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      <text>9.2.1.6 Position vectors</text>
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      <text><![CDATA[<h4>Position vectors</h4><p>Now consider the special case when \(\underline{ r}\) represents the vector from the origin \(O\) to the point \(P(a,b)\). This vector is known as the <b>position vector</b> of \(P\) and is shown in the figure below.</p><p><br></p>
<div class="HELM_keypoint">
<h4>Key Point<br></h4>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/9-2-1-1KP2.svg" alt="A set of x-y with the point P(a,b) marked on it and the position vector r pointing from the origin to P" class="img-responsive atto_image_button_text-bottom" width="368" height="199">
</figure>
<p>The position vector of \(P\) with coordinates \((a,b)\) is \(\underline{r} = \overrightarrow{OP} = a\underline{i} + b\underline{j} \) </p>
</div>
<p>Unlike most vectors, position vectors cannot be freely translated. Because they indicate the position of a point they are <span><strong>fixed</strong></span> vectors in the sense that the tail of a position vector is always located at the origin.</p>
<hr>
<h4 class="HELM_example">Example<br></h4>
<p> State the position vectors of the points with coordinates:
</p><ol class="HELM_parts_inline">
<li>\(P(2,4)\)</li>
<li>\(Q(-1,5)\)</li>
<li>\(R(-1,-7)\)</li>
<li>\(S(8,-4)\)</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>\(2\underline{i} + 4\underline{j}\) </li>
<li>\(-\underline{i} + 5\underline{j}\) </li>
<li>\(-\underline{i} - 7\underline{j}\) </li>
<li>\(8\underline{i} - 4\underline{j}\) </li>
</ol>
<p>&nbsp;</p>
<hr>
<h4 class="HELM_example">Example<br></h4>
<p> Sketch the position vectors:</p>
<ol class="HELM_parts_inline">
<li>\(\underline{r}_{1} = 3\underline{i}+4\underline{j}\)</li>
<li>\(\underline{r}_{2} = -2\underline{i}+5\underline{j}\)</li>
<li>\(\underline{r}_{3} = -3\underline{i}-2\underline{j}\)</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<p> The vectors are shown below. Note that all position vectors start at the origin.&nbsp;</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/9-2-1Ex3PlotPosVector.svg" alt="Three vectors pointing from the origin to the three points specified in the question." class="img-responsive atto_image_button_text-bottom" width="281" height="235">
</figure>
<hr>
<p> The <b>modulus</b> of any vector \(\underline{r}\) is equal to its length. As we have noted earlier, the modulus of \(\underline{ r}\) is usually denoted by \(|\underline{ r}|\). When \(\underline{ r}=a\underline{i}+b\underline{ j}\) the modulus can be obtained using Pythagoras’ theorem. If \(\underline{r}\) is the position vector of point \(P\) then the modulus is, clearly, the distance of \(P\) from the origin.</p>
<div class="HELM_keypoint">
<h4>Key Point<br></h4>
<p>If \(\underline{ r}=a\underline{i}+b\underline{ j}\) then \(|\underline{r}| = \sqrt{a^2 + b^2}\)</p>
</div>
<hr>
<h4 class="HELM_example">Example<br></h4>
<p>Find the modulus of each of the vectors shown in the previous Example.</p>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>  \(|\underline{r}_{1}| = |3\underline{i}+4\underline{j}| = \sqrt{3^2+4^2} = \sqrt{25} = 5\)</li>
<li>  \(|\underline{r}_{2}| = \sqrt{(-2)^2+5^2} = \sqrt{4+25} = \sqrt{29}\)</li>
<li>  \(|\underline{r}_{3}| = \sqrt{(-3)^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\)</li>
</ol>]]></text>
<file name="9-2-1-1KP2.svg" path="/" encoding="base64">[encoded data, 34532 characters]</file>
<file name="9-2-1Ex3PlotPosVector.svg" path="/" encoding="base64">[encoded data, 28452 characters]</file>
<file name="9-2-1Ex3PlotPosVectors.svg" path="/" encoding="base64">[encoded data, 28452 characters]</file>
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