The Distance Between a Plane and a Point

# The Distance Between a Plane and a Point

Sometimes it may be important to find the distance $D$, between some point $P_0$ and a plane in the form $ax + by + cz + d = 0$. The formula for the distance between a plane $\Pi: ax + by + cz + d = 0$ and a point is $P(x_0, y_0, z_0)$ is as follows:

(1)
\begin{align} D = \frac{\mid ax_0 + by_0 + cz_0 + d \mid}{\sqrt{a^2 + b^2 + c^2}} \end{align}

We will now go forth to prove this formula.

 Theorem 1: Let $\Pi$ be a plane in $\mathbb{R}^3$ and $P_0(x_0, y_0, z_0)$ be a point in $\mathbb{R}^3$. Then the distance $D$ between $\Pi$ and $P_0$ is given by $D = \frac{\mid ax_0 + by_0 + cz_0 + d \mid}{\sqrt{a^2 + b^2 + c^2}}$.
• Proof: Let $Q(x_1, y_1, z_1)$ be any point on the plane $\Pi$ and let $P_0 = (x_0, y_0, z_0)$ be our point of interest. Let $\vec{QP_0} = (x_0 - x_1, y_0 - y_1, z_0 - z_1)$. Recall that this plane can be defined with a vector that is perpendicular to the plane known as a normal. Let $n = (a, b, c)$ be a normal to our plane $\Pi$.
• The orthogonal projection of $\vec{QP_0}$ onto $\vec{n}$ is equal to the distance between the plane $\Pi$ and $P_0$ as illustrated below:
• Thus it follows that $D = \| \mathrm{proj}_{\vec{n}} \vec{QP_{0}} \|$. When simplified and letting $d = -ax_1 - by_1 - cz_1$ we obtain:
(2)
\begin{align} D = \| \mathrm{proj}_{\vec{n}} \vec{QP_{0}} \| \\ D = \frac{\| \vec{QP_0} \cdot \vec{n} \|}{ \| \vec{n} \|} \\ D = \frac{ \mid (x_0 - x_1, y_0 - y_1, z_0 - z_1) \cdot (a, b, c) \mid}{\sqrt{a^2 + b^2 + c^2}} \\ D = \frac{\mid a(x_0 - x_1) + b(y_0 - y_1) + c(z_0 - z_1) \mid}{\sqrt{a^2 + b^2 + c^2}} \\ D = \frac{\mid ax_0 + by_0 + cz_0 + d \mid}{\sqrt{a^2 + b^2 + c^2}} \\ \blacksquare \end{align}

## Example 1

What is the distance between the point $P(1, 2, 3)$ and the plane $\Pi: 2x + 2y -3z +3= 0$?

We can answer this question by directly applying our formula such that we obtain:

(3)
\begin{align} D = \frac{\mid 2(1) + 2(2) -3(3) + 3 \mid}{\sqrt{2^2 + 2^2 + (3)^2}} \\ D = \frac{\mid -2 \mid}{\sqrt{17}} \\ D = \frac{ 2 }{\sqrt{17}} \end{align}

## Example 2

What is the distance between the point $P(-2, -7, -12)$ and the plane $\Pi: 4x - 8y -3z + 5 = 0$?

Once again, we can apply the formula to obtain the answer.

(4)
\begin{align} D = \frac{\mid 4(-2) -8(-7) -3(-12) + 5 \mid}{\sqrt{4^2 + (-8)^2 + (-3)^2}} \\ D = \frac{\mid -8 +56 +36 + 5 \mid}{\sqrt{16 + 64 + 9}} \\ D = \frac{\mid 89 \mid}{\sqrt{89}} \\ D = \frac{89}{\sqrt{89}} \end{align}