Re: [math-fun] perpendicular planes
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?
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
No prizes for guessing who had followed precisely this chain of reasoning, less than a month ago albeit in reverse order, in the course of constructing the line mutually perpendicular to two given lines ... DUH! By the way, L,M need not be non-intersecting, but they must be non-parallel: otherwise P,Q are undefined, and there are then two distinct solutions (unless L = M). WFL On 3/19/11, Dan Asimov <dasimov@earthlink.net> wrote:
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?
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
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Fred lunnon