The Distance Between Parallel Planes

The Distance Between Parallel Planes

We will first define what it means for two lines to be parallel, and then learn how to compute the distance between such planes.

 Definition: Two planes $\Pi_1$ and $\Pi_2$ are said to be Parallel if the normal of $\Pi_1$ ($n_1 = (a_1, b_1, c_1)$ is a scalar multiple $k$ of the normal of $\Pi_2$ ($n_2 = (ka_2, kb_2, kc_2)$).

For example, consider the planes $\Pi_1: 2x + 4y + 6z + 1 = 0$ and $\Pi_2: 4x + 8y + 12z + 6 = 0$. We can easily pull off the norms of these two planes to get that $n_1 = (2, 4, 6)$ and $n_2 = (4, 8, 12)$. Clearly $2n_1 = n_2$, so $\Pi_1 \parallel \Pi_2$.

Finding the distance between two parallel planes is relatively easily. First, suppose we have two planes $\Pi_1$ and $\Pi_2$. If we select an arbitrary point on either plane and then use the other plane's equation in the formula for the distance between a point and a plane, then we will have obtained the distance between both planes.

For example, consider the planes $\Pi_1: 2x + 3y + 4z -3 = 0$ and $\Pi_2: -4x -6y -8z + 8 = 0$. These planes are parallel. First let's select an arbitrary point off the first plane such as $(0, 0, \frac{4}{3})$. We will now use the formula $D = \frac{\mid ax_0 + by_0 + cz_0 + d \mid}{\sqrt{a^2 + b^2 + c^2}}$ in order to calculate the distance between both planes:

(1)
\begin{align} D = \frac{\mid ax_0 + by_0 + cz_0 + d \mid}{\sqrt{a^2 + b^2 + c^2}} \\ D = \frac{\mid -4(0) + -6(0) + -8(3/4) + 8 \mid}{\sqrt{(-4)^2 + (-6)^2 + (-8)^2}} \\ D = \frac{\mid -6 + 8 \mid}{\sqrt{(16 + 36 + 64)}} \\ D = \frac{\mid 2\mid}{\sqrt{116}} \\ D = \frac{2}{\sqrt{116}} \end{align}