Distance Between Parallel Planes

Distance between parallel planes

For parallel planes, first rescale both equations to use the same normal \(\mathbf n\). If they become \(\mathbf r\cdot\mathbf n=d_1\) and \(\mathbf r\cdot\mathbf n=d_2\), then \(\displaystyle D=\frac{|d_2-d_1|}{|\mathbf n|}\). Equal constants give coincident planes.

Derivation

Consider parallel planes \(\Pi_1:\mathbf r\cdot\mathbf n=k_1\) and \(\Pi_2:\mathbf r\cdot\mathbf n=k_2\), written with the same normal \(\mathbf n\). Let \(A\) and \(B\) be points on \(\Pi_1\) and \(\Pi_2\), with position vectors \(\mathbf a\) and \(\mathbf b\).

Because \(A\) and \(B\) lie on their respective planes, \(\mathbf a\cdot\mathbf n=k_1\) and \(\mathbf b\cdot\mathbf n=k_2\).

The perpendicular distance is the magnitude of the projection of \(\overrightarrow{AB}=\mathbf b-\mathbf a\) onto the unit normal: \(\left|(\mathbf b-\mathbf a)\cdot\dfrac{\mathbf n}{|\mathbf n|}\right|=\dfrac{|k_2-k_1|}{|\mathbf n|}\).

For example, rewrite \(\Pi_2:\mathbf r\cdot\begin{pmatrix}-1\\-2\\-1\end{pmatrix}=10\) as \(\mathbf r\cdot\begin{pmatrix}1\\2\\1\end{pmatrix}=-10\), so it has the same normal as \(\Pi_1:\mathbf r\cdot\begin{pmatrix}1\\2\\1\end{pmatrix}=4\).

Hence the distance in this example is \(\dfrac{|-10-4|}{\sqrt{1^2+2^2+1^2}}=\dfrac{14}{\sqrt6}\) units.