PROOF

(i) Call the plane p. For any direction vector D = Q - P of p, we have

N · D = N · (Q - P) = N · Q - N · P = d - d = 0.

Let M be parallel to N, say M = sN.

M · D = (sN) · D = s(N · D) = 0.

Hence M ⊥ D and M is normal to p.

Now suppose M is normal to p. Let C and D be two nonparallel direction vectors of p. Then both M and N are perpendicular to C and D. Therefore M and N are parallel to C × D and hence parallel to each other.

(ii) Set d = N · P. The plane p with the equation N · X = d has position vector P and normal vector N by (i).

To show p is unique let q be any plane with position vector P and normal vector N. q has a vector equation

M · X = e.

By (i), N is parallel to M, say N = sM. Then the following equations are equivalent for all X:

N · X = N · P = d.

(sM) · X = (sM) · P.

s(M · X) = s(M · P).

M · X = M · P = e.

It follows that q equals p.