ and $$b=-6$$ ; this comes from swapping $$6$$ and $$5$$ and negating one of them. Then, to find $$c$$ in

$$ 5x-6y=c $$

just plug in a known point on the line, e.g., $$(5,3)$$ :

$$ 5(5)-6(3)=7 $$

Therefore, one equation is $\boxed{5x-6y=7}$ . Other equations for this line are nonzero scalar multiples of this equation.

(7 pts) 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=5+12t\\y=4+10t\end{cases}

\begin{cases}x=11+12t\\y=8+10t\end{cases}

\begin{cases}x=-1-18t\\y=-2-15t\end{cases}

\begin{cases}x=11+5t\\y=8-6t\end{cases}

Solution

\begin{cases}x=5+12t\\y=4+10t\end{cases}

\begin{cases}x=11+12t\\y=8+10t\end{cases}

\begin{cases}x=-1-18t\\y=-2-15t\end{cases}

\begin{cases}x=11+5t\\y=8-6t\end{cases}

Select the second and third options. A parametrization describes $\ell$ when its starting point lies on $$5x-6y=7$$ and its nonzero direction is a scalar multiple of the original direction.

1.
Invalid: The direction is correct, but $$(5,4)$$ is not on $\ell$ : $5(5)-6(4)=1\ne7$ .
2.
Valid: $$5(11)-6(8)=7$$ , and the direction is twice the original direction.
3.
Valid: $$5(-1)-6(-2)=7$$ , and the direction is $$-3$$ times the original direction.
4.
Invalid: The starting point is on $\ell$ , but the direction is perpendicular to the original direction, rather than parallel to it.

(4 pts) Let $\ell’$ be the line perpendicular to $\ell$ that passes through $$(7,-6)$$ . Write an equation for $\ell’$ in the form $$ax+by=c$$ .

Solution

Because $\ell’$ is perpendicular to $\ell$ , the direction of $\ell$ is a normal to $\ell’$ . Thus $\ell’$ has an equation $$6x+5y=c$$ . Substituting its point $$(7,-6)$$ gives

$$ c=6(7)+5(-6)=12 $$

One equation is $\boxed{6x+5y=12}$ .

Problem 4 (11 pts)

Let $\vec u_1$ and $\vec u_2$ form an orthonormal basis of $\mathbb R^2$ . Suppose

\vec w=6\vec u_1+2\vec u_2.

(4 pts) Find $|\vec w|$ .

Solution

Orthonormality gives $|\vec u_1|=|\vec u_2|=1$ and $\vec u_1\cdot\vec u_2=0$ . Therefore,

\|\vec w\|^2=(6\vec u_1+2\vec u_2)\cdot(6\vec u_1+2\vec u_2) =36+24(0)+4=40

So $\boxed{|\vec w|=\sqrt{40}=2\sqrt{10}}$ .

(3 pts) Find $\vec w\cdot\vec u_1$ .

Solution

Distribute the dot product and use orthonormality:

\vec w\cdot\vec u_1 =6(\vec u_1\cdot\vec u_1)+2(\vec u_2\cdot\vec u_1) =6(1)+2(0)=\boxed{6}

(4 pts) Suppose $\vec v$ is another vector in $\mathbb R^2$ . Given that the angle between $\vec v$ and $\vec u_1$ is $35^\circ$ , which of the following could be the angle between $\vec v$ and $\vec u_2$ ? Select all that apply.

35^\circ

55^\circ

90^\circ

125^\circ

145^\circ

Solution

35^\circ

55^\circ

90^\circ

125^\circ

145^\circ

The basis vectors $\vec u_1$ and $\vec u_2$ are perpendicular. We’re given that the angle between $\vec v$ and $\vec u_1$ is $35^\circ$ —this could mean $\vec v$ is $35^\circ$ “below” $\vec u_1$ or $35^\circ$ “above” $\vec u_1$ .

The angles with $\vec u_2$ are $90^\circ-35^\circ=55^\circ$ and $90^\circ+35^\circ=125^\circ$ . These two pictures cover all possibilities up to rotation and reflection; the length of $\vec v$ does not affect its angles.

Problem 5 (12 pts)

Consider the vectors

\vec u_1=\frac1{9}\begin{bmatrix}4\\4\\7\end{bmatrix},\qquad \vec u_2=\frac1{9}\begin{bmatrix}-8\\1\\4\end{bmatrix},\qquad \vec w=\begin{bmatrix}4\\2\\3\end{bmatrix}

The vector $\vec w$ is in $\operatorname{span}(\vec u_1,\vec u_2)$ .

(6 pts) Verify that $\vec u_1$ and $\vec u_2$ are orthogonal and each has length $$1$$ .

Solution

Compute the dot product and both lengths:

\begin{align*} \vec u_1\cdot\vec u_2&=\frac{4(-8)+4(1)+7(4)}{81}=\frac{-32+4+28}{81}=0\\ \|\vec u_1\|&=\frac{\sqrt{4^2+4^2+7^2}}9=\frac{\sqrt{81}}9=1\\ \|\vec u_2\|&=\frac{\sqrt{(-8)^2+1^2+4^2}}9=\frac{\sqrt{81}}9=1 \end{align*}

The zero dot product verifies orthogonality, and the two lengths verify that both vectors are unit vectors.

(6 pts) Write $\vec w$ as a linear combination of $\vec u_1$ and $\vec u_2$ .

Solution

Write $\vec w=a\vec u_1+b\vec u_2$ . Taking dot products with the orthonormal vectors isolates the coefficients:

\begin{align*} a=\vec w\cdot\vec u_1&=\frac{4(4)+2(4)+3(7)}9=\frac{45}9=5\\ b=\vec w\cdot\vec u_2&=\frac{4(-8)+2(1)+3(4)}9=\frac{-18}9=-2 \end{align*}

Therefore, $\boxed{\vec w=5\vec u_1-2\vec u_2}$ . We can check the result directly:

5\vec u_1-2\vec u_2 =\frac19\begin{bmatrix}20+16\\20-2\\35-8\end{bmatrix} =\begin{bmatrix}4\\2\\3\end{bmatrix}=\vec w

Problem 6 (16 pts)

Let $$P$$ be the plane through $$(6,-5,8)$$ with direction vectors

\vec u=\begin{bmatrix}2\\5\\3\end{bmatrix},\qquad \vec v=\begin{bmatrix}4\\5\\1\end{bmatrix}

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

Solution

Let $\vec n=\begin{bmatrix}a\\b\\c\end{bmatrix}$ be a normal. It must be perpendicular to both directions, so

\begin{cases}2a+5b+3c=0\\4a+5b+c=0\end{cases}

Subtract the first equation from the second to get $$2a-2c=0$$ , or $$c=a$$ . Substitute into the first:

2a+5b+3a=0\quad\Longrightarrow\quad b=-a

Choose $$a=1$$ , giving $\vec n=\begin{bmatrix}1\\-1\\1\end{bmatrix}$ . The plane therefore has an equation $$x-y+z=d$$ . Its point $$(6,-5,8)$$ determines the constant:

$$ d=6-(-5)+8=19 $$

Thus $\boxed{P:\ x-y+z=19}$ . The two given directions are not scalar multiples, so they span a plane, and our normal is perpendicular to both.

(8 pts) For each vector-parametric equation below, select whether it is a valid description of $$P$$ and give a brief justification. In every option, $s,t\in\mathbb R$ .

Hint: Use the equation you found in the previous part.

i)
(2 pts) $\displaystyle\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\-5\\8\end{bmatrix}+s(\vec u+\vec v)+t\vec v$
Valid Invalid

Solution
Valid Invalid
Valid. The starting point is on $$P$$ , and $\vec u+\vec v$ and $\vec v$ span the same directions as $\vec u$ and $\vec v$ : recover $\vec u=(\vec u+\vec v)-\vec v$ . Thus they still span the entire plane.
ii)
(2 pts) $\displaystyle\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-4\\6\end{bmatrix}+s\vec u+t\vec v$
Valid Invalid

Solution
Valid Invalid
Valid. The new starting point satisfies $$9-(-4)+6=19$$ , so it lies on $$P$$ . The original two directions still span the plane; moving the starting point within $$P$$ does not change the plane.
iii)
(2 pts) $\displaystyle\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\-5\\9\end{bmatrix}+s\vec u+t\vec v$
Valid Invalid

Solution
Valid Invalid
Invalid. The starting point does not satisfy the plane’s equation:
$6-(-5)+9=20\ne19$
This parametrization describes a parallel plane, $$x-y+z=20$$ .
iv)
(2 pts) $\displaystyle\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\-5\\8\end{bmatrix}+s(2\vec u-3\vec v)+t(4\vec u-6\vec v)$
Valid Invalid

Solution
Valid Invalid
Invalid. The directions are scalar multiples:
$4\vec u-6\vec v=2(2\vec u-3\vec v)$
They provide only one independent direction, so this describes a line within $$P$$ , rather than the whole plane.

Problem 7 (10 pts)

The equations

P_1:\ 5x+3y=10,\qquad P_2:\ 2x+6y+4z=20

describe two planes in $\mathbb R^3$ . Write the set of common solutions to these two equations in vector-parametric form.

Solution

Method 1: Find a direction vector and a position vector. The intersection line lies in both planes, so its direction vector must be parallel to both planes. Thus it must be perpendicular to both normal vectors,

\vec n_1=\begin{bmatrix}5\\3\\0\end{bmatrix},\qquad \vec n_2=\begin{bmatrix}2\\6\\4\end{bmatrix}.

Write the direction vector as $\vec v=\begin{bmatrix}a\\b\\c\end{bmatrix}$ . Requiring both dot products to be zero gives

\vec v\cdot\vec n_1=5a+3b=0,\qquad \vec v\cdot\vec n_2=2a+6b+4c=0.

The first equation gives $$b=-5a/3$$ . Substituting into the second,

2a+6\left(-\frac{5a}{3}\right)+4c=0 \quad\Longrightarrow\quad -8a+4c=0 \quad\Longrightarrow\quad c=2a.

Therefore $\vec v=a\begin{bmatrix}1\\-5/3\\2\end{bmatrix}$ . We need a nonzero direction vector; choosing $$a=3$$ gives $$b=-5$$ , $$c=6$$ , and $\vec v=\begin{bmatrix}3\\-5\\6\end{bmatrix}$ .

To find a position vector, find any one point that lies on both planes, so we know it is on their intersection line. Set $$y=0$$ . Then $$P_1$$ gives $$5x=10$$ , so $$x=2$$ , and $$P_2$$ gives $$4+4z=20$$ , so $$z=4$$ . Indeed,

5(2)+3(0)=10,\qquad 2(2)+6(0)+4(4)=20.

Thus we can use $\begin{bmatrix}2\\0\\4\end{bmatrix}$ as a position vector, giving

\boxed{\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}2\\0\\4\end{bmatrix} +t\begin{bmatrix}3\\-5\\6\end{bmatrix},\qquad t\in\mathbb R.}

The normals are not parallel, so the planes intersect in a line; this point and nonzero direction vector describe that entire line.

Method 2: Solve directly for the common points.

First solve for $$y$$ in terms of $$x$$ using the first equation:

5x+3y=10\quad\Longrightarrow\quad y=\frac{10-5x}{3}

Substitute this into the second equation to solve for $$z$$ :

\begin{align*} 2x+6\left(\frac{10-5x}{3}\right)+4z&=20\\ 2x+20-10x+4z&=20\\ z&=2x \end{align*}

Since $$x$$ can be any real number, the common solutions are

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}0\\10/3\\0\end{bmatrix} +x\begin{bmatrix}1\\-5/3\\2\end{bmatrix},\qquad x\in\mathbb R

This is already a valid answer. To avoid fractions, multiply the direction vector by $$3$$ , which does not change the line. Choosing the point corresponding to $$x=2$$ as our starting point gives

\boxed{\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}2\\0\\4\end{bmatrix} +t\begin{bmatrix}3\\-5\\6\end{bmatrix},\qquad t\in\mathbb R}

Equivalently, we have reparameterized by setting $$x=2+3t$$ . As $$t$$ ranges over all real numbers, $$x$$ does too, so we still describe the entire intersection line.

Problem 8 (12 pts)

Let

\vec v=\begin{bmatrix}7\\5\\6\end{bmatrix},\qquad \ell=\operatorname{span}\left(\begin{bmatrix}4\\-5\\7\end{bmatrix}\right),\qquad P:\ 4x-5y+7z=0

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

Solution

Let $\vec w=\begin{bmatrix}4\\-5\\7\end{bmatrix}$ , the vector spanning $\ell$ . Use the projection formula:

\begin{align*} \operatorname{proj}_{\ell}(\vec v) &=\frac{\vec v\cdot\vec w}{\vec w\cdot\vec w}\vec w\\ &=\frac{7(4)+5(-5)+6(7)}{4^2+(-5)^2+7^2}\begin{bmatrix}4\\-5\\7\end{bmatrix}\\ &=\frac{45}{90}\begin{bmatrix}4\\-5\\7\end{bmatrix} =\boxed{\begin{bmatrix}2\\-5/2\\7/2\end{bmatrix}} \end{align*}

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

Solution

The coefficients of $$P$$ tell us that the vector $\begin{bmatrix}4\\-5\\7\end{bmatrix}$ is perpendicular to $$P$$ . But $\ell$ is the span of $\begin{bmatrix}4\\-5\\7\end{bmatrix}$ , so $$P$$ and $\ell$ are perpendicular, and $\vec v$ ’s projections onto $\ell$ and onto $$P$$ sum to $\vec v$ exactly. Thus,

\begin{align*} \operatorname{proj}_{P}(\vec v) &=\vec v-\operatorname{proj}_{\ell}(\vec v)\\ &=\begin{bmatrix}7\\5\\6\end{bmatrix}-\begin{bmatrix}2\\-5/2\\7/2\end{bmatrix} =\boxed{\begin{bmatrix}5\\15/2\\5/2\end{bmatrix}} \end{align*}

As a check, this vector lies in $$P$$ :

4(5)-5\left(\frac{15}2\right)+7\left(\frac52\right)=0