Proving that the line of intersection between two planes has a direction vector perpendicular to both of the lines. Assume two arbitrary planes are non-parallel then the cross product of the respective normal give us the direction vector of the line through their intersection. I look for a geometric/ algebraic proof for why that must be the condition.
 A: We have two non-parallel planes $\mathscr{P}_1$ and $\mathscr{P}_2$ with normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ respectively. Let $\mathscr{L}$ be the line of intersection of $\mathscr{P}_1$ and $\mathscr{P}_2$.

Let $P_0$ be a point on $\mathscr{L}$ and suppose that $\mathbf{v}$ is a vector parallel to $\mathscr{L}$. Note that $\mathbf{v}$ is a vector in both $\mathscr{P}_1$ and $\mathscr{P}_2$. This means that $\mathbf{v}\cdot\mathbf{n}_1=0$ and $\mathbf{v}\cdot\mathbf{n}_2=0$. That is, $\mathbf{v}$ is a vector orthogonal to both $\mathbf{n}_1$ and $\mathbf{n}_2$. Hence $\mathbf{v}$ is parallel to the cross-product $\mathbf{u}=\mathbf{n}_1\times\mathbf{n}_2$. 
A: Let $e$ be the intersection line of planes $P_1$ and $P_2$, and let their normals be $n_1$ and $n_2$.
Then $n_1$ is perpendicular to all lines of $P_1$, in particular, to $e$. Similarly, $n_2$ is perpendicular to $e$.
Hence, the direction of $e$ can be obtained as a vector orthogonal to both $n_1$ and $n_2$, which means (a scalar multiple of) $n_1\times n_2$.
To actually find $e$, we also need a point on $e$ (i.e. we have to identify one point -anyhow- which lies on both planes $P_1$ and $P_2$).
