Great circle through two points on a sphere

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.

Cross product

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 vb 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).

One thought on “Great circle through two points on a sphere

  1. Very nice write-up! I just wanted to add that you don’t need to go into angles in that last parameterization. Instead you can subtract the projection of w onto v from w. That is u=w-(w*v)v where * is the dot product. Then v*u=v*w-w*v=0 (Since v is on the unit sphere v*v=1).

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