I'm working in a four dimensional Euclidean space. I have two planes, each represented by a bivector. How do I find the intersection of these two planes? This will be either 0, a vector, or a bivector.
I could do this if I could find the smallest subspace that contains both planes.
Suppose your $2$-vectors are $a\wedge b$ and $c\wedge d$, with $a,b,c,d$ column vectors. Consider the matrix
$$ M = \begin{bmatrix} | & | & | & | \\ a & b & c & d \\ | & | & | & | \end{bmatrix}. $$
Note $\{a,b\}$ and $\{c,d\}$ are linearly independent. If the column rank of $M$ is $2$, then $\mathrm{span}\{a,b\}=\mathrm{span}\{c,d\}$ are the same plane. If the column rank of $M$ is $3$, then we can compute a $u$ in its kernel, unique up to scalar multiples, and then we can say that $u_1a+u_2b=-u_3c-u_4d$ is in the intersection of these planes. And if the column rank of $M$ is $4$ then the planes do not intersect.
