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Topic: Distance Between Lines in R^3 (fwd)
Replies: 15   Last Post: Sep 13, 2013 1:25 PM

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Thomas Nordhaus

Posts: 433
Registered: 12/13/04
Distance between lines in R^n (Was: Distance Between Lines in R^3

Posted: Jul 27, 2013 6:40 AM
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Am 26.07.2013 16:26, schrieb Bart Goddard:
> William Elliot <> wrote in

>> How do we find the shortest distance between two lines L, L' in R^3
> Each line has a direction vector. The cross product n of the two
> direction vectors is perpendicular to both lines. Pick a point
> on each line, P and Q. The projection of the vector from P to Q
> onto the vector n is the distance.

That's nice and simple. However the situation in R^3 seems to be special:

Let L: s -> v0 + s*v, L': t -> w0 + t*w. Suppose v and w aren't parallel
then there are exactly two unit vectors n1 and n2 that are perpendicular
to v and w. Furthermore n1 = -n2. You can compute them easily via the
vector (cross) product in R3: n1,n2 = +- (vxw)/norm(vxw).
Then the distance D(L,L') between the lines L and L' is given by

(A): D(L,L') = |<vxw,v0-w0>/(norm(vxw))|, <,> the scalar product.

Again, what makes this special is that in R^3 you only have one line
through the origin which is perpendicular to the plane spanned by v and
w. In R^n this is a hyperplane of dimension n-2.

On the other hand, using variational principles it is easy to compute
D(L,L') for all n>=2 just using the euclidean structure, i.e. scalar
products. I got the following general result (using Matlab-Notation for
matrices and vectors):

Let L: s -> v0 + s*v, L': t -> w0 + t*w be non-parallel lines.
Let M=[<w,w>,<v,w> ; <v,w>,<v,v>]/(<v,v><w,w> - <v,w>^2),
[S;T] = M*[<v,w0-v0>;<w,v0-w0>].

Then D(L,L') = norm(v0+S*v-(w0+T*w)).

(I derived this in the accompanying thread). This even works for n=2,
giving D(L,L')=0.

Question: Is there a generalization of the vector-product solution (A)
to higher dimensional euclidean space R^n, n>3 using some sort of
generalized vector-product?

Thomas Nordhaus

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