Skew Planes?

Question

Skew lines cannot exist in two dimensions but can in three dimensions. How many dimensions are required for skew planes to be possible? Please give reasons for your answer.

Answer 1

Certainly 5d is enough: take P1 as the plane with x = y = z = 0, w & v free, P2 the plane with w = v = 0, z = 1, and x & y free.

Could you get away with 4d? No. Intuitively, each plane requires 2 degrees of freedom, and there's "no room" left over for two nonparallel planes not to intersect. The best you can do is a one-point intersection (x = y = 0 and z = w = 0).

Answer 2

Any plane in any dimension can be defined by a parametric sum of 3 vectors, as follows:

v1 + a(v1-v2) + b(v1-v3)

If two planes have a point in common, then we'd have the following equation true for some a, b, c, d:

v11 + a(v11-v12) + b(v11-v13) = v21 + c(v21-v22) + d(v21-v23)

If a solution for a, b, c, d exists, then the planes are not skew. So, what's the minimum dimension for the vectors so that a solution is not guaranteed? We need to set up 5 equations for 4 unknowns, making it possible for a non-existence of a solution for a, b, c, d. this means that the vectors must be 5-tuples, meaning that in 5-dimensional space, 2 planes may be skew and lacking a common point.