Chance in 4D That Two Random 2D Planes Intersect?

[PatrickPowers]

I got this far:

Without loss of generality we can fix one plane to be described by [w,x,0,0]. That is, if one plane is fixed but the other is random then the relation between the two planes is still as random as ever. That relation is all we care about. The two planes intersect if and only if there is a point in the random plane where y=z=0.

The random plane is chosen by taking three points at random. We take one point as the origin [a,b,c,d] then subtract it from the other two points to get vectors. A point is in the plane if it is in the span of the two vectors plus the origin.

Going to the vectorized random plane with [0,0,0,0] as the origin, that means this plane has to include a point where y=-c and z=-d. Excluding the probability zero case where y is a constant we can find points in the random plane where y=-c. What's the probability that one or more of these points also has z=-d?

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After this what I did was shaky.

[Hugh]

I asked o3-mini-high to respond and this is what it came up with, I hope it helps!

Patrick, your reasoning is on the right track. Once you fix one plane as all points with y = z = 0, the condition for intersection becomes finding a point in the random plane that also has y = z = 0. Writing the random plane in affine form as

  p + s·u + t·v  (with p = [a, b, c, d]),

the intersection condition reduces to solving

  c + s·u₃ + t·v₃ = 0
  d + s·u₄ + t·v₄ = 0.

Here u₃, u₄ and v₃, v₄ are the components of u and v in the (y, z) directions. Since u and v are chosen randomly, with probability 1 the 2×2 matrix
  A = [u₃ v₃; u₄ v₄]
is invertible. In that case, for any given (c, d) the linear system has a unique solution (s, t). In other words, aside from a degenerate situation (which has probability zero), the random plane will always contain exactly one point with y = z = 0.

Thus, the probability that the random plane meets the fixed plane is 1 (almost surely). This is a nice illustration of how in 4D two generic 2-planes, whose dimensions add up exactly to the ambient space’s dimension, will intersect in a unique point.

[PatrickPowers]

OK now I get it. Usually the random plane will have points for all possible values of (y,z). The exceptions are when y or z is a constant. One non-zero constant is enough, so it is possible to have planes that don't intersect but are nevertheless not parallel in that the minimal distance between them is not a constant.

This has to do with the 4D helicopter with two rotors. You have to have two rotors, in the wz and xz planes. In non-geometrically-ideal engineering you can't arrange things so the planes won't theoretically intersect, but I expect you can offset the two rotors enough that that intersection is always further away than the extent the blades. Then there's no need to synchronize the blades. It's even possible to have a single hub, just arrange the drive system to circle around the point where the planes intersect, avoiding it.

[PatrickPowers]

While two random 2D planes can't pass through one another, two solid 2D squares of flexible material can do it fairly easily, spontaneously. Looking grim for 4D cloth.