Prove that three planes with normal vectors which are linearly independent intersect in exactly one point.

*Proof.* Let the normals of the three planes be given by

Then, the Cartesian equations of the three planes are given by

Since the normals are independent we know that they span the zero vector uniquely (by definition of independence). BY Theorem 12.7 (page 463 of Apostol) this implies that they span every vector in . Hence, the vector equation has a unique solution . Therefore, the system of equations has a unique solution. Hence, there is exactly one point on the intersection of three planes with linearly independent normal vectors

*Related*

How do we know that the question asks for the planes in V_3 and not in general in V_n (n > 2)? It seems that it holds only in V_3.

Let X be a intersect point:

N1.X=d1

N2.X=d2

N3.X=d3

we can get (N1 + N2 + N3) . X = d1 + d2 + d3

Suppose that there exist two intersect points (X1 and X2) =>

(N1 + N2 + N3) . X1 = d1 + d2 + d3

(N1 + N2 + N3) . X2 = d1 + d2 + d3

=>

(N1 + N2 + N3) . (X1 – X2) = 0

and N1, N2 and N3 linearly independent

=>

X1 – X2 = 0 => X1 = X2

so, there can only be one intersect point.

Let X be a intersect point:

N1.X=d1

N2.X=d2

N3.X=d3

we can get (N1 + N2 + N3) . X = d1 + d2 + d3

Suppose that there exist two intersect points (X1 and X2) =>

(N1 + N2 + N3) . X1 = d1 + d2 + d3

(N1 + N2 + N3) . X2 = d1 + d2 + d3

=>

(N1 + N2 + N3) . (X1 – X2) = 0

and N1, N2 and N3 linearly independent

=>

X1 – X2 = 0 => X1 = X2

so, there can only be one intersect point.