I believe @wfmcgee has answered @PatrickPowers’s question in the general case, where a = u\wedge v and b = w\wedge x are linearly independent hence have a unique point (4-vector) a \wedge b in common.
When a \wedge b= 0, however, it is possible that they have a line (3-vector) in common, or that they are even identical and hence represent the same 2-vector. Then the question is, how can this common element be found?
I frankly haven’t worked out the answer for the 4D case that Patrick has posted. However it is closely related to a similar case in 3D that I encountered several years ago. The same question recently came up on the bivector discord channel. It has the advantage that it can be more easily visualized than this 4D example. I’ll present it here and invite readers to see whether they can find an analogous solution in 4D.
Let a and b be bivectors as above, but now they represent lines in 3-space. When a \wedge b = 0, they are linearly dependent and either have a point (3-vector) in common or they are identical. How to find this point C?
It turns out there are several ways to answer this (as is happily often the case in PGA). I present two of these solutions here, one from @enki and one from me.
@enki provided this solution on discord: (((a\wedge e_0)\vee b) \cdot b)\wedge a. That is, find the common plane c = (a\wedge e_0)\vee b of a and b, then take the inner product c \cdot b, producing the plane containing b perpendicular to c, and finally intersect that plane with a to produce C.
An alternative is: let P be any point that is not contained in the common plane c of a and b. Assuming that we have found such a P (see below), then a \vee P is a plane containing a but not containing b, and (a \vee P) \wedge b will be the desired common point C.
How to find P? Recall n := \langle ab \rangle_2 is the line perpendicular to both a and b (when it’s non-zero). We obtain a point not lying in c by P:= n \wedge e_0, producing the solution C = (((\langle ab \rangle_2)\wedge e_0)\vee a)\wedge b. (When n=0, the two lines are the same.)
Exercises: 1. Show that when the arguments a and b are normalized, then the solutions for C found above satisfy \|C\| = sin(\alpha) where \alpha is the angle between a and b.
2. Devise an analogous solution to the 4D problem presented by @PatrickPowers in the original post on this thread.