I have been taken to task for failing to point out that the formula given for the distance between two Euclidean subspaces, in particular lines L,M in 3-space, has an (unavoidable) defect: it fails when the subspaces are parallel. I'll post a fuller discussion if anybody requests it, but to cut a long story short: Algorithm for distance between parallel lines L,M in 3-space: Check that above formula for d^2(L, M) yields 0/0; Intersect either line with plane at infinity giving (same) point at infinity, then dualise giving (same) perpendicular plane: F = (<L•o>_3)* , G = (<M•o>_3)* ; Intersect both lines with this plane giving finite points: P = <L•F>_3 , Q = <M•G>_3 ; Compute the distance between these points: d(L,M)^2 = d(P,Q)^2 = ||<PºQ>_2|| / ||<P•Q>_0|| . In terms of Pluecker coordinates, P = [||L||, L_2 L^3 – L_3 L^2, L_3 L^1 – L_1 L^3, L_1 L^2 – L_2 L^1], Q = [||M||, M_2 M^3 – M_3 M^2, M_3 M^1 – M_1 M^3, M_1 M^2 – M_2 M^1], and denominator ||L||^2 ||M||^2 , where ||L|| = (L^1)^2 + (L^2)^2 + (L^3)^2, ||M|| = (M^1)^2 + (M^2)^2 + (M^3)^2 ; despite which, the numerator turns out to comprise 216 octic terms, and cannot substantially be reduced modulo the Grassmann relations L_1 L^1 + L_2 L^2 + L_3 L^3 = M_1 M^1 + M_2 M^2 + M_3 M^3 = 0 on the line coordinates. Somehow I don't think this would have been what Henry Baker had in mind ... Fred Lunnon On 3/22/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
In Euclidean 3-space, let P,Q be points with projective coordinates [P^0, P^1, P^2, P^3], [Q^0, Q^1, Q^2, Q^3]; for a finite point P, we may take P^0 = 1 and P^1, P^2, P^3 familiar Cartesian x-,y-,z-components.
Then the line L through P, Q has Pluecker coordinate [L_1, L_2, L_3, L^1, L^2, L^3] where L_1 = P^2 Q^3 - P^3 Q^2, L_2 = P^3 Q^1 - P^1 Q^3, L_3 = P^1 Q^2 - P^2 Q^1, L^1 = P^0 Q^1 - P^1 Q^0, L^2 = P^0 Q^2 - P^2 Q^0, L^3 = P^0 Q^3 - P^3 Q^0.
Now given two lines L,M, the square of the distance d between L and M equals
( L_1 M^1 + L_2 M^2 + L_3 M^3 + L^1 M_1 + L^2 M_2 + L^3 M_3 )^2
-------------------------------------------------------------------------------------------------- ( (L^2 M^3 - L^3 M^2)^2 + (L^3 M^1 - L^1 M^3)^2 + (L^1 M^2 - L^2 M^1)^2 )
[where (...)^2 denote squares rather than superscripts!]
If L•M denotes the product of L and M in the DCQ Clifford algebra Cl(3,0,1), the numerator and denominator above are the grade-4 part squared (the "wedge product") and the magnitude of the of the grade-2 part respectively:
d^2 = ||<L•M>_4*|| / ||<L•M>_2||
One way to derive this expression is to construct the parallel planes (called S,T earlier) joining L to the point at infinity along M, and M to the point at infinity along L resp; then subtract their normalised vectors, leaving [d,0,0,0].
When finding the point on a line or plane nearest to a given point, there is a much simpler method available: reflect the point in the line or plane, then halve the distance from the point to its reflection.
[I didn't include this formula in the appendix to the bicycle spokes paper, since it wasn't actually needed there; also it is rather more elaborate than the others.]
Fred Lunnon
On 3/20/11, Henry Baker <hbaker1@pipeline.com> wrote:
Perhaps closely related is the problem of the shortest distance between two non-parallel ("skew") lines in 3 (or more) dimensions.
I googled this & found this wikipedia entry:
http://en.wikipedia.org/wiki/Skew_lines
I'm not happy about the ugliness of these expressions. Yes, the "triple product" is a bit more satisfying, but I seem to recall seeing an expression somewhere that was much more satisfying -- something along the lines of the problem of the "distance of a point to a line", where the distance is obtained by plugging the point into the standard form of the line in terms of direction cosines:
directed distance from point (x0,y0) to sin(theta)*x+cos(theta)*y+C is sin(theta)*x0+cos(theta)*y0+C.
http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-...
I checked about 10 pages of Google's results, but couldn't find anything better than the wikipedia-type calculation for skew lines.
At 11:29 AM 3/19/2011, Fred lunnon wrote:
Given non-parallel lines L,M in Euclidean 3-space, there exist unique planes S,U meeting in L, and T,V in M, with both S,T perpendicular to both U,V.
An algebraic proof of this turned out to be surprisingly nontrivial --- is there a more intuitive synthetic demonstration?
[The appropriate generalisation of this this apparently tedious little lemma provides one crucial link in the classification of isometries for a quadratic inner-product space. It may well be a (special case of) some well-known theorem concerning bases of vector subspaces, though I didn't recognise it as such.]
Fred Lunnon