Determine a vector equation of the line of intersection of the planes p1: $3x - y + 4z - 2 = 0$ and p2: $x + 6y + 10z + 8 = 0$.
I know that I can find the cross product of the normals of these vectors (direction vector), but for a line, i need a direction vector AND a point, how would I get the point? Or is there another way to do it?
Thanks in advance!
Edit:
I got:
[2 3 -1 | 4] ~ [2 3 -1| -4]
[1 0 3 | 5] [0 0 -7 | -14] R1 - 2R2
therefore, t = 2
then how do I find x and y?







You just need to choose a point that satisfies both $3x - y + 4z - 2 = 0$, and $x + 6y + 10z + 8 = 0$, knowing that any such point will be on the line of intersection. Choosing $x=2$ and solving the resulting system of equations \begin{align} -y+4z+4=0 \\ 6y+10z+10=0 \end{align} for $y$ and $z$, we see that $y=0$ and $z=-1$, so $(2,0,-1)$ is a point on the line of intersection.