# 2.5 Equations of lines and planes in space  (Page 7/19)

 Page 7 / 19

Find the measure of the angle between planes $x+y-z=3$ and $3x-y+3z=5.$ Give the answer in radians and round to two decimal places.

$1.44$ rad

When we find that two planes are parallel, we may need to find the distance between them. To find this distance, we simply select a point in one of the planes. The distance from this point to the other plane is the distance between the planes.

Previously, we introduced the formula for calculating this distance in [link] :

$d=\frac{\stackrel{\to }{QP}·\text{n}}{‖\text{n}‖},$

where $Q$ is a point on the plane, $P$ is a point not on the plane, and $\text{n}$ is the normal vector that passes through point $Q.$ Consider the distance from point $\left({x}_{0},{y}_{0},{z}_{0}\right)$ to plane $ax+by+cz+k=0.$ Let $\left({x}_{1},{y}_{1},{z}_{1}\right)$ be any point in the plane. Substituting into the formula yields

$\begin{array}{cc}\hfill d& =\frac{|a\left({x}_{0}-{x}_{1}\right)+b\left({y}_{0}-{y}_{1}\right)+c\left({z}_{0}-{z}_{1}\right)|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}\hfill \\ & =\frac{|a{x}_{0}+b{y}_{0}+c{z}_{0}+k|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}.\hfill \end{array}$

We state this result formally in the following theorem.

## Distance from a point to a plane

Let $P\left({x}_{0},{y}_{0},{z}_{0}\right)$ be a point. The distance from $P$ to plane $ax+by+cz+k=0$ is given by

$d=\frac{|a{x}_{0}+b{y}_{0}+c{z}_{0}+k|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}.$

## Finding the distance between parallel planes

Find the distance between the two parallel planes given by $2x+y-z=2$ and $2x+y-z=8.$

Point $\left(1,0,0\right)$ lies in the first plane. The desired distance, then, is

$\begin{array}{cc}\hfill d& =\frac{|a{x}_{0}+b{y}_{0}+c{z}_{0}+k|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}\hfill \\ & =\frac{|2\left(1\right)+1\left(0\right)+\left(-1\right)\left(0\right)+\left(-8\right)|}{\sqrt{{2}^{2}+{1}^{2}+{\left(-1\right)}^{2}}}\hfill \\ & =\frac{6}{\sqrt{6}}=\sqrt{6}.\hfill \end{array}$

Find the distance between parallel planes $5x-2y+z=6$ and $5x-2y+z=-3.$

$\frac{9}{\sqrt{30}}$

## Distance between two skew lines

Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product.

The symmetric forms of two lines, ${L}_{1}$ and ${L}_{2},$ are

$\begin{array}{}\\ \\ {L}_{1}:\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}\hfill \\ {L}_{2}:\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}.\hfill \end{array}$

You are to develop a formula for the distance $d$ between these two lines, in terms of the values ${a}_{1},{b}_{1},{c}_{1};{a}_{2},{b}_{2},{c}_{2};{x}_{1},{y}_{1},{z}_{1};\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{x}_{2},{y}_{2},{z}_{2}.$ The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.

1. First, write down two vectors, ${\text{v}}_{1}$ and ${\text{v}}_{2},$ that lie along ${L}_{1}$ and ${L}_{2},$ respectively.
2. Find the cross product of these two vectors and call it $\text{N}.$ This vector is perpendicular to ${\text{v}}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{\text{v}}_{2},$ and hence is perpendicular to both lines.
3. From vector $\text{N},$ form a unit vector $\text{n}$ in the same direction.
4. Use symmetric equations to find a convenient vector ${\text{v}}_{12}$ that lies between any two points, one on each line. Again, this can be done directly from the symmetric equations.
5. The dot product of two vectors is the magnitude of the projection of one vector onto the other—that is, $\text{A}·\text{B}=‖\text{A}‖‖\text{B}‖\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,$ where $\theta$ is the angle between the vectors. Using the dot product, find the projection of vector ${\text{v}}_{12}$ found in step $4$ onto unit vector $\text{n}$ found in step 3. This projection is perpendicular to both lines, and hence its length must be the perpendicular distance $d$ between them. Note that the value of $d$ may be negative, depending on your choice of vector ${\text{v}}_{12}$ or the order of the cross product, so use absolute value signs around the numerator.
6. Check that your formula gives the correct distance of $|-25|\text{/}\sqrt{198}\approx 1.78$ between the following two lines:
$\begin{array}{}\\ \\ {L}_{1}:\frac{x-5}{2}=\frac{y-3}{4}=\frac{z-1}{3}\hfill \\ {L}_{2}:\frac{x-6}{3}=\frac{y-1}{5}=\frac{z}{7}.\hfill \end{array}$
7. Is your general expression valid when the lines are parallel? If not, why not? ( Hint: What do you know about the value of the cross product of two parallel vectors? Where would that result show up in your expression for $d?\right)$
8. Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersect if they are not parallel and they are in the same plane. Hence, consider the direction of $\text{n}$ and ${\text{v}}_{12}.$ What is the result of their dot product?
9. Consider the following application. Engineers at a refinery have determined they need to install support struts between many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these struts at the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure, they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute them to the installation crews without spending valuable time making measurements.
The rectangular frame structure has the dimensions $4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}15.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}10.0\phantom{\rule{0.2em}{0ex}}\text{m}$ (height, width, and depth). One sector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposed corner (the one farthest away at the top); call this ${L}_{1}.$ A second pipe enters and exits at the two different opposite lower corners; call this ${L}_{2}$ ( [link] ).

Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector $\text{n},$ define a vector that spans two points on each line, and finally determine the minimum distance between the lines. (Take the origin to be at the lower corner of the first pipe.) Similarly, you may also develop the symmetric equations for each line and substitute directly into your formula.

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