Point–Plane Distance

19. Point–Plane Distance

For a plane in $ℝ^{3}$ defined by $𝐴 𝑥 + 𝐵 𝑦 + 𝐶 𝑧 = 𝐷$ and a point $(𝑥_{0}, 𝑦_{0}, 𝑧_{0})$ , the perpendicular distance from the point to the plane is:

𝑑 = \frac{| 𝐴 𝑥_{0} + 𝐵 𝑦_{0} + 𝐶 𝑧_{0} - 𝐷 |}{\sqrt{𝐴^{2} + 𝐵^{2} + 𝐶^{2}}}

The denominator is the norm of the plane’s normal vector $\vec{𝑛} = (𝐴, 𝐵, 𝐶)$ .

The signed distance (drop the absolute value) tells you which side of the plane the point is on:

𝑑_{signed} = \frac{𝐴 𝑥_{0} + 𝐵 𝑦_{0} + 𝐶 𝑧_{0} - 𝐷}{\sqrt{𝐴^{2} + 𝐵^{2} + 𝐶^{2}}}

Example

Plane: $𝑥 - 2 𝑦 + 3 𝑧 = 5$ . Point: $(2, 3, 1)$ .

\begin{aligned} 𝑑 & = \frac{1 \cdot 2 - 2 \cdot 3 + 3 \cdot 1 - 5}{\sqrt{1^{2} + {(- 2)}^{2} + 3^{2}}} \\ = \frac{2 - 6 + 3 - 5}{\sqrt{1 + 4 + 9}} \\ = \frac{- 6}{\sqrt{14}} \approx - 1.60 \end{aligned}

The negative sign means the point lies on the opposite side of the plane from the normal direction.

19.1. Generalization

The same formula generalizes to a hyperplane in $ℝ^{𝑛}$ defined by $\vec{𝑎} \cdot \vec{𝑥} = 𝑏$ :

𝑑 = \frac{| \vec{𝑎} \cdot \vec{𝑝} - 𝑏 |}{‖} \vec{𝑎} ‖

19.2. Connections

Planes — definition and normal vector
Projection — same idea: project the displacement vector onto the unit normal
Hyperplane — $𝑛$ -dimensional generalization