Foot of Perpendicular from a Point to a Plane

The foot of the perpendicular from P(x₁,y₁,z₁) to plane ax+by+cz+d=0. The perpendicular from P has direction (a,b,c) (the normal); substitute the parametric point into the plane equation to find the parameter.

Let $P(x_1,y_1,z_1)$ and plane $ax+by+cz+d=0$.

The perpendicular from $P$ to the plane has direction $(a,b,c)$ — the normal direction. Parametric form:

$$(x, y, z) = (x_1+at,\; y_1+bt,\; z_1+ct)$$

From the distance derivation: $t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$.

The foot $F(x_F, y_F, z_F)$:

$$\frac{x_F-x_1}{a} = \frac{y_F-y_1}{b} = \frac{z_F-z_1}{c} = -\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$$

Let $k = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$. Then:

$$x_F = x_1+ak, \quad y_F = y_1+bk, \quad z_F = z_1+ck$$

Verification: $F$ must satisfy the plane equation — substitute $(x_F,y_F,z_F)$ and confirm it equals zero.