Point-Plane Distance

Given a Plane

$$ax+by+cz+d = 0$$ (1)

and a point $(x_0,y_0,z_0)$, the Normal to the Plane is given by

$${\bf v} = \left[{\matrix{a\cr b\cr c\cr}}\right],$$ (2)

and a Vector from the plane to the point is given by

$${\bf w} = \left[{\matrix{x-x_0\cr y-y_0\cr z-z_0\cr}}\right].$$ (3)

Projecting ${\bf w}$ onto ${\bf v}$,

$$\vert{\rm proj}_{\bf v}{\bf w}\vert = {\vert{\bf v}\cdot {\bf w}\vert\over \vert{\bf v}\vert}$$

$$= {\vert a(x-x_0)+b(y-y_0)+c(z-z_0)\vert\over\sqrt{a^2+b^2+c^2}}$$

$$= {\vert ax+by+cz-ax_0-by_0-cz_0\vert\over \sqrt{a^2+b^2+c^2}}$$

$$= {\vert ax_0+by_0+cz_0+d\vert\over \sqrt{a^2+b^2+c^2}}.$$ (4)