It is a known fact that the shortest path between tow points in sphere is a part of the great circle. but I don't know the proof of this claim so I tried to rigorously prove it myself.
My attempt: Claim 1 Let two points $a,b$ be in the opposite poles of the sphere with radius $r$ assume the distance is less than $\pi r$ by the symmetry of the sphere any two points in the opposite side of the sphere must have their distance less than $\pi r$ i.e the radius is less than $r$ which contradicts the hypothesis (I am not sure this part is correct).
Claim: any two points can be part of a Great circle.
Let two points $a,b$ be in the opposite poles and let $c$ be in the middle of distance between them in the great sphere if the distance between $c,a$ is less than half of distance in the great sphere then by the symmetry of the sphere the distance between $c,b$ which contradicts Claim 1.
the second method can be generalised for all rational numbers but I am having a trouble generalising this claim for irrationals
I am sure there is a simple rigorous proof of this fact but I can't find it.
First, I'll define the following unit vectors:
$u_0 = \dfrac{b - a}{\| b - a\|}$
$u_1 = \dfrac{a + b}{\| a + b \| }$
$u_2 = \dfrac{a \times b }{ \| a \times b \| }$
then the axis of rotation that takes $a$ to $b$ is given by
$ U = \cos \alpha \ u_1 + \sin \alpha \ u_2 $
unit vectors $v_1$ and $v_2$ are orthogonal to $U$, where
$ v_1 = \sin \alpha \ u_1 - \cos \alpha \ u_2 $
$ v_2 = u_0 $
Now vector $a$ can written in the basis $ u_0 , u_1 , u_2 $ as
$ a = r( - \sin \phi \ u_0 + \cos \phi \ u_1) $
$ b = r( \sin \phi \ u_0 + \cos \phi \ u_1) $
where $ 2 \phi $ is the angle between $a $ and $ b$ .
The projection of $a$ onto $v_1$ is
$ a_1 = ( a \cdot v_1 ) v_1 = r (\sin \alpha \cos \phi) v_1 $
and onto $v_2$ is
$ a_2 = (a \cdot v_2 ) v_2 = r ( - \sin \phi ) v_2 $
Similarly, the projections of $ b$ onto $v_1$ and $v_2$ are
$ b_1 = r (\sin \alpha \cos \phi) v_1 $
$ b_2 = r \sin \phi v_2 $
Therefore, the angle of rotation is given by
$ \psi = 2 \tan^{-1} \left( \dfrac{ \sin \phi }{ \sin \alpha \cos \phi } \right) = 2 \tan^{-1} \left( \dfrac{\tan \phi }{ \sin \alpha } \right) $
The length of the projection of $a$ onto the plane containing $v_1 $ and $v_2$ is
$ \| \text{Proj}(a) \| = \| \text{Proj}(b) \| = r \sqrt{\sin^2 \alpha \cos^2 \phi + \sin^2 \phi} $
Therefore, the distance travelled in rotating from $a$ to $b$ is
$ s(\alpha) = 2 r \sqrt{\sin^2 \alpha \cos^2 \phi + \sin^2 \phi} \tan^{-1} \left( \dfrac{\tan \phi }{ \sin \alpha } \right) $
Note that $ s(0) = \displaystyle \lim_{\alpha \to 0+} s(\alpha) = r \ \pi \sin(\phi) $, while $s(\dfrac{\pi}{2}) = 2 r \phi $
We can differentiate $s(\alpha)$, but it is going to be involved, so instead, I've plotted the function $s(\alpha)$ over $\alpha \in [0, \pi]$ for a typical $\phi = \dfrac{\pi}{6} $ and for $ r = 1 $.
This is shown in this Sage Worksheet
Clearly the minimum (of the arc length) occurs at $\alpha = 90^\circ$.
Hence, the axis that minimizes the arc length is given by
$ U = u_2 $
With this axis, the circular arc resulting from rotating $a$ into $b$ is part of a great circle, whose normal vector (the vector perpendicular to its plane) is $U$.