Prove that the intersection of a line and a plane such that the line is not parallel to the plane contains one and only one point.

*Proof.* Denote the line by and the plane by . Let be the set of points

Since is not parallel to w know that its direction vector is not in the span of and . Further, by definition of a plane, we know the vectors and are linearly independent. Hence, are linearly independent. Then, any point in the intersection must be a solution to the system of equations

By the linear independence of we know this system has exactly one solution . Hence, contains exactly one point

Suppose there are two intersect points: X1 X2

X1 , X2 contained in the line L => X1 – X2 = kA (A is the direction vector of line)

X1, X2 contained in the plain M => X1 – X2 = mS + nT

X1 not equal with X2 so k != 0.

can get now A = m’S + n’T => A in the spanning plain of S and T => Line L is parallel to plain M

which contradicts with L not parallel to M, so there can not be more than one solution.

Now to prove that if direction vector A of line L not parallel to M there exist one solution at least:

Let x0 be one point contained in line L => X=x0 + kA for X in line L.

Let p0 be one point contained in plain M => M: (X-p0) . N = 0 (N is normal vector of M)

we can solve that two equation to get “k” =>

(x0 + kA – p0) . N = 0 => (x0 – p0) . N + k(A . N) = 0

A. N != 0 because A is not parallel to M. Suppose A . N = 0 => p0 + A is a solution of

M: (X – P0) . N = 0 => p0 + A = p0 + mS + nT (where S, T linearly independent) => A is in the span of plain (mS + nT) => L is parallel to M which contradicts L is not parallel to M

so we can get a solution k = – (x0 – p0) . N / (A . N).

Hence there can only be exist one intersect point if line L is not parallel to plain M.