Vector equation of a line defined by two planes

Question

"Given two planes:

$ 3x+4y-z=2 $
$ -2x+y+2z=6 $

Find and equation of the line defined by the intersection of these planes.

I have found the normal vectors, and cross-producted them. But I am not sure what to do next.

Thanks

Answer 1

You have

$ 3x+4y-z=2 $
$ -2x+y+2z=6 $

So let's set $x=0\implies 4y-z=2, y+2z=6$ so $9y=10\implies y=\frac{10}9, z=\frac{22}9$ So because a line $r$ is equal to $\vec a+t\vec b$

You have $r=\begin{bmatrix}0\\\frac{10}9\\\frac{22}9\end{bmatrix}+t\vec b$ while $\vec b$ is the result of the cross product