> Actually, here's a thought... What if a take the dot product of point C 
> against the vector from A to B? Then I take the dot product of the point 
> where the line intersects the plane. If the answer has the same sign but 
> is nearer to zero, then I can do the same check with the other pair(s) 
> of points...
> 
> Maybe that could work...

Uh... no... that's NOT going to work...

What I want is a vector *perpendicular* to AB, but still in the same 
plane as the triangle... :-S

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