Module 3: The Distance Between Lines, Planes, and a Line and a Plane

Suppose we wanted to find the distance between two lines: $\ell_1$ and $\ell_2$. As before, our goal is to find another line which is perpendicular to both of these. But this can be hard due to the sheer number of directions which are perpendicular to a given line. Instead, check out the following image which shows how this distance is related to the distance between two parallel planes! So we will begin this section by first figuring out how to measure the distance between two planes. Then we will show how we can generate these planes given a pair of non-intersecting lines.

Let $Ax+By+Cz=D_1$ and $AX+By+Cz=D_2$ be two parallel planes. (Why are the same values for $A,B,C$ used in both equations?) Note that all we need to do to find the distance between these planes is to choose one point in one plane (any random point will do) and then the problem reduces to finding the distance between a point and a plane which was covered in the last section! An easy way to do choose a point is to make two of $x,y,z$ equal to zero.

Example: Find the distance between the planes $x+2y = 1$ and $x+2y = 2$. (Note that in this example $C=0$.)

In this case, we will choose $y=z=0$ in the first plane to find $x=1$. (Note: we should not choose $x=y=0$ since this makes the nonsense equation $0=1$. In general, if the coefficient of a variable is zero, make the variable itself a zero.) So the point $(1,0,0)$ lies on the first plane and we need to find the distance between this point and the plane $x+2y=2$. Using the technique from the last section we find this distance is $\frac{1}{\sqrt{5}}$.

Now let's move to the case where we have a line ($\ell$) given by $\vec{x}_0 + t \vec{v}$ and a plane ($P_1$) given by $Ax+By+Cz=D$. Now if the line and plane are NOT parallel, then the line will intersect the plane and so the distance between them must be zero. Therefore we should first verify that the line and the plane are indeed parallel. There are several ways to do this, but the easiest is taking the plane's normal vector and showing that it must have zero dot product with the direction vector of the line: $\vec{v} \cdot \begin{pmatrix}A\\B\\C\\\end{pmatrix}=0$. If this is not true, then the line must intersect the plane. This is illustrate in the following diagram:

So in the case that the line and plane are parallel then a key is to find a plane $(P_2)$ which contains $\ell$. Or rather, it's not, since ANY POINT IN THE LINE IS A POINT IN THE PLANE WHICH CONTAINS IT! So just choose a point in the line and follow the argument from before! The easiest point to use is when $t$ is zero ($\vec{x}_0$) in the parametric equation for $\ell$. See the following diagram:

The final case is the distance between two lines. There are three cases to consider:

CASE 1: The lines have the SAME direction vector. That is the lines

$\vec{x}_0 + t \vec{v} \mbox{ and } \vec{y}_0 + s \vec{w}$

have the property that $\vec{v} \parallel \vec{w}$. In this case the distance between them will be the distance between one of the lines and a point on the other as in the following image.

The situation of the distance between a point and a line has been handled in the prior section.

CASE 2: The lines are skew. This is the case where the lines are non-parallel and non-intersecting. This case we will elaborate on.

CASE 3: The lines intersect. In this case, the distance between them is zero. The method outlined below will show if the distance between the lines must be zero.

In the case of lines $\ell_1$ and $\ell_2$ given by

$\vec{x}_1 + t \vec{v}_1 \mbox{ and } \vec{x}_2 + t \vec{v}_2$

being skew, we need two parallel planes. One for each line. Note that the generic form for a plane is:

$A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$

where $\vec{n} = \begin{pmatrix}A\\B\\C\\\end{pmatrix}$ is the normal to the plane and $(x_0,y_0,z_0)$ is a point in the plane. Since the planes are parallel to both lines, then we must have $\vec{n}$ orthogonal to both $\vec{v}_1$ and $\vec{v}_2$. So let us choose $\vec{n} = \vec{v}_1 \times \vec{v}_2$! Then the rest is simple:

1. Construct a plane parallel to both lines which contains $\ell_2$. This is the plane that has the point $\vec{x}_2$ and normal vector $\vec{v}_1 \times \vec{v}_2$. Call this plane $P$.
2. Find the distance between a point on $\ell_1$ (say $\vec{x}_1$) and $P$.

Example:

Let us find the distance between the lines

$\begin{pmatrix}1\\1\\0\\\end{pmatrix} + t \begin{pmatrix}1\\6\\2\\\end{pmatrix} and \begin{pmatrix}1\\5\\-2\\\end{pmatrix} + t \begin{pmatrix}2\\15\\6\\\end{pmatrix}.$

We begin by looking for a plane which contains one of the lines. This will have normal vector:

$\begin{pmatrix}1\\6\\2\\\end{pmatrix}\times \begin{pmatrix}2\\15\\6\\\end{pmatrix}= \begin{pmatrix}6\\-2\\3\\\end{pmatrix}$

Using point $\begin{pmatrix}1\\1\\0\\\end{pmatrix}$ from the first line we can create the plane:

$6(x-1) -2(y-1) + 3z=0.$

Now we just need to find the distance from the point $\begin{pmatrix}1\\5\\-2\\\end{pmatrix}$ and the plane $6(x-1) -2(y-1) + 3z=0$. Using the method from the prior section we have that this distance is $2$.