Thanks for the correction. Yes, it seems I omitted (so conveniently) that asin is multivalued, so there can be up to two solutions.

However, using the law of cosines (instead of sines) fixes this inconvenience and also eliminates the need to use trigonometric functions.
Also, it'll be the solution with the relative velocity that has the larger magnitude (trivial).
Let's say <Va,Vba>/|Va|/|Vba|=cos(a)
So: Vba = sqrt(Va^2(cos(a)^2-1) + Vb^2) - Va*cos(a)

So still, no need to find the actual time of collision.

I think I understand. You're using triangle similarity:



First you find the magnitude of 'd' by using either the Law of Sines or Cosines. Then you can calculate the ratio of 'D / d', which you can then use to modulate Va or Vb to find the distance each object has to travel in order to reach the interception point.
This is a very smart solution. Thanks a lot!

EDIT: I see now you're finding the angle between Vba and Vb with the Law of Sines, which is just as well. The angle object B needs to be turned by will be 'alpha + beta' (as it's external to the triangle, and relative to A's path).

Yep, you got it exactly right!

I just thought that all you were really interested is the velocity of B, so I stopped there. But if you need the intersection point itself then this is the way to obtain it.