Distance from a Point to a Plane

Perpendicular distance from point P(x₁, y₁, z₁) to plane ax+by+cz+d=0. The formula is the 3D analogue of the 2D point-to-line distance.

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

Method: The normal to the plane has direction $(a,b,c)$. The line through $P$ perpendicular to the plane:

$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = t$$

Parametric point: $(x_1+at, y_1+bt, z_1+ct)$. Substituting into the plane equation:

$$a(x_1+at)+b(y_1+bt)+c(z_1+ct)+d = 0$$ $$(ax_1+by_1+cz_1+d) + t(a^2+b^2+c^2) = 0$$ $$t = -\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$$

Distance $= |t| \cdot \sqrt{a^2+b^2+c^2}$:

$$d = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$$

Distance between parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$:

$$d = \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$$

(Distance from origin to each plane, then take the difference — or use the above formula with a point from one plane.)