Chapter 2.5, Problem 9SP

Figure 2.73 Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes?

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

$\begin{array}{l}{L}_1:\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\\ L_2:\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\end{array}$

You are to develop a formula for the distance $d$ between these two lines, in terms of the values

$a_1,b_1,c_1;b_2,c_2;x_1,y_1,z_1;$ and $x_2,y_2,z_2.$ 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.

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 $4.0\times 15.0\times 10.0\text{m}$ (height, width, and depth). One sector has a pipe entering the lower comer of the standard frame unit and exiting at the diametrically opposed comer (the one farthest away at the top); call this $L_{\text{1}}.$ A second pipe enters and exits at the two different opposite lower comers; call this $L_2$ (Figure 2.74).

Figure 2.74 Two pipes cross through a standard frame unit. Write down the vectors along the lines representing those pipes, find the cross product between them from which to create the unit vector $n,$ 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 cooler of the ?rst pipe.) Similarly, you may aim develop the symmetric equations for each line and substitute directly into your formula.