> Inverse trigonometric operations are slow.
> 
> Try this search:
> 
> http://www.google.com/search?q=efficient+line+triangle+intersection

Was about to try something even more complex and longwinded. But the 
very first link gives me this:

Plane thru A, B, C is given by
   P(u, v) = A + u*AB + v*AC
Line thru J, K is given by
   L(t) = J + t*JK
Intersection of P and L is I=L(t) where
   t = N . (A - P) / N . (Q - P)
u,v coordinate of I is
   u = AI . (N x AC) / AB . (N x AC)
   v = AI . (N x AB) / AC . (N x AB)
(N = normal of the plane).

Finally, the point lies within the triangle iff
   u >= 0
   v >= 0
   u+v <= 1

Add to that a check in case the line and triangle are perpendicular - 
and anoter check because I want my line to be finite length (check that 
0<=t<=1) and we're done! Ha!

Andrew.

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