I can't see why the method you gave would work. Of course, that doesn't necessarily mean it doesn't work.
What I can see working is the following:
Parameterize the two lines A and B as usual x = ta + A and x = tb + B (anything in bold is a vector) where a,b are normalized.
The closest points will be joined by a vector perpendicular to both lines, call this n = a x b .
Define a plane with normal perpendicular to both n and b and passing through the point B . The closest point on line A to line B will then be the intersection between this plane and the line A. Lets call this point r .
r is in the plane defined above so
(r - B ) . (n x b ) = 0
and r is also on the line A so
r = t a + A
for some t.
Substituting the second equation in the first gives
t = ( (B - A ) . (n x b ) ) / (a . (n x b ) )
and then substitute this back to find r . It's then pretty simple to find the corresponding point on line B.
This is probably the same to the method you just put up. But hope it helps anyway.
[edited by - sQuid on November 3, 2002 10:43:42 PM]
Point on a line nearest to another line
The projection method finds the shortest distant constrained to be in the plane that is projected into... i.e. it''s not guaranteed to be *the* shortest path, just a path that might be short 
"3D Game Engine Design" by David Eberly
The minimum distance occurs when grad(Q) = (0,0)
Q(s, t) = | L0(s) - L1(t) |
where L0 & L1 are the parameteric equations for the lines
"3D Game Engine Design" by David Eberly
The minimum distance occurs when grad(Q) = (0,0)
Q(s, t) = | L0(s) - L1(t) |
where L0 & L1 are the parameteric equations for the lines
struct Line{float M[3];float B[3];};float DistanceSquare(Line line1, Line line2, float& s, float& t){diff = line0.B - line1.B;a = dot( line0.M, line0.M );b = -dot( line0.M, line1.N );d = dot( line1.M, line1.M );d = dot( line0.M, diff );f = dot( diff, diff );det = |a*b - b*b|;//note this is a fuzzy-floating float compare (use an epsilon etc...)if(det > 0) { e = -dot( line1.M, diff ); invdet = 1/det; //Q(s,t) is the closest pair of points s = (b*e - c*d) * invdet; t = (b*d - a*e) * invdet; //this is the distance squared return s*(a*s + b*t 2*d) + t*( b*s + c*t + 2*e) + f; }else {//lines are parallel s = -d/a; t = 0; return d*s + f; }} 
Author
The results from the method I posted are either correct or just close, I can''t tell. There''s something strange going on with some part of the collision so I''ll try your method, sQuid.
But is it really necessary to say (a x b) x b? That''s what n x b is. Is there a simpler way to put that?
~CGameProgrammer( );
But is it really necessary to say (a x b) x b? That''s what n x b is. Is there a simpler way to put that?
~CGameProgrammer( );
Author
But I already know the distance between the two lines, Magmai. I wrote it in the main post in this thread. And it works in all cases
At least, it appears to.
~CGameProgrammer( );
At least, it appears to.
~CGameProgrammer( );
Author
Well sQuid, your method produces identical results as the method I posted. And because both methods involve two cross-products and two dot-products (if you use a variable for n x b and a x b) there appears to be no advantage to using either over the other.
~CGameProgrammer( );
~CGameProgrammer( );
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