First Midterm Midpoint Assessment

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These are meant to be questions to help you gauge where you are in terms of your level of understanding of the material so far. This is the official half way point between the start of the semester and the first midterm, so I thought this would be a good time for it. They are split up into definitions/general philosophy, quick answer problems, medium questions, and then harder problems.

Some problems have hints.

Definitions and General Philosophy

A unit vector $\textbf {u}$ is a vector that has length $\answer {1}$ .

Which of the following is true of $\text {proj}_{\textbf {w}} \textbf {v}$ , the projection of $\textbf {v}$ onto $\textbf {w}$ ? Select all that apply.

If the projection is not the zero vector, then it must be parallel to $\textbf {w}$ . If the projection is not the zero vector, then it must be in the direction of $\textbf {w}$ . If the projection is not the zero vector, then it must be in the direction of $\textbf {v}$ . If the angle between $\textbf {v}$ and $\textbf {w}$ is $\pi /2$ , then the projection is the zero vector.

To define a line $L$ in any dimension, we need a point and a:

A vector perpendicular to $L$ . A vector parallel to $L$ .

To define a plane $P$ , we need a point and a:

A vector perpendicular to $P$ . A vector in the plane $P$ .

The normalization of a nonzero vector $\textbf {v}$ is a vector $\hat {\textbf {v}}$ of length $\answer {1}$ which is in the samein the opposite direction of the vector $\textbf {v}$ .

Problems

Let $\textbf {v}$ be the vector $\textbf {v} = \left \langle 2,1,3 \right \rangle$ . Then:

$\| \textbf {v}\| = \answer {\sqrt {14}}$ .
$2 \textbf {v} = \left \langle \answer {4}, \answer {2}, \answer {6} \right \rangle$ .
$\textbf {v} \cdot \textbf {v} = \answer {14}$ .
$\textbf {v} \times \textbf {v} = \left \langle \answer {0}, \answer {0}, \answer {0} \right \rangle$ .
$\text {proj}_{\textbf {v}} \textbf {v} = \left \langle \answer {2}, \answer {1}, \answer {3} \right \rangle$ .
The normalization $\hat {\textbf {v}} = \left \langle \answer {\frac {2}{\sqrt {14}}},\answer {\frac {1}{\sqrt {14}}},\answer {\frac {3}{\sqrt {14}}} \right \rangle$ .

Suppose $\textbf {r}(t) = \left \langle 2t^2,3t-1,\sin (2t) \right \rangle$ . Then the tangent vector to $\textbf {r}(t)$ at the point $t=2$ is $\textbf {r}'(2) = \left \langle \answer {8}, \answer {3}, \answer {2\cos (4)} \right \rangle .$

The equation of the plane through the point $(1,2,1)$ and perpendicular to the vector $\textbf {n} = \left \langle 1,-2,1 \right \rangle$ is (give the answer in the form $ax+by+cz=d$ ): $\answer {x-2y+z}=-2.$

A vector equation for the line through the points $(1,1,2)$ and $(2,3,1)$ is given by $\textbf {r}(t) = \left \langle 1+t,\answer {1+2t},\answer {2-t} \right \rangle .$

If $\textbf {v} = \left \langle 2,1,1 \right \rangle$ and $\textbf {w} = \left \langle 1,0,-2 \right \rangle$ , then $\textbf {v} \times \textbf {w} = \left \langle \answer {-2},\answer {5},\answer {-1} \right \rangle .$

An equation for the plane through the point $(1,2,-1)$ which is perpendicular to the line given by symmetric equations
$\displaystyle \frac {x-1}{3} = y-2 = \frac {5-z}{5}$ is (give the answer in the form $ax+by+cz = d$ ): $\answer {3x+y-5z} = 10.$

What relationship does the normal vector for the plane have to the direction vector for the line?

The direction vector for the line is the normal vector for the plane. Note that the direction vector is $\left \langle 3,1,-5 \right \rangle$ . We also know a point on the plane. Use these two things to write down the plane’s equation.

Are the lines $\textbf {r}_1(t) = \left \langle 1+t,2-2t,3-t \right \rangle$ and $\textbf {r}_2(t) = \left \langle 2-3t,t,2t \right \rangle$ parallel, skew, or intersecting?

Parallel Skew Intersecting

To see if they are parallel, check if their direction vectors are parallel.

To see if they are intersecting or skew, we set $\textbf {r}_1(t) = \textbf {r}_2(s)$ and try to solve for $t$ and $s$ .

The direction vectors are $\left \langle 1,-2,-1 \right \rangle$ and $\left \langle -3,1,2 \right \rangle$ , which are not parallel vectors, so the lines are not parallel. To determine whether they intersect or are skew, we set $\textbf {r}_1(t) = \textbf {r}_2(s)$ and try to solve for $s$ and $t$ . We get three equations: $1+t = 3-2s$ $2-2t = s$ $3-t = 2s.$ Solving the last two equations for $s$ and $t$ gives $t = \answer {\frac {1}{3}}$ and $s = \answer {\frac {4}{3}}$ . However, these do not satisfy the first equation. Therefore, the lines are skew.

An equation for the plane which contains both the point $(1,1,0)$ and the line given by $\textbf {r}(t) = \left \langle 2+3t,5-t,3 \right \rangle$ is given by (give the answer in the form $ax+by+cz=d$ ): $\answer {-3x-9y+13z} = -12.$

Get two vectors in the plane and cross them to get the normal vector.

One vector to use is the direction vector for the line. Another vector you can use is a vector from the line to the extra point $(1,1,0)$ . Choose any point on the line for this second vector.

The equation should be $-3x-9y+13z=-12$ .

The volume of the parallelepiped formed by the vectors $\textbf {u} = \left \langle 1,-1,0 \right \rangle$ , $\textbf {v} = \left \langle 2,3,0 \right \rangle$ , and $\textbf {w} = \left \langle 2,3,5 \right \rangle$ is equal to $\answer {25}$ .

The volume is $|\textbf {u} \cdot (\textbf {v} \times \textbf {w})|$ .

The arc length of the portion of the curve $\textbf {r}(t) = \left \langle \cos (t),\sin (t),t \right \rangle$ which is above the $xy$ -plane and below the paraboloid $z = 9-x^2-y^2$ is equal to $\answer {8\sqrt {2}}$ .

Notice that $\| \textbf {r}'(t) \| = \answer {\sqrt {2}}$ . Now we need endpoints for $t$ . The $xy$ -plane is where $z = \answer {0}$ , so if the curve is to be above the $xy$ -plane, the equation for the curve tells us $t \geq \answer {0}$ . We can also find where the helix hits the paraboloid by plugging in the expressions for $x$ , $y$ , and $z$ into the paraboloid’s equation. Doing this gives $t = \answer {8}$ , so this is our upper bound for $t$ . Now use the arc-length formula.

Find an equation of a plane whose intersection with the plane $2x-y+z=0$ is the line $\textbf {r}(t) = \left \langle t+1,2t,-2 \right \rangle$ . (Nothing to input here, see the hint if you want a possible solution.)

To get a point on the plane, let $t=0$ in the line, so $(1,0,-2)$ (you can let $t$ be anything to get a point on the plane). One possible normal vector comes from crossing the normal vector of the given plane and the direction vector for the line. This gives $\left \langle 2,-1,1\right \rangle \times \left \langle 1,2,0\right \rangle = \left \langle -2,1,5 \right \rangle .$ In reality, any normal which is perpendicular to the direction vector for the line will do.

The point on the plane $x+2y-z = 3$ closest to the point $(1,1,1)$ is $\left (\answer {\frac {7}{6}},\answer {\frac {4}{3}},\answer {\frac {5}{6}}\right )$ .

To get the closest point, form the line from $(1,1,1)$ in the direction of the normal vector. This gives $\textbf {r}(t) = \left \langle 1,\answer {1},\answer {1} \right \rangle + t\left \langle 1,\answer {2},\answer {-1} \right \rangle = \left \langle t+1,2t+1,-t+1 \right \rangle$ . To get the intersection with the plane, plug in $x$ , $y$ , and $z$ into the plane’s equation, to get $t = \answer {1/6}$ . Now plug this into the line’s equation to get the closest point.

True/False and Multiple Choice

The graph of $f(x,y) = 8+2\sqrt {x^2+y^2}$ is a:

Paraboloid Cone Sphere

The level curves of the function $f(x,y) = \sqrt {x+y}$ are:

Square root graphs Parabolas Lines Circles

True/False: For a function $z = f(x,y)$ , if $c_1 \neq c_2$ , then the two level curves $c_1 = f(x,y)$ and $c_2 = f(x,y)$ can intersect.

True False

True/False: The map $\textbf {r}(t) = \left \langle \cos (2t),\sin (2t),\sin (2t) \right \rangle$ for $0 \leq t \leq \pi$ parametrizes the intersection of $x^2+z^2=1$ with the plane $y=z$ .

True False

True/False: The integral $\int _a^b \textbf {r}'(t)dt$ gives the arc length of the portion of the curve $\textbf {r}(t)$ for $a \leq t \leq b$ .

True False

True/False: If $\textbf {u}$ and $\textbf {v}$ are nonzero vectors with $\textbf {u} \cdot \textbf {v} > 0$ and $\text {proj}_{\textbf {v}} \textbf {u} = \text {proj}_{\textbf {u}} \textbf {v}$ , then it must be true that $\textbf {u} = \textbf {v}$ .

True False

Which of the following vector valued functions parametrize the curve $x = y^2$ for $-3 \leq y \leq 3$ . Select all that apply.

$\textbf {r}(t) = \left \langle t^2,t \right \rangle$ for $-3 \leq t \leq 3$ $\textbf {r}(t) = \left \langle t,\sqrt {t} \right \rangle$ for $0 \leq t \leq 3$ $\textbf {r}(t) = \left \langle t^4,t^2\right \rangle$ for $-3 \leq t \leq 3$

The bounds matter.