I am considering the following points on a sphere with $\rho=1$, $A=(0,1,0)$, $B=(-1,0,0)$ e $C=\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$. I need to find the geodesics $\gamma_{AB}$, $\gamma_{BC}$ e $\gamma_{AC}$. My professor used a material with general theorems and prepositions and with no examples. I am looking for a way to start maybe with $\gamma_{AB}$.
I thought first to write $A$ and $B$ in polar coordinates, but still I can find no formula to apply these points on and find this geodesic.
ANY tips would be highly appreciated.
Let $\vec{a} = (0, 1, 0)$, and $\vec{b} = (-1, 0, 0)$.
These two points and origin (center of the sphere) are on a plane that passes through origin, and whose normal is $$\vec{n} = \vec{a} \times \vec{b} = (0, 0, 1)$$ because vector cross product yields a vector that is perpendicular to both $\vec{a}$ and $\vec{b}$, i.e. $\vec{n} \perp \vec{a}$ and $\vec{n} \perp \vec{b}$, by definition.
This plane contains both geodesics (minimum and maximum distance between the two points) between $\vec{a}$ and $\vec{b}$, because the intersection with the sphere is the great circle.
Furthermore, $$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{\left\lVert\vec{a}\right\rVert \left\lVert\vec{b}\right\rVert}$$ where $\theta$ is the angular separation between $\vec{a}$ and $\vec{b}$.
If you rotate $\vec{a}$ around vector $\vec{n}$ by angle $\theta$, you'll arrive at $\vec{b}$. (Here, $\lVert\vec{a}\rVert = \lVert\vec{b}\rVert = 1$; more generally, you'll rotate $\vec{a}$ parallel to $\vec{b}$ using this scheme.)