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.
Hi. I am working in 4D Minkowski, but the answer should be the same. The meet of two planes is the point of intersection. The product of two bivectors produces a scalar, a bivector and a 4-vector (the point). For example, if the bivectors are u^v and w^x, all being mutually orthonormall, the point of intersection is u^v^w^x, which is the 4-vector representing the origin.
@PatrickPowers @wfmcgee
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.
That’s fascinating. If you pick any point P not on a x b=2, then the hyperplane a V P does not contain b, and the common line is indeed (a V P ) wedge b, after normalization.
That’s fascinating. If you pick any point P not on a x b=2, then the hyperplane a V P does not contain b, and the common line is indeed (a V P ) wedge b, after normalization.
This is my third try. Sorry. Gunn’s formula is fine. My proof is bad. Given two planes PL1 and PL2 in 4D with a common line L, P1 and P2 points different and not on L, PL1=P1VL, PL2=P2VL, then, for any point P, the Common Factor Theorem dual (Browne, page 103) gives
(PVPL1)^PL2=(PVP1VL)^(P2VL)=(PVP1VP2VL)^L. The first factor is a scalar (which Browne requires for the theorem to hold) , (the grade in P4GA is 4+4+4+3-15=0), which, if the factor is not zero, after normalization proves the assertion. The factor is non-zero if P does not lie on the (assumed non-zero) hyperplane P1VP2VL=P1VPL2.
@wfmcgee Many thanks for the reference to the Common Factor Theorem. I had heard of it but never really studied it; now I see that this is exactly the justification for my ad hoc algorithm and appears to work also for the original 4D problem.
Note: choosing P using the metric methods described in my post yields a result whose norm gives the sine of the angle between the two given lines (or planes). If I’m not mistaken, a general P will not produce provide this information.