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