From a Point to a Plane, distance, etc

The usual form of the 3-d equation of a general plane is:

ax + by + cz = d\mbox{ with }\mathbf{v}=[a,b,c]\mbox{ a vector normal to the plane}\,

Suppose we have an external point \mathbf{p}=[x_1,y_1,z_1] and wish to know the shortest distance to the plane and where on the plane the closest approach point, \mathbf{q}, is. The shortest distance line to the plane is perpendicular to the plane. To see this draw some other line, \mathbf{pr}, to a point, \mathbf{r}, on the plane and make the right triangle \mathbf{pqr}. The distance along the line \mathbf{pr} is the hypotenuse of the right triangle and is therefore longer than \mathbf{pq}.

A convenient method to find the point \mathbf{q} on the plane at the foot of the perpendicular is to write the equation of a line through \mathbf{p} which is parallel to \mathbf{v}, the normal to the plane. Using a simple scalar variable t\, as a parameter for the line and solving for the intersection with the plane, we have:

[x,y,z] = [x_1,y_1,z_1] + t[a,b,c]\,\mbox{ parametric equation of the line}

inserting the parameterized [x,y,z] into the equation for the plane:

ax_1 + a^2t + by_1 + b^2t + cz_1 +c^2t = d\,

t= \frac{d-ax_1-by_1-cz_1}{a^2 + b^2 + c^2}\mbox{ at the intersection with the plane}

This formulation is good for a computer program and it is good to check that the denominator is not zero. If the denominator is zero the plane was no good to begin with.

\mathbf{q} = [x_1,y_1,z_1] + t[a,b,c] = [x_1,y_1,z_1] + \left(\frac{d-ax_1-by_1-cz_1}{a^2 + b^2 + c^2}\right)[a,b,c]

We can use the distance formula ( to find the distance from \mathbf{p} to \mathbf{q}:

\left|\mathbf{q}-\mathbf{p}\right|=\left|\frac{d-ax_1-by_1-cz_1}{a^2 + b^2 + c^2}\right|\left|[a,b,c]\right|\ =\ \left|\frac{d-ax_1-by_1-cz_1}{\sqrt{a^2 + b^2 + c^2}}\right|