Suppose we have two planes, P = t[a1,a2,a3,a4] + s[b1,b2,b3,b4] + [p1,p2,p3,p4] and Q = r[c1,c2,c3,c4] + m[d1,d2,d3,d4] + [q1,q2,q3,q4], given in vector form in 4 dimensions and now we have to find the line of intersection.
I tried setting both vector equations equal to each other and solve for t, s, r, and m, but since there are four variables and four equations, the system has only one solution, and so I got a point. Did I mess it up? How else can I find the line of intersection without using the cross product of the normal vectors?
Best Answer
You probably want to adjust your intuition a bit: the intersection can be just a point, and indeed this is the typical case.
In $\mathbb R^4$, a two-dimensional plane represents two linear (technically “affine”) constraints. So intersecting two planes results in four constraints, which (assuming the constraints are independent) yields a single point.
Contrast to the situation in $\mathbb R^3$ where a plane only determines 1 constraint, so two planes determine 2 constraints, leaving enough freedom (in the generic case) to define a line.
So it’s your 3D intuition tricking you into thinking that two planes intersect in a line. It’s similar to how two lines in $\mathbb R^2$ intersect in a point, but in $\mathbb R^3$ the usual case is that two lines don’t intersect at all.