mr_e_man wrote:The planes intersect at a single point if and only if the bivectors' wedge product is a non-zero quadvector (representing the 4D space).
The planes coincide if and only if the bivectors' wedge product and cross product (commutator) are both zero; equivalently, if their geometric product is a scalar.
If the planes intersect at a line, then the bivectors' cross product gives a third bivector, perpendicular to that line. We want the 3D dual of this bivector, which is the vector representing the line. Call the first two bivectors A and B. Find some vector c in the plane of A and not in the plane of B; so c∧A = 0 ≠ c∧B = T, that last term being a non-zero trivector representing the 3D space containing the two planes. Normalize T if necessary, and multiply by it to take the dual: (A×B) T.
(The only questionable part here is finding c....)
Yes, it is that questionable part that has been stopping me. I have saved the two vectors used to create A. Zero, one, or both of those vectors are in the plane of B. In the case of one then I have to use trial and error to determine which of the two vectors is in the plane of B. It seems strange to have to use trial and error in an algebra but maybe that's just the way it is.
By the way, this is why the term "bivector" bothers me. Information is lost when two vectors are used to form a bivector. Suppose I didn't have those two vectors used to create A. Then it seems to me that I'd be reduced to choosing a random basis vector, projecting it into A, then testing it to see whether this was non-zero. If so, then next test whether it is in B. I was embarrassed to come up with such a crude solution. I thought there must be a better way. Maybe projective GA will do this for me.