Geometic Algebra -- Intersection of Planes in 4D.

[PatrickPowers]

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.

[mr_e_man]

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....)

[PatrickPowers]

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.

[Challenger007]

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.

In my opinion, this is a fairly simple way to check the correctness of the given calculations.

[PatrickPowers]

A solution that doesn't require trial and error may be found at https://en.wikipedia.org/wiki/Geometric_algebra#Extensions_of_the_inner_and_exterior_products

[mr_e_man]

I don't see a solution there.

The formula ((AI)∧(BI))I, where I is the unit quadvector, doesn't work: A is a bivector, so its 4D dual AI is also a bivector, and (AI)∧(BI) is a quadvector, whose dual is a scalar. No vectors here.

In fact this looks like a proof that there is no solution of the type you want. All we have here are even-grade multivectors, which form a sub-algebra; there is no way to get a vector (an odd-grade multivector) by adding, multiplying, negating, inverting, and grade-projecting these. But maybe there's some kind of square root that takes an even multivector and gives an odd multivector... No, there are too many degrees of freedom; choosing one square root is as arbitrary as choosing one vector in the plane of A.