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<main class="main"><div id="content" class="pretext-content"><section xmlns:svg="http://www.w3.org/2000/svg" class="section" id="section-32"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">4.8</span> <span class="title">Geometric applications</span>
</h2>
<section class="introduction" id="introduction-13"><p id="p-970">The dot product and cross product in \(\R^3\) allow us to compute many different geometric properties of lines and planes.</p></section><section class="subsection" id="subsection-79"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">4.8.1</span> <span class="title">The equation of the line of intersection of two planes</span>
</h3>
<p id="p-971">Suppose we are given the equation of two planes and want the equation of the line of intersection. If the two equations are</p>
<div class="displaymath">
\begin{align*}
a_1x+b_1y+c_1z+d_1\amp=0 \text{ and}\\
a_2x+b_2y+c_2z+d_2\amp=0
\end{align*}
</div>
<p class="continuation">then the line of intersection will be the set of all points that satisfy <em class="emphasis">both</em> equations. We can rewrite the two equations as</p>
<div class="displaymath">
\begin{align*}
a_1x+b_1y+c_1z \amp =-d_1 \text{ and}\\
a_2x+b_2y+c_2z \amp =-d_2
\end{align*}
</div>
<p class="continuation">and use the augmented matrix</p>
<div class="displaymath">
\begin{equation*}
\left[
\begin{array}{ccc|c}
a_1\amp b_1\amp c_1\amp -d_1\\
a_2\amp b_2\amp c_2\amp -d_2
\end{array}
\right]
\end{equation*}
</div>
<p id="p-972">When this matrix is put in reduced row echelon form, there will be one free variable, which we may call \(t\text{.}\) Writing the solution as a vector and rearranging give the equation of a line.</p>
<p id="p-973">Here is an example to show how this technique works: the equations of the planes are</p>
<div class="displaymath">
\begin{gather*}
x+y+z=-2\\
2x+y-z=2
\end{gather*}
</div>
<p class="continuation">Here is the picture of the two planes and the line of intersection:</p>
<figure class="figure figure-like" id="figure-49"><div class="image-box" style="width: 70%; margin-left: 15%; margin-right: 15%;"><div class="asymptote-box" style="padding-top: 92.7125506072875%"><iframe src="images/image-51.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">4.8.1<span class="period">.</span></span><span class="space"> </span></figcaption></figure><p id="p-974">The augmented matrix for the system is</p>
<div class="displaymath">
\begin{equation*}
\left[
\begin{array}{ccc|c}
2\amp1\amp-1\amp2\\
1\amp1\amp1\amp-2
\end{array}
\right]
\end{equation*}
</div>
<p class="continuation">which has a reduced row echelon form of</p>
<div class="displaymath">
\begin{equation*}
\left[
\begin{array}{ccc|c}
1\amp0\amp-2\amp-4\\
0\amp1\amp3\amp6
\end{array}
\right]
\end{equation*}
</div>
<p class="continuation">Hence \(z\) is a free variable and can be assigned the value \(t\text{.}\) This gives \(z=t\text{,}\) \(y=6-3t\) and \(x=-4+2t\text{,}\) which may be written as</p>
<div class="displaymath">
\begin{equation*}
(x,y,z)=(-4+2t,6-3t,t)=(-4,6,0) + t(2,-3,1)
\end{equation*}
</div>
<p id="p-975">As another example, consider the two planes with equations written as</p>
<div class="displaymath">
\begin{gather*}
x+y=2\\
x+y-z=3
\end{gather*}
</div>
<figure class="figure figure-like" id="figure-50"><div class="image-box" style="width: 70%; margin-left: 15%; margin-right: 15%;"><div class="asymptote-box" style="padding-top: 49.6%"><iframe src="images/image-52.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">4.8.2<span class="period">.</span></span><span class="space"> </span></figcaption></figure><p id="p-976">Then the augmented matrix is</p>
<div class="displaymath">
\begin{equation*}
\left[
\begin{array}{ccc|c}
1\amp1\amp0\amp2\\
1\amp1\amp-1\amp3
\end{array}
\right]
\end{equation*}
</div>
<p class="continuation">with reduced row echelon form</p>
<div class="displaymath">
\begin{equation*}
\left[
\begin{array}{ccc|c}
1\amp1\amp0\amp2\\
0\amp0\amp1\amp-1
\end{array}
\right].
\end{equation*}
</div>
<p class="continuation">This means that \(y\) is a free variable and may be assigned the value \(t\text{,}\) from which follows \(z=-1\text{,}\) \(y=t\) and \(x=2-t\text{.}\) In other words,</p>
<div class="displaymath">
\begin{equation*}
(x,y,z)=(2-t,t,-1)= (2,0,-1)+t(-1,1,0)
\end{equation*}
</div>
<p class="continuation">which is the equation of the line \(L\text{.}\)</p></section><section class="subsection" id="IntersectionLinePlane"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">4.8.2</span> <span class="title">Intersection of a Line and a Plane</span>
</h3>
<p id="p-977">Suppose we have a line \(L\) with equation \((x_0,y_0,z_0)+t\vec {n_0}\) and a plane with equations \(ax+by+cz=0\text{.}\) Is there a point in both the line and the plane, and, if so, what is it? As usual when discussing planes, let \(\vec n=(a,b,c)\text{.}\) Then a point \(\vec x=(x,y,z)\) is in the plane if and only if \(\vec x\cdot \vec n=0\text{.}\) If \(\vec x\) is also on the line, then \(\vec x=(x_0,y_0,z_0)=t\vec{n_0}\) for some \(t\text{.}\) Then</p>
<div class="displaymath">
\begin{equation*}
((x_0,y_0,z_0)+t\vec{n_0})\cdot\vec n=0\\
(x_0,y_0,z_0)\cdot \vec n= -t( \vec{n_0}\cdot\vec n)
\end{equation*}
</div>
<p class="continuation">and, if \(\vec{n_0}\cdot\vec n\not=0\text{,}\)</p>
<div class="displaymath">
\begin{equation*}
t=-\frac{(x_0,y_0,z_0)\cdot \vec n}{\vec{n_0}\cdot\vec n}
\end{equation*}
</div>
<p id="p-978">Here is the picture of the situation. The points on the line are determined by using all the different values of \(t\text{.}\) The point of intersection is the particular \(t\) where \(t=-\frac{(x_0,y_0,z_0)\cdot \vec n}{\vec{n_0}\cdot\vec n}\text{.}\)</p>
<figure class="figure figure-like" id="figure-51"><div class="image-box" style="width: 75%; margin-left: 12.5%; margin-right: 12.5%;"><div class="asymptote-box" style="padding-top: 72.5925925925926%"><iframe src="images/image-53.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">4.8.3<span class="period">.</span></span><span class="space"> </span></figcaption></figure><p id="p-979">What if \(\vec{n_0}\cdot\vec n=0\text{?}\) Then a line with direction vector \(n\) (that is, orthogonal to the plane) itself is orthogonal to \(L\text{.}\) If \((x_0,y_0,z_0)\) is not in the plane, the line is then parallel to the plane and never intersects it. On the other hand, if \((x_0,y_0,z_0)\) is in the plane, the line is then actually contained in the plane, so every point on the line intersects it.</p>
<p id="p-980">In the following picture, the red line had direction vector \(\vec n=(a,b,c)\) and hence is orthogonal to the plane. Since the line \(L\) has direction vector \(\vec{n_0}\text{,}\) and \(\vec{n_0}\cdot\vec{n}=0\text{,}\) \(L\) is also orthogonal to the red line. That makes the plane and the line \(L\) parallel.</p>
<figure class="figure figure-like" id="figure-52"><div class="image-box" style="width: 75%; margin-left: 12.5%; margin-right: 12.5%;"><div class="asymptote-box" style="padding-top: 75.8465011286682%"><iframe src="images/image-54.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">4.8.4<span class="period">.</span></span><span class="space"> </span></figcaption></figure></section><section class="subsection" id="subsection-81"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">4.8.3</span> <span class="title">Distance Between Two Lines</span>
</h3>
<p id="p-981">If two lines intersect, the distance between them is obviously \(0\text{.}\) Otherwise, there are two cases:</p>
<ul class="disc">
<li id="li-329"><p id="p-982">The lines lie in a plane, in which case the lines are <em class="emphasis">parallel</em>.</p></li>
<li id="li-330"><p id="p-983">The lines do not lie in a plane, in which case the lines are <em class="emphasis">skew</em>.</p></li>
</ul>
<section class="paragraphs" id="paragraphs-17"><h5 class="heading"><span class="title">Parallel lines.</span></h5>
<p id="p-984">We start with two parallel lines with equations</p>
<ul class="disc">
<li id="li-331"><p id="p-985">\(L_1\) with equation \((x,y,z)=\vec{X_1} + t_1\vec{n_1}\)</p></li>
<li id="li-332"><p id="p-986">\(L_2\) with equation \((x,y,z)=\vec{X_2} + t_2\vec{n_2}\)</p></li>
</ul>
<figure class="figure figure-like" id="ParallelLineDistance"><div class="image-box" style="width: 75%; margin-left: 12.5%; margin-right: 12.5%;"><div class="asymptote-box" style="padding-top: 70.6043956043956%"><iframe src="images/image-55.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">4.8.5<span class="period">.</span></span><span class="space"> </span></figcaption></figure><p id="p-987">The distance \(d\) between the two lines satisfies \(d=\|\vec X_2-\vec X_1 \|\sin(\theta)\text{.}\) Recall from <a class="xref" data-knowl="./knowl/CrossProductLength.html" title="Theorem 4.5.5: \(\|\vec x\times\vec y\|=\|\vec x\|\,\|\vec y\|\sin\theta\)">Theorem 4.5.5</a> that the length of \(\vec x\times \vec y\) is \(\|\vec x\| \|\vec y\| \sin(\theta)\text{.}\) Hence</p>
<div class="displaymath">
\begin{equation*}
d
=\|\vec X_2-\vec X_1\|
\frac {\|(\vec X_2-\vec X_1) \times t\vec n_1\|} {\|\vec X_2-\vec X_1\| \|t \vec n_1\|}
= \frac {\|(\vec X_2-\vec X_1) \times t\vec n_1\|} {\|t \vec n_1\|}
= \frac {\|(\vec X_2-\vec X_1) \times \vec n_1\|} {\|\vec n_1\|}
\end{equation*}
</div></section><section class="paragraphs" id="paragraphs-18"><h5 class="heading"><span class="title">Skew lines.</span></h5>
<p id="p-988">We start with the equations of two skew lines:</p>
<ul class="disc">
<li id="li-333"><p id="p-989">\(L_1\) has equation \((x,y,z)=\vec{X_1}+t_1\vec{n_1}\text{,}\) \(-\infty \lt t \lt\infty\)</p></li>
<li id="li-334"><p id="p-990">\(L_2\) has equation \((x,y,z)=\vec{X_2}+t_2\vec{n_2}\) \(-\infty \lt t \lt\infty\)</p></li>
</ul>
<p class="continuation">We want the (shortest) distance \(d\) between the lines.</p>
<figure class="figure figure-like" id="figure-54"><div class="image-box" style="width: 75%; margin-left: 12.5%; margin-right: 12.5%;"><div class="asymptote-box" style="padding-top: 81.8181818181818%"><iframe src="images/image-56.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">4.8.6<span class="period">.</span></span><span class="space"> </span></figcaption></figure><p id="p-991">The direction vector for \(L_1\) is \(\vec{n_1}\) while that of \(L_2\) is \(\vec{n_2}\text{.}\) Let us say that the shortest line segment joining \(L_1\) to \(L_2\) is on the line \(L\text{.}\) The length of the line segment, \(d\text{,}\) is the distance we wish to compute. In addition, let \(t_1\) and \(t_2\) be the particular values so both \(\vec{X_1}+t_1\vec{n_1}\) and \(\vec{X_2}+t_2\vec{n_2}\) lie on \(L\text{.}\)</p>
<p id="p-992">Now \(L\) is perpendicular to both \(L_1\) and \(L_2\) , so the direction vector of \(L\) is orthogonal to both \(\vec{n_1}\) and \(\vec{n_2}\text{.}\) How can such a direction vector be produced? With the cross product, of course! And so we let \(\vec{n_1}\times\vec{n_2}\) be the direction vector of \(L\text{.}\) We then have, for some \(t_3\text{,}\)</p>
<div class="displaymath" id="p-993">
\begin{align*}
\vec{X_1}+t_1\vec{n_1} + t_3(\vec{n_1}\times\vec{n_2})
\amp=\vec{X_2}+t_2\vec{n_2}\\
t_1\vec{n_1} - t_2\vec{n_2} + t_3(\vec{n_1}\times\vec{n_2})
\amp=\vec{X_2}-\vec{X_1}\tag{$*$}
\end{align*}
</div>
<p id="p-994">In principle the problem is solved. We have three unknowns \(t_1\text{,}\) \(t_2\) and \(t_3\text{,}\) and each of the three coordinates gives an equation, so we have three equations in three unknowns. There is a quicker way: take the dot product of both sides of (\(*\)) above with \(\vec{n_1}\) and then with \(\vec{n_2}\text{.}\) This gives</p>
<div class="displaymath">
\begin{align*}
\|\vec{n_1}\|^2 t_1 -(\vec{n_1}\cdot\vec{n_2})t_2
\amp =(\vec{X_2}-\vec{X_1})\cdot \vec{n_1}\\
(\vec{n_1}\cdot\vec{n_2}) t_1 -\|\vec{n_2}\|^2t_2
\amp =(\vec{X_2}-\vec{X_1})\cdot \vec{n_2}
\end{align*}
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