Chapter 2.5, Problem 5SP

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.

5. The dot product of two vectors is the magnitude of the projection of one vector onto the other—that is, $A\cdot B=\Vert A\Vert \Vert B\Vert \cos \theta ,$ where $\theta$ is the angle between the vectors. Using the dot product, find the projection of vector $v_{12}$ found in step $4$ onto unit vector $n$ 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 $v_{12}$ at the order of the cross product, so use absolute value signs around the numerator.