Point Plane Distance

From CGAFaq

Given a plane containing the point $P$ and having unit--length normal $N$ , a plane equation is $N \cdot (X - C) = 0$ or $N \cdot X = \lambda$ for a suitably chosen scalar $\lambda$ . If in fact a plane point $C$ is known, then $\lambda = N \cdot C$ .

The distance from a point $Q$ to the plane is the length of the projection of $Q - C$ onto a normal line to the plane. Any point $Q$ may be written as

$Q = C + a N + R$

where $R$ is the projection of $Q-C$ onto the plane. Necessarily $R \cdot N = 0$ . The coefficient $a$ is

$a = N \cdot (Q-C) = N \cdot Q - \lambda$

The length of the projection, which is the distance from $Q$ to the plane, is

$d = |a| = |N \cdot Q - \lambda|$

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Another method can be used to calculate the point/plane distance that does not require selecting an arbitrary point on the plane. This method takes advantage of the properties of a plane in normal form.

Recall that the equation of a plane can be written as $ax + by + cz + d = 0$ where a, b, c, d are constants and (x, y, z) represents any point on the plane. We can write this more simply as the plane vector P = (a, b, c, d).

Let N be the vector (a, b, c), extracted from the equation of the plane. N represents a vector pointing along the normal of the plane. The plane vector can be converted into normal form by dividing by the length of N. That is:

$P' = P / sqrt(a^2 + b^2 + c^2)$

P' = (a', b', c', d') is the equation of the plane in normal form. Note that N' = (a', b', c') is now unit length. Additionally, |d'| is the minimum distance from the origin to the plane. More particularly, -d' * N' is the closest point on the plane to the origin.

Now, the distance from a point Q = (x, y, z) to a plane with normal N' passing through the origin is found by projecting Q onto N'

$Q \cdot N'$