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I'm new here so sorry if this is a really silly question but I can't solve it myself.

If I have two points on a unit sphere, A and B, and the shortest path from A to B over the surface of the sphere, then how do I calculate the direction of this line at point B (as a vector)?

Do I need to compute this using quaternions? I have read a bit about it but find it very hard to understand how to use them. Alternatively, if rotation matrices can be used I would be very happy too.

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It seems to me that your problem is discussed in this thread. –  Jyrki Lahtonen May 31 '12 at 15:35
@Jyrki Thanks for the link. Looks like it does, and it gives a good deal of background too. I just find the answer here easier to follow though, but that is a matter of taste I guess. –  Paul May 31 '12 at 15:50
the answers there are "essentially" the same as the answer given by Peter H. here. He manages to do it with less words though :-). OTOH he doesn't need to discuss the conversions from one coordinate system to another. –  Jyrki Lahtonen May 31 '12 at 15:55

2 Answers 2

up vote 2 down vote accepted

The shortest paths connecting two points on a sphere are arcs of great circles, so the direction you are looking for is the tangent to the great circle containing $A$ and $B$. This tangent

  • lies in the plane of the great circle, i.e. is orthogonal to the plane normal given by $A \times B$
  • is orthogonal to the location vector of $B$

Therefore the direction that points from $B$ towards $A$ is the cross product of these two vectors, given by $B \times (A \times B)$.

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Wow, looking at it that way it is brilliantly simple and very understandable. Thanks. –  Paul May 31 '12 at 15:46
On a side note, how does the magnitude of this vector relate to the derivative? –  Paul May 31 '12 at 16:11

This works provided the points are not antipodes. Write an equation for the line connecting the two points; that is of the form $r(t) = (a + bt){\rm\bf i} + (c + dt){\rm\bf j} + (e + ft){\rm\bf k}. $ Now divide out by the norm of this function and you have it.

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