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}