Given two points on a unit sphere, there is a unique great circle passing through the two points. This post will show two ways to find a parameterization for this circle.
Both approaches have their advantages. The first derivation is shorter and in some sense simpler. The second derivation is a little more transparent and generalizes to higher dimensions.
Let the two points on our sphere be v and w. If these two vectors are orthogonal, then the great circle connecting them is given by
cos(t) v + sin(t) w.
The problem is that w might not be orthogonal to v. So the task is to find a new vector u in the plane spanned by v and w that is perpendicular to v.
The cross product of two vectors in three dimensions is perpendicular to both. So
z = v × w
is orthogonal to v and w. So
y = z × v
is perpendicular to v and lives in the same plane as w. So y is the vector we’re looking for, except that it might not have unit length. (In fact, it’s probably too short. It will have length equal to sin θ where θ is the angle between v and w. If sin θ were 1, then v and w were orthogonal and we wouldn’t need to go through this exercise.)
So we need to normalize y, setting
u = y / || y ||.
This solution is quick and simple, but it obscures the dependence on w. It also only works in 3 dimensions because cross product is only defined in 3 dimensions.
If you look back, we used the fact that we’re working in ℝ³ when we argued that y was in the plane spanned by v and w. In more dimensions, we could find a vector z perpendicular to v and w, and a vector y perpendicular to z but not in the plane of v and w.
More general solution
We need to find a unit vector u in the space spanned by v and w that is orthogonal to v. Since u is in the space spanned by v and w,
u = a v + b w,
for some a and b, and so a parameterization for our circle is
cos(t) v + sin(t) (a v + b w).
We just need to find a and b. An advantage of this approach over the approach above is that the vector w is explicit in the parameterization.
Also, v and w could be vectors on some high-dimensional unit sphere. Even if v and w live in 100 dimensions, the subspace they span is two-dimensional and everything here works.
Since u is orthogonal to v, we have
u · v = (a v + b w) · v = a + b cos θ = 0
where the angle θ between v and w is given by
cos θ = v · w.
We can obtain another equation for a and b from the fact that u is a unit vector:
a² + b² + 2 ab cos θ = 1.
The solution to our system of equations for a and b is
a = ± cot θ
b = ∓ csc θ
and so an equation for our circle is
cos(t) v + sin(t) (cot θ v – csc θ w).