I'll use 3-D coordinates with the $z$-axis passing through the poles, so we start at $(\cos \theta, 0, \sin \theta)$. The optimal path $\gamma$ must be a geodesic, so $\gamma$ is an arc of a great circle through $(\cos \theta, 0, \sin \theta)$. Let $\nu$ be a vector in $\mathbb{R}^3$ othogonal to $\gamma$. Since $(\cos \theta, 0, \sin \theta) \in \gamma$, we have $(\cos \theta, 0, \sin \theta) \cdot \nu = 0$.

Let the end point of this geodesic have longitude $\psi$. Then $\gamma$ must be the shortest path to the line of longitude $(-\sin \psi, \cos \psi, 0)^{\perp}$. So $\gamma$ and $(-\sin \psi, \cos \psi, 0)^{\perp}$ must meet orthogonally, which means that $(-\sin \psi, \cos \psi, 0) \cdot \nu = 0$. 

So $\nu$ is a multiple of 
$$(\cos \theta, 0, \sin \theta) \times (-\sin \psi, \cos \psi, 0) = 
(-\cos (\psi) \sin (\theta),-\sin (\psi) \sin (\theta),\cos (\psi) \cos (\theta )).$$

The point of intersection of the geodesic $\nu^{\perp}$ with the line of longitude $(-\sin \psi, \cos \psi, 0)^{\perp}$ is a scalar multiple of
$$(-\sin \psi, \cos \psi, 0) \times \nu= (\cos ^2(\psi) \cos (\theta),\sin (\psi) \cos (\psi) \cos (\theta),\sin (\theta)).$$

One can compute the angle between $(\cos \theta, 0, \sin \theta)$ and $(\cos ^2(\psi) \cos (\theta),\sin (\psi) \cos (\psi) \cos (\theta),\sin (\theta))$, in order to find out how long the path is, but it doesn't seem to simplify in any nice way.