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      <text>9.3.3.1 Resolving one vector along another</text>
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      <text><![CDATA[<h4><b>Resolving one vector along another</b></h4><p>The scalar product can be used to find the component of a vector in the direction of another vector. Consider the figure below which shows two arbitrary vectors <span class="math inline">\(\underline{a}\)</span> and <span class="math inline">\(\underline{n}\)</span>. Let <span class="math inline">\(\underline{\hat{n}}\)</span> be a <span><strong>unit vector</strong></span> in the direction of <span class="math inline">\(\underline{n}\)</span>.</p>

<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/block4fig4.svg" alt="projection of a onto n" role="presentation" class="img-responsive atto_image_button_text-bottom" width="312" height="150">
</figure>

<p>Study the figure carefully and note that a perpendicular has been drawn from <span class="math inline">\(P\)</span> to meet <span class="math inline">\(\underline{n}\)</span> at <span class="math inline">\(Q\)</span>. The distance <span class="math inline">\(OQ\)</span> is called the projection of <span class="math inline">\(\underline{a}\)</span> onto <span class="math inline">\(\underline{n}\)</span>. Simple trigonometry tells us that the length of the projection is <span class="math inline">\(|\underline{a}|\cos\theta\)</span>. Now by taking the scalar product of <span class="math inline">\(\underline{a}\)</span> with the unit vector <span class="math inline">\(\underline{\hat{n}}\)</span> we find&nbsp;</p><p><br></p><p>&nbsp;

<span class="math inline">\(\qquad\underline{a}\cdot\underline{\hat{n}} =|\underline{a}|\,|\hat{\underline{n}}|\,\cos\theta =|\underline{a}|\cos\theta\qquad\qquad\hbox{(since } |\underline{\hat{n}}|=1)\)</span>&nbsp;</p><p><br></p>


<p>&nbsp;We conclude that</p>


<div class="HELM_keypoint">
<h4>Key Point</h4>
<p style="text-align: center; "><span class="math display">\[\underline{a}\cdot \underline{\hat{n}} \hbox{ is the component of } \underline{a} \hbox{ in the direction of } \underline{n}\]</span></p>
</div>

<hr>
<h4 class="HELM_example ">Example</h4><p>The figure above shows a plane containing the point <span class="math inline">\(A\)</span> which has position vector <span class="math inline">\(\underline{a}\)</span>. The vector <span class="math inline">\(\underline{\hat{n}}\)</span> is a unit vector perpendicular to the plane (such a vector is called a unit normal vector). Find an expression for the perpendicular distance, <span class="math inline">\(\ell\)</span>, of the plane from the origin.</p><br>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/9-3-fig-4-normal-to-plane.svg" alt="" width="273" height="121" class="img-responsive atto_image_button_text-bottom">
</figure>
<h4 class="HELM_solution">Solution</h4>
<p>From the diagram we note that the perpendicular distance <span class="math inline">\(\ell\)</span> of the plane from the origin is the projection of <span class="math inline">\(\underline{a}\)</span> onto <span class="math inline">\(\underline{\hat{n}}\)</span> and, using the previous Key point, is thus <span class="math inline">\(\underline{a}\cdot \hat{\underline{n}}\)</span>.</p><p></p>]]></text>
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<file name="block4fig4.svg" path="/" encoding="base64">[encoded data, 15732 characters]</file>
<file name="block4fig5.svg" path="/" encoding="base64">[encoded data, 28504 characters]</file>
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