In article <email@example.com>, Thomas Nordhaus <firstname.lastname@example.org> wrote:
> Am 28.07.2013 23:01, schrieb Ken Pledger: > > In article <email@example.com>, > > Thomas Nordhaus <firstname.lastname@example.org> wrote: > > > >> Am 26.07.2013 05:33, schrieb William Elliot: > >>> How do we find the shortest distance between two lines L, L' in R^3 > >>> ? > >>> > >>> 2. > >>> http://at.yorku.ca/cgi-bin/bbqa?forum=calculus;task=show_msg;msg=0792 > >>> > >> > >> How about doing it straight forward from scratch? Let P1(s) = v0 +s*v, > >> P2(t) = w0 + t*w be arbitrary points on the line L, L' resp. > >> .... > > > > > > Yes. Then just make the vector (P1(s) - P2(t)) perpendicular to > > both lines: > > That's what I meant "by scratch": You start by minimizing distances > between points on the lines (variational principle) and end up with > orthogonality conditions:
Establishing such orthogonality in real vector spaces does require an inner product, but does not require either differentiation or integration, so what part, or parts, of calculus do you claim it does require? > > > > > v.((v0 +s*v) - (w0 + t*w)) = 0 > > w.((v0 +s*v) - (w0 + t*w)) = 0. > > > > Therefore > > (v.v)s - (v.w)t = - v.v0 + v.w0 > > (w.v)s - (w.w)t = - w.v0 + w.w0. > > > > Solve those linear equations for s and t, > > then find ||(P1(s) - P2(t))||. > > > > > That's all. It's a traditional method in old text-books which aren't > > read much any more, and it doesn't need any calculus. > > I think implicitly it does.
Think again! > > > > > Ken Pledger. > > > > > -- > Thomas Nordhaus --