Finding the intersection of two planes, given their normal vectors Two vectors $v_1=(-1,1,0), v_2=(1,0,1)$ span a plane $P$ in $\Bbb R^3$.
Two vectors $w_1=(0,1,0), w_2=(1,1,0)$  span a plane $Q$ in $\Bbb R^3$.
I am asked to show that $P$ and $Q$ are different and to find $P ∩ Q$.
I have computed the normal vectors to each plane:
normal to $P$ is $n_3∗ (0,-1,1)^t$
normal to $Q$ is $n_3∗(0,0,1)^t$
If the normal vectors are different, then the planes must be different.
How do I find $P ∩ Q$?
Thank you for your help.
 A: Here is a trick, since you have already computed the normal vectors.
$Normal_P$ and $Normal_Q$ are orthogonal to the planes $P$ and $Q$, respectively. Hence they are both orthogonal to the intersection of the planes $P$ and $Q$. But the intersection of 2 planes is a line, and it is a line spanned by a vector that is orthogonal to $N_P$ and $N_Q$.
The cross product of the normals will give you a vector orthogonal to $N_P$ and $N_Q$, which is necessarily nonzero, and which lies on the intersection. This is enough to describe the line that is the intersection.
A: Being orthogonal to a vector$~\vec p$ is a linear homogeneous equation for an unknown vector$~\vec v$. If (as in this example) one is in dimension$~3$ and $\vec p\neq\vec0$, then the solution set is a plane, say$~P$. If you add a similar condition for another vector $\vec q$ independent of$~\vec p$, then that condition is also a linear homogeneous equation, which defines another plane$~Q$, and the solution for the system of equations is the intersection $P\cap Q$, which is a line. Just write down and solve the system of linear homogeneous equations to find a more direct description of that line.
A: Hint: Let $n_1$ and $n_2$ be normal vectors. If $n_1\neq cn_2$ then intersetion is a line with direction vector $n_1\times n_2$.
