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<h1><a href="/">In principio erat Verbum </a></h1>
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<li><a href="/category/expository.html">Expository</a></li>
<li class="active"><a href="/category/math-212.html">Math 212</a></li>
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<h1 class="entry-title">
<a href="/quiz-1-solution.html" rel="bookmark"
title="Permalink to Quiz 1 Solution">Quiz 1 Solution</a></h1>
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<abbr class="published" title="2015-09-18T18:39:00-04:00">
Published: Fri 18 September 2015
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<br />
<abbr class="modified" title="2015-11-18T00:00:00-05:00">
Updated: Wed 18 November 2015
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<address class="vcard author">
By <a class="url fn" href="/author/tingran-gao.html">Tingran Gao</a>
</address>
<p>In <a href="/category/math-212.html">Math 212</a>.</p>
<p>tags: <a href="/tag/math-212.html">Math 212</a> <a href="/tag/quiz.html">Quiz</a> </p>
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<p><strong>Problem 1.</strong> Let <span class="math">\(\vec{a}=\left(6,4,2\right),\quad \vec{b}=\left(-9,6,3\right),\quad \vec{c}=\left(-3,6,3\right)\)</span>.</p>
<p>(a) Compute <span class="math">\(2\vec{a}-\vec{b}+\vec{c}\)</span> and <span class="math">\(-3\vec{a}+2\vec{b}+4\vec{c}\)</span>.</p>
<p><strong>Solution.</strong>
</p>
<div class="math">\begin{align*}
2\vec{a}-\vec{b}+\vec{c} &= \left(12+9-3, 8-6+6, 4-3+3\right)=\left(18, 8, 4\right),\\
-3\vec{a}+2\vec{b}+4\vec{c} &= \left(-18-18-12, -12+12+24, -6+6+12\right)=\left(-48, 24, 12\right).
\end{align*}</div>
<p>(b) Do <span class="math">\(\vec{a},\vec{b},\vec{c}\)</span> lie on the same plane? Show your computation.</p>
<p><strong>Solution.</strong> Yes. Check the volume of the parallelepiped spanned by <span class="math">\(\vec{a},\vec{b},\vec{c}\)</span>:
</p>
<div class="math">\begin{align*}
\vec{a}\times\vec{b}\cdot\vec{c}
=\begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \notag
\end{vmatrix}
=\begin{vmatrix}
6 & 4 & 2 \\
-9 & 6 & 3 \\
-3 & 6 & 3 \notag
\end{vmatrix}=0.
\end{align*}</div>
<p>
(If you notice that the second column is a scalar multiple of the third column, you can deduce that the determinant must be <span class="math">\(0\)</span> — without any computation.)</p>
<hr />
<p><strong>Problem 2.</strong> Find the distance from the point <span class="math">\(\left(0,2,1\right)\)</span> to the plane <span class="math">\(2x-3y+5z-1=0\)</span>. <em>(Hint: The only formula you need is the dot product. Keep <span class="math">\(\sqrt{38}\)</span> as it is.)</em></p>
<p><strong>Solution.</strong> Let <span class="math">\(P\)</span> be an arbitrary point on the plane, say <span class="math">\(P=\left(-1,-1,0\right)\)</span>. The distance from <span class="math">\(A=\left(0,2,1\right)\)</span> equals to the absolute value of the component of the vector <span class="math">\(\overrightarrow{AP}\)</span> along the normal vector of the plane.
<div style="text-align:center" markdown=1>
<img alt="pointPlaneDistance" src="/images/pointPlaneDistance.png" />
</div>
A normal vector of the plane can be chosen as <span class="math">\(\vec{n}=\left(2,-3,5\right)\)</span>. Moreover, <span class="math">\(\overrightarrow{AP}=\overrightarrow{OP}-\overrightarrow{OA}=\left(-1,-3,-1\right)\)</span>. Thus the desired distance is
</p>
<div class="math">$$\left|\mathrm{comp}_{\vec{n}}\,\overrightarrow{AP}\right|=\frac{\left|\overrightarrow{AP}\cdot\vec{n}\right|}{\left|\vec{n}\right|}=\frac{\left|-2+9-5\right|}{\sqrt{38}}=\frac{2}{\sqrt{38}}.$$</div>
<hr />
<p><strong>Problem 3.</strong> Consider a point <span class="math">\(M\)</span> in the same plane as the triangle <span class="math">\(ABC\)</span>. Show that there exist real numbers <span class="math">\(\lambda,\mu,\nu\)</span> such that
</p>
<div class="math">$$\overrightarrow{OM}=\lambda \overrightarrow{OA}+\mu\overrightarrow{OB}+\nu\overrightarrow{OC}$$</div>
<p>
and
</p>
<div class="math">$$\lambda + \mu + \nu = 1.$$</div>
<p>
<em>(Hint: <span class="math">\(\overrightarrow{AM}\)</span> lies in the plane spanned by <span class="math">\(\overrightarrow{AB}\)</span> and <span class="math">\(\overrightarrow{AC}\)</span>.)</em></p>
<div style="text-align:center">
<p><img alt="triangle" src="/images/triangle.png" /></p>
</div>
<p><strong>Proof.</strong> Since <span class="math">\(M\)</span> in the same plane as the triangle <span class="math">\(ABC\)</span>, <span class="math">\(\overrightarrow{AM}\)</span> lies in the plane spanned by <span class="math">\(\overrightarrow{AB}\)</span> and <span class="math">\(\overrightarrow{AC}\)</span>. Thus there exists real numbers <span class="math">\(s,t\)</span> such that
</p>
<div class="math">\begin{equation}\label{eq:linearDependent}
\overrightarrow{AM} = s\,\overrightarrow{AB} + t\,\overrightarrow{AC}.
\end{equation}</div>
<p>
Note that
</p>
<div class="math">\begin{align*}
\overrightarrow{AM} &= \overrightarrow{OM}-\overrightarrow{OA},\\
\overrightarrow{AB} &= \overrightarrow{OB}-\overrightarrow{OA},\\
\overrightarrow{AC} &= \overrightarrow{OC}-\overrightarrow{OA},
\end{align*}</div>
<p>
thus \eqref{eq:linearDependent} reduces to
</p>
<div class="math">$$\left(\overrightarrow{OM}-\overrightarrow{OA}\right) = s\left(\overrightarrow{OB}-\overrightarrow{OA}\right) + t\left(\overrightarrow{OC}-\overrightarrow{OA}\right).$$</div>
<p>
Re-grouping terms involving <span class="math">\(\overrightarrow{OA}\)</span>, <span class="math">\(\overrightarrow{OB}\)</span>, <span class="math">\(\overrightarrow{OC}\)</span>, one has
</p>
<div class="math">$$\overrightarrow{OM}=\left(1-s-t\right) \overrightarrow{OA}+s\,\overrightarrow{OB}+t\,\overrightarrow{OC}.$$</div>
<p>
Set
</p>
<div class="math">$$\lambda = 1-s-t,\quad \mu=s,\quad \nu=t,$$</div>
<p>
then
</p>
<div class="math">$$\overrightarrow{OM}=\lambda \overrightarrow{OA}+\mu\overrightarrow{OB}+\nu\overrightarrow{OC}$$</div>
<p>
and
</p>
<div class="math">$$\lambda + \mu + \nu = 1.$$</div>
<p>
<span class="math">\(\blacksquare\)</span></p>
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<li><a href="http://www.math.duke.edu/">Department of Mathematics</a></li>
<li><a href="http://www.math.duke.edu/courses/math_everywhere/">Math Everywhere @ Duke</a></li>
<li><a href="http://www.wolframalpha.com/">Wolfram Alpha</a></li>
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