
Re: Distance Between Lines in R^3 (fwd)
Posted:
Jul 26, 2013 5:06 AM


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/cgibin/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.
Let f(s,t) = P1(s)P2(t)^2 = <P1(s)P2(t),P1(s)P2(t)> where <,> is the dotproduct. Compute the partials D1f, D2f w.r.t. s and t:
D1(s,t) = 2<P1'(s),P1(s)>  2<P1'(s),P2(t)> D2(s,t) = 2<P2'(t),P2(t)>  2<P2'(t),P1(s)>
this yields
D1(s,t) = 2<v,v0+s*v  w0t*w> D1(s,t) = 2<w,w0+t*w  v0s*v>.
You want that D1 and D2 vanish simultaneously. This results in the equations:
(A): s*<v,v>  t*<v,w> = <v,w0v0> = 0 s*<v,w> + t*<w,w> = <w,v0w0> = 0
This can be written as a matrixequation:
(B): M*[s,t] = [<v,w0v0>,<w,v0w0>]' where M is a 2x2 matrix and []' is a 2x1 columnvectors.
M is invertible provided its determinant is nonzero:
det(M) = <v,v><w,w>  <v,w>^2
***** (Case I, det(M)= 0) *****
Now, by the CauchySchwarz inequality det(M) is zero iff v and w are collinear, i.e. there is a real number a !=0 such that w = a*v. Then
det(M) <v,v><a*v,a*v>  <v,a*v>^2 = 0.
This refers to the case that L and L' are parallel. In this case going back to (A) results in:
(sat)*<v,v> = <v,w0v0> (as+a^2*t)*<v,v> = a<v,v0w0>
which are the same equations. Therefore one has to solve
(sat)*<v,v> = <v,w0v0>. So s = s(t) = at + <v,w0v0>/<v,v>.
Next we compute f(s(t),t), the square of the distance. This value must be independent of t. Choosing t=0 one obtains:
f(s(0),0) = <v0+s(0)vw0,v0+s(0)vw0>. The distance D between L and L' therefore is
D =  v0  w0 + (<v,w0v0>/v^2)*v 
***** (Case II, det(M)!= 0) *****
In this case you have to solve (A) for s and t. The unique solution (S,T) is then given by:
[S,T] = M^(1) * [<v,w0v0>,<w,v0w0>]'
The distance D between L and L' is then given by
 v0+S*v  w0T*w.
Hope I got it right!  Thomas Nordhaus

