Distance of a Point from a Plane

Theorem 6.20 (vector form). The perpendicular distance from a point with position vector $\vec u$ to the plane $\vec r\cdot\vec n=p$ is

Proof sketch. Drop the perpendicular from $A(\vec u)$ to the plane, meeting it at $F$; the line $AF$ is $\vec r=\vec u+t\vec n$ (parallel to the normal). Substituting into the plane's equation gives the parameter $t_1$ at $F$, namely $t_1=\dfrac{p-\vec u\cdot\vec n}{|\vec n|^2}$; then $\vec{FA}=-t_1\vec n$ has length $\delta=|t_1||\vec n|=\dfrac{|\vec u\cdot\vec n-p|}{|\vec n|}$. The position vector of the foot $F$ itself is $\vec r_1=\vec u+t_1\vec n=\vec u+\dfrac{p-\vec u\cdot\vec n}{|\vec n|^2}\vec n$.

(b) Cartesian form. For a point $(x_1,y_1,z_1)$ and plane $ax+by+cz=p$:

$\delta=\frac{|ax_1+by_1+cz_1-p|}{\sqrt{a^2+b^2+c^2}}.$ …