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      <text>9.2.2.1 Three-dimensional coordinate frames</text>
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      <text><![CDATA[<h4>Three-dimensional coordinate frames<br></h4>
<p>The real world is three-dimensional and in order to solve many engineering problems it is necessary to develop expertise in the mathematics of three-dimensional space. An important application of vectors is their use to locate points in three dimensions. When two distinct points are known we can draw a unique straight line between them. Three distinct points which do not lie on the same line form a unique plane. Vectors can be used to describe points, lines, and planes in three dimensions. These mathematical foundations underpin much of the technology associated with computer graphics and the control of robots. In this part of the quiz we shall introduce the vector methods which underlie these applications.</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/9-2-2three-DCoord.svg" alt="First panel shows a 3D set of coordinate axes with mutually perpendicular x,y,z axes.  Middle panel shows a point P(a,b,c) marked on the axes with an arrow pointing from the origin to it representing its position vector. Third panel shows the axes with the unit vectors i,j and k along the x,y and z axes respectively." class="img-responsive atto_image_button_text-bottom" width="630" height="283">
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<p>The figure above shows a three-dimensional coordinate frame. Note that the third dimension requires the addition of a third axis, the \(z\)-axis. Although these three axes are drawn in the plane of the paper you should remember that we are now thinking of three-dimensional situations. Just as in two-dimensions the \(x\) and \(y\) axes are perpendicular, in three dimensions the \(x\),\(y\) and \(z\) axes are all perpendicular to each other. We say they are <b> mutually perpendicular</b>. There is no reason why we could not have chosen the \(z\)-axis in the opposite direction. However, it is conventional to choose the directions shown in the figure above.</p>
<p>Any point in the three dimensional space can be defined in terms of its \(x\), \(y\) and \(z\) coordinates. Consider the point \(P\) with coordinates \((a,b,c)\) as shown in panel (b) of the figure above. The vector from the origin to the point \(P\) is known as the position vector of \(P\), denoted \(\overrightarrow{OP}\) or \(\underline{r}\). To arrive at \(P\) from \(O\) we can think of moving \(a\) units in the \(x\) direction, \(b\) units in the \(y\) direction and \(c\) units in the \(z\) direction.</p>
<p>A <span><strong>unit vector</strong></span> pointing in the positive direction of the \(z\)-axis is denoted by \(\underline{ k}\). Noting that \(\overrightarrow{OQ}=a\underline{ i}+b\underline{ j}\) and that \(\overrightarrow{QP}=c\underline{ k}\) we can state \[\underline{ r}=\overrightarrow{OP} = 
\overrightarrow{OQ}+\overrightarrow{QP} =  a\underline{ i} +b\underline{ j} + c\underline{ k}\]</p>
<p>We conclude that the position vector of the point with coordinates \((a,b,c)\) is \(\underline{ r}=a\underline{ i}+b\underline{j}+c\underline{ k}\). (We might, for convenience, sometimes use a subscript notation. For example we might refer to the position vector \(\underline{ r}\) as \(\underline{ r}=r_x\underline{i}+r_y\underline{j}+r_z\underline{ k}\) in which \((r_{x}, r_{y}, r_{z})\) have taken the place of \((a,b,c)\).)</p>
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<h4>Key Point<br></h4>
<p>If \(P\) has coordinates \((a,b,c)\) then its position vector is \[\underline{r} = \overrightarrow{OP} = a\underline{i} +b\underline{j} + c\underline{k}\]</p>
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<p> The modulus of the vector \(\overrightarrow{OP}\) is equal to the distance \(OP\), which can be obtained by Pythagoras' theorem:</p>
<div class="HELM_keypoint">
<h4>Key Point<br></h4>
<p>If \(\underline{r} =  a\underline{ i} +b\underline{ j} + c\underline{ k}\) then \[|\underline{r}| = \sqrt{a^2+b^2+c^2}\]</p>
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<h4 class="HELM_example">Example<br></h4>
<p>Points \(A\), \(B\) and \(C\) have coordinates \((-1,1,4)\), \((8,0,2)\) and \((5,-2,11)\) respectively. </p>
<ol class="HELM_parts_inline">
<li>Find the position vectors of \(A\), \(B\) and \(C\).</li>
<li>Find \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) </li>
<li>Find \(|\overrightarrow{AB}|\) and \(|\overrightarrow{BC}|\)</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
  <li>Denoting the position vectors of \(A\), \(B\) and \(C\) by \(\underline{a}\), \(\underline{b}\) and \(\underline{c}\) respectively, we find \[\underline{a}  = -\underline{i} + \underline{j} + 4\underline{k},\] \[\underline{b} = 8\underline{i} + 2\underline{k},\] \[\underline{c}  = 5\underline{i} - 2\underline{j} + 11\underline{k}\]</li>
  <li> \(\overrightarrow{AB} = \underline{b}  - \underline{a} =  9\underline{i} - \underline{j} - 2\underline{k}.\) \(\overrightarrow{BC} = \underline{c}  - \underline{b} =  -3\underline{i} - 2\underline{j} + 9\underline{k}.\)</li>
  <li> \(|\overrightarrow{AB}| = \sqrt{9^2+(-1)^2+(-2)^2}=\sqrt{86}\). \(|\overrightarrow{BC}| = \sqrt{(-3)^2+(-2)^2+9^2} = \sqrt{94}\). </li>
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