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      <text>9.4.4 Using determinants to evaluate a vector product</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h4><b>Using determinants to evaluate a vector product</b></h4><p>Evaluation of a vector product using the formula in a previous Key point is very cumbersome. A more convenient and easily remembered method is to use determinants. As we hope you will remember, for a <span class="math inline">\(3\times 3\)</span> determinant,</p>

<p>\[
\left|
\begin{matrix}
a &amp; b &amp; c \\
d &amp; e &amp; f \\
g &amp; h &amp; i
\end{matrix}
\right| = a\left|
\begin{matrix}
e &amp; f\\
h &amp; i\\
\end{matrix}\right|
-
b\left|
\begin{matrix}
d &amp; f\\
g &amp; i\\
\end{matrix}\right|
+c\left|
\begin{matrix}
d &amp; e\\
g &amp; h\\
\end{matrix}\right|\]</p>

<p>The vector product of two vectors <span class="math inline">\(\underline{a}=a_1\underline{i}+a_2\underline{j}+a_3\underline{k}\)</span> and <span class="math inline">\(\underline{b}=b_1\underline{i}+b_2\underline{j}+b_3\underline{k}\)</span> can be found by evaluating the determinant:</p>


<p>\[
\underline{a}\times\underline{b}= \left|
\begin{matrix}
\underline{i} &amp; \underline{j}&amp; \underline{k}\\
a_1 &amp; a_2 &amp; a_3 \\
b_1 &amp; b_2 &amp; b_3 \\
\end{matrix}\right|\]</p>

<p><br></p><p>in which <span class="math inline">\(\underline{i}\)</span>, <span class="math inline">\(\underline{j}\)</span> and <span class="math inline">\(\underline{k}\)</span> are (temporarily) treated as if they were scalars.</p><p><br></p>

<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>If&nbsp;<span class="math inline">\(\underline{a}=a_1\underline{i}+a_2\underline{j}+a_3\underline{k}\)&nbsp;</span>and <span class="math inline">\(\underline{b}=b_1\underline{i}+b_2\underline{j}+b_3\underline{k}\)&nbsp;</span>then <span class="math display">\[\underline{a}\times \underline{b} = \left|
\begin{matrix}
\underline{i} &amp; \underline{j} &amp; \underline{k} \\
a_1 &amp; a_2 &amp; a_3 \\
b_1 &amp; b_2 &amp; b_3
\end{matrix}
\right|=\underline{i}(a_2b_3-a_3b_2)
-\underline{j}(a_1b_3-a_3b_1)+\underline{k}(a_1b_2-a_2b_1)\]</span></p>
</div>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the vector product of <span class="math inline">\(\underline{a}=3\underline{i}-4\underline{j} +2\underline{k}\)</span> and <span class="math inline">\(\underline{b}=9\underline{i}-6\underline{j}+2\underline{k}\)</span>.</p>

<h4 class="HELM_solution">Solution</h4>
<p>We have <span class="math inline">\( \underline{a}\times\underline{b}=\left|\begin{matrix} \underline{i} &amp; \underline{j} &amp; \underline{k}\\
3 &amp; -4 &amp; 2 \\
9 &amp; -6 &amp; 2 \\\end{matrix}\right|\)&nbsp;</span>which, when evaluated, gives</p>

<p><span class="math inline">\(\qquad \underline{a}\times \underline{b}=\underline{i}(-8-(-12)) - \underline{j}(6-18) +
\underline{k}(-18-(-36)) =4\underline{i} +12\underline{j} +18\underline{k}\)</span></p>


<hr>
<h4 class="HELM_example">Example</h4>
<p>The area <span class="math inline">\(A_T\)</span> of the triangle shown in the Figure below is given by the formula <span class="math inline">\(A_T=\dfrac{1}{2} bc\sin \alpha\)</span>. Show that an equivalent formula is <span class="math inline"> \(A_T = \dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\)</span>.</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/block5fig7.svg" alt="triangle with vertices A,B and C and standard notation for sides and angles" role="presentation" class="img-responsive atto_image_button_text-bottom" width="121" height="112">
</figure>

<h4 class="HELM_solution">Solution</h4>
<p>We use the definition of the vector product <span class="math inline">\(\quad |\overrightarrow{AB}\times \overrightarrow{AC}|= |\overrightarrow{AB}|\,| \overrightarrow{AC}| \sin \alpha.\)</span></p>
<p>Since <span class="math inline">\(\alpha\)</span> is the angle between <span class="math inline">\(\overrightarrow{AB}\)</span> and <span class="math inline">\(\overrightarrow{AC}\)</span>, and <span class="math inline">\(|\overrightarrow{AB}|=c\)</span> and <span class="math inline">\(|\overrightarrow{AC}|=b\)</span>, the required result follows immediately: <span class="math inline">\(\qquad \dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\dfrac{1}{2}c\cdot b\cdot\sin\alpha\)</span>.</p><p><br></p><h5>Moments</h5>

<p>The <span><strong>moment</strong></span> (or <span><strong>torque</strong></span>) of the force <span class="math inline">\(\underline{F}\)</span> about a point <span class="math inline">\(O\)</span> is defined as&nbsp;</p><p><span class="math inline">\(\underline{M}_o=\underline{r}\times\underline{F}\)</span>&nbsp;</p><p><br></p><p>where <span class="math inline">\(\underline{r}\)</span> is a position vector from <span class="math inline">\(O\)</span> to <span><em>any</em></span> point on the line of action of <span class="math inline">\(\underline{F}\)</span> as shown in the following Figure.</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/workbook9bl58a.svg" alt="" role="presentation" class="img-responsive atto_image_button_text-bottom" width="214" height="95"><br></p><p style="text-align: center; ">
</figure>

<p>It may seem strange that any point on the line of action may be taken but it is easy to show that exactly the same vector <span class="math inline">\({\underline M}_o\)</span> is always obtained.<br>
By the properties of the cross product the direction of <span class="math inline">\(\underline{M}_o\)</span> is perpendicular to the plane containing <span class="math inline">\(\underline{r}\)</span> and <span class="math inline">\(\underline{F}\)</span> (i.e. out of the screen). The magnitude of the moment is</p>

<p><span class="math inline">\(\qquad |\underline{M}_0|=|\underline{r}||\underline{F}|\sin\theta\)</span>.&nbsp;</p><p>From the figure above, <span class="math inline">\(|\underline{r}|\sin\theta= OD\)</span>. Hence <span class="math inline">\(|\underline{M}_o|=OD|\underline{F}|\)</span>. This would be the same no matter which point on the line of action of <span class="math inline">\(\underline{F}\)</span> was chosen.</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the moment of the force given by <span class="math inline">\(\underline{F}=3\underline{i}+4\underline{j}+5\underline{k}\)</span> (N) acting at the point <span class="math inline">\((14,-3,6)\)</span> about the point <span class="math inline">\(P(2,-2,1)\)</span>.</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/workbook9bl58b.svg" alt="" role="presentation" class="img-responsive atto_image_button_text-bottom" width="453" height="139">
</figure>

<h4 class="HELM_solution"><br></h4><h4 class="HELM_solution">Solution</h4>
<p>The vector <span class="math inline">\(\underline{r}\)</span> can be any vector from the point <span class="math inline">\(P\)</span> to any point on the line of action of <span class="math inline">\(\underline{F}\)</span>. Choosing <span class="math inline">\({\underline r}\)</span> to be the vector connecting <span class="math inline">\(P\)</span> to <span class="math inline">\((14,-3,6)\)</span> (and measuring distances in metres)</p><p>&nbsp;we have: <span class="math inline">\(\qquad
\underline{r}=(14-2)\underline{i}+(-3-(-2))\underline{j}+(6-1)\underline{k}=12\underline{i}-\underline{j}+5\underline{k}\)</span>.&nbsp;</p><p><br></p><p>The moment is <span class="math inline">\(\quad \underline{M}_\rho=\underline{r}\times \underline{F}=\left|
\begin{matrix}
\underline{i}&amp;\underline{j}&amp;\underline{k}\\
12&amp;-1&amp;5\\
3&amp;4&amp;5\end{matrix}\right|=
-25\underline{i}-45\underline{j}+51\underline{k}\)</span> (N m)</p>]]></text>
<file name="block5fig7.svg" path="/" encoding="base64">[encoded data, 20704 characters]</file>
<file name="workbook9bl58a.svg" path="/" encoding="base64">[encoded data, 15808 characters]</file>
<file name="workbook9bl58b.svg" path="/" encoding="base64">[encoded data, 52088 characters]</file>
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