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Test 1
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<h2 class="justme">There was no cross product on this one</h2>
<h3>Name-<br>
Math 213, Fall 2016, Quiz - Review of Chapter 9</h3>
<ol>
<li>
Consider the following regular hexagon.  (Regular means that all sides are of equal length and that all interior angles are equal).<br>
<img src="0g/T1.3s.jpg"><br>
<ol type="a"><li>Give the vector components for $\myv b$.  <br><br><br>
<li>Compute $\myv a\cdot\myv b$. <br><br><br>
<li>In the diagram, draw $\text{proj}_\myv a \myv b$, which is the vector projection of $\myv b$ onto $\myv a$ and label it.
<li>In the diagram, draw the vector $\myv s=\myv a+\myv b+\myv c$ and label it.
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<li>
Find an equation of the plane through the point $(2,-3,6)$ with normal vector $\myv n = \myc{2,1, -4}$
<p class="answer">Using the components of the normal vector, the plane could have the equation $2x+y-4z=c$.  Substituting in the coordinates of the one point we know...
$$c=2(2)+1(-3)-4(6)=4-3-24=-23$$
So the equation is:
$$2x+y-4z=-23.$$

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Given the points (1,4,-5) and (3,-1,5)
<ol type="a">
<li>
Find a parametric equation for the line that passes through the points.
<p class="answer">Let the first point be the position vector $\myv a$, and the second point have the position vector $\myv b$. The difference vector between these two points is
$$\myv b - \myv a =2\uv i-5\uv j+10\uv k.$$
So, the parametric equation could be
$$\myv a + t (\myv b-\myv a) = (1,4,-5) +t\myc{ 2,-5, 10}.$$

<li>
Find the point where the line intersects the $xz$-plane.</li>
<p class="answer">The intersection happens when $y=0$. Find $t$:
$$0=4-5t; \ \ \Rightarrow t=\frac 45.$$
The point of intersection is...
$$(1,4,-5)+\frac 45(2,-5,10)=(2.8,0,3).$$


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Match the equations (not all will have a match) with their graphs.<br>

<img src="0g/Qsurf.jpg" style="width:70%;"><br>
<ol type="a">
<li>$r=3$ (Cylindrical)
<li>$8x+2y+3z=0$
<li>$\theta=\pi/3$ (Cylindrical)
<li>$\phi=\pi/6$ (Spherical)
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