Let's see: We assume that L, M do not intersect. Then there are unique points P on L and Q on M such that the distance ||P-Q|| is the least distance between any point of L and any point of M. Then the segment PQ must be perpendicular to both L and M (and conversely, and segment from L to M that's perpendicular to both lines has the least distance between them). Such a segment us unique. Then the planes mentioned below are the planes U through L and PQ, V through M and PQ, S through L and perpendicular to L, and T through M perpendicular to M. These can be found analytically by letting L: {y0 + sy} and M: {z0 + tz} for fixed vectors y0, y, z0, z and employing standard vector operations. (E.g., if w is an arbitrary segment from L to M, and r is the unique (up to +-) unit vector perpendicular to L and M, given by r = y x z, then R := <r,w> r is a vector whose length is the least distance from L to M in the direction of segment PQ, etc. --Dan Fred asks: << 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?
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