Home » Blog » Prove that the intersection of a line and plane which are not parallel contains exactly one point

Prove that the intersection of a line and plane which are not parallel contains exactly one point

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 L and the plane by M. Let M be the set of points

    \[ M = \{ P + sB + tC \}. \]

Since L is not parallel to M w know that its direction vector A is not in the span of B and C. Further, by definition of a plane, we know the vectors B and C are linearly independent. Hence, A,B,C are linearly independent. Then, any point X = (x_1, x_2, x_3) in the intersection L \cap M must be a solution to the system of equations

    \begin{align*}  a_1 x_1 + a_2 x_2 + a_3 x_3 &= d_1 \\  b_1 x_1 + b_2 x_2 + b_3 x_3 &= d_2 \\  c_1 x_1 + c_2 x_3 + c_3 x_3 &= d_3. \end{align*}

By the linear independence of A,B,C we know this system has exactly one solution (x_1, x_2, x_3). Hence, L  \cap M contains exactly one point. \qquad \blacksquare

One comment

  1. Anonymous says:

    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.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):