Prove that there is exactly one point in the intersection of a given line and given plane

by RoRi

April 29, 2016

Consider the line $L$ through the point $(1,1,1)$ and parallel to the vector $(2,-1,3)$ . Then, let $M$ be the plane through the point $(1,1,-2)$ and spanned by the vectors $(2,1,3)$ and $(0,1,1)$ . Prove that there is exactly one point in $L \cap M$ , and determine the point.

Proof. First, we have that $L$ and $M$ are the sets of points

$\begin{align*} L &= \{ (1,1,1) + r(2,-1,3) \} = \{ (1+2r, 1-r, 1+3r) \} \\ M &= \{ (1,1,-2) + s(2,1,3) + t(0,1,1) \} = \{ (1+2s, 1+s+t, -2+3s+t) \}. \end{align*}$

Then, the points $(x,y,z)$ in $M$ are those which satisfy

$x = 1+2s, \qquad y = 1+s+t, \qquad z = -2 + 3s + t.$

This gives us the Cartesian equation

$x+y-z = 4.$

The points $(x,y,z)$ on $L$ are those such that

$x = 1+2s, \qquad y = 1-s, \qquad z = 1+3s.$

Thus, if $(x,y,z) \in M \cap L$ then we have

$\begin{align*} && (1+2s) + (1-s) - (1+3s) &= 4 \\ \implies && -2s &= 3 \\ \implies && s &= -\frac{3}{2}. \end{align*}$

Hence,

$x = -2, \qquad y = \frac{5}{2}, \qquad z = -\frac{7}{2}$

is the unique point in $L \cap M. \qquad \blacksquare$

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