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9.3 - The Regular Hexagon
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<h2>9.3 - The regular Hexagon</h2>

<img src="0g/9.3.hex.jpg">

<p>Hints...
<ul><li>In #3 think carefully about the positions of the vectors and the way the angle between two vectors is defined in <a href="../gco/mv09.3.php#dotp">the dot product</a>.
<li>In #6, try sketching the sum visually by drawing the vectors head to tail...
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Answers:
<ol>
<li>$|\myv a| = \sqrt{1^2+0}=1$.  And since all the sides of the regular
hexagon are the same length,<span class="answer">$|\myv a|=|\myv b|=|\myv c|=1$</span>.
<li>The angle is <span class="answer">120${}^o$ or $\pi/3$ radians</span>
<li>If we put $\myv a $ and $\myv b$ tail-to-tail, the angle between them is $60^o$, so $\myv a\cdot\myv b=ab\cos 60^0=1*1*\frac 12 =$ <span class="answer">$\frac 12$</span>.
<li>The angle between $\myv a$ and $\myv c$ is 120${}^o$, and the cosine of that angle is -1/2.  So, the dot product is <span class="answer">$-\frac 12$</span>.
<li><img src="0g/9.3.hex.51.jpg"><br>The projection of $\myv b$ onto $\myv a$ is a vector which has magnitude $b\cos 60^o=1*\frac12=$1/2, and points in the direction of $\myv a$. Now, the length of $\myv a$ is 1, so it is already a unit vector, so I could writes this as  <span class="answer">$\text{proj}_{\myv a}\myv b=\frac 12 \myv a$ or $\frac12 \uv a$</span>
<br><br><img src="0g/9.3.hex.52.jpg"><br>
$\text{proj}_{\myv b}\myv c = $ <span class="answer">$\frac 12\uv b$</span>.
<li><img src="0g/9.3.hex.6.jpg"><br>
The component of a vector is a <i>scalar</i>.  From the sketch, than answer is <span class="answer">1</span>
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