Geometric applications

Section 5.8 Geometric applications

The dot product and cross product in $\R^3$ allow us to compute many different geometric properties of lines and planes.

Subsection 5.8.1 The equation of the line of intersection of two planes

Suppose we are given the equation of two planes and want the equation of the line of intersection. If the two equations are

\begin{equation*} a_1x+b_1y+c_1z+d_1=0 \rlap{\text{ and}}\\ a_2x+b_2y+c_2z+d_2=0 \end{equation*}

then the line of intersection will be the set of all points that satisfy “both” equations. We can rewrite the two equations as

\begin{equation*} a_1x+b_1y+c_1z=-d_1 \rlap{\text{ and}}\\ a_2x+b_2y+c_2z=-d_2 \end{equation*}

and use the augmented matrix

\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*}

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.

Here is an example to show how this technique works: the equations of the planes are

\begin{equation*} x+y+z=-2\\ 2x+y-z=2 \end{equation*}

Here is the picture of the two planes and the line of intersection:

Figure 5.8.1.

The augmented matrix for the system is

\begin{equation*} \left[ \begin{array}{ccc|c} 2\amp1\amp-1\amp2\\ 1\amp1\amp1\amp-2 \end{array} \right] \end{equation*}

which has a reduced row echelon form of

\begin{equation*} \left[ \begin{array}{ccc|c} 1\amp0\amp-2\amp-4\\ 0\amp1\amp3\amp6 \end{array} \right] \end{equation*}

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

\begin{equation*} (x,y,z)=(-4+2t,6-3t,t)=(-4,6,0) + t(2,-3,1) \end{equation*}

As another example, consider the two planes with equations written as

\begin{equation*} x+y=2\\ x+y-z=3 \end{equation*}

Figure 5.8.2.

Then the augmented matrix is

\begin{equation*} \left[ \begin{array}{ccc|c} 1\amp1\amp0\amp2\\ 1\amp1\amp-1\amp3 \end{array} \right] \end{equation*}

with reduced row echelon form

\begin{equation*} \left[ \begin{array}{ccc|c} 1\amp1\amp0\amp2\\ 0\amp0\amp1\amp-1 \end{array} \right]. \end{equation*}

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,

\begin{equation*} (x,y,z)=(2-t,t,-1)= (2,0,-1)+t(-1,1,0) \end{equation*}

which is the equation of the line $L\text{.}$

Subsection 5.8.2 Intersection of a Line and a Plane

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

\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*}

and, if $\vec{n_0}\cdot\vec n\not=0\text{,}$

\begin{equation*} t=-\frac{(x_0,y_0,z_0)\cdot \vec n}{\vec{n_0}\cdot\vec n} \end{equation*}

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 $t=-\frac{(x_0,y_0,z_0)\cdot \vec n}{\vec{n_0}\cdot\vec n}\text{.}$

Figure 5.8.3.

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.

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.

Figure 5.8.4.

Subsection 5.8.3 Distance Between Two Lines

If two lines intersect, the distance between them is obviously $0\text{.}$ Otherwise, there are two cases:

The lines lie in a plane, in which case the lines are parallel.
The lines do not lie in a plane, in which case the lines are skew.

Parallel lines.

We start with two parallel lines with equations

$L_1$ with equation $(x,y,z)=\vec{X_1} + t_1\vec{n_1}$
$L_2$ with equation $(x,y,z)=\vec{X_2} + t_2\vec{n_2}$

Figure 5.8.5.

Skew lines.

We start with the equations of two skew lines:

$L_1$ has equation $(x,y,z)=\vec{X_1}+t_1\vec{n_1}\text{,}$ $-\infty \lt t \lt\infty$
$L_2$ has equation $(x,y,z)=\vec{X_2}+t_2\vec{n_2}$ $-\infty \lt t \lt\infty$

We want the (shortest) distance $d$ between the lines.

Figure 5.8.6.

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 $L\text{.}$

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 $t_3\text{,}$

\begin{equation*} \vec{X_1}+t_1\vec{n_1} + t_3(\vec{n_1}\times\vec{n_2}) =\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}) =\vec{X_2}-\vec{X_1}\tag{$*$} \end{equation*}

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

\begin{equation*} \|\vec{n_1}\|^2 t_1 -(\vec{n_1}\cdot\vec{n_2})t_2 =(\vec{X_2}-\vec{X_1})\cdot \vec{n_1}\\ (\vec{n_1}\cdot\vec{n_2}) t_1 -\|\vec{n_2}\|^2t_2 =(\vec{X_2}-\vec{X_1})\cdot \vec{n_2} \end{equation*}