I am seeking the unit vector $v=(x,y,z)$ tangent at $p$ to the geodesic on sphere connecting $p=(a,b,c)$ and $q=(d,e,f)$. The curve is orientated from $p$ to $q$.
Once the tangent vector is ortogonal to the point vectors, so I must have
$$\begin{cases}ax+by+cz=0\\ x^2+y^2+z^2=1\end{cases}$$
I need more two equations to get the vector, could you help me?
The aim is to prove that the application $(p,q)\mapsto v$ is continuous.
Thank you in advance.
On
Another way to get $v$ is to first consider the vector $w: = q-p$, which points in the right direction, but it's not tangent. One can check that its projection onto the tangent space $T_p \mathbb{S}^2$ $$ v := w - (w \cdot p) p $$ is the tangent vector we want.
The outcome is the same if we just consider the projection of $q$ onto $T_p \mathbb{S}^2$: $v = q - (q \cdot p) p$.
HINT: There is no calculus or differential geometry here. If you want the great circle through $p$ passing through $q$ ($q\ne -p$), it lies in the plane with normal vector $p\times q$. The tangent vector to a great circle will lie in that plane and be orthogonal to the position vector. How do you find a vector in that plane orthogonal to $p$?