Closest point on line segment of a great circle If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, \theta_A,\phi_A)$ and $(R, \theta_B,\phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points. 
Given a third point $P$ at $(R, \theta_P,\phi_P)$, how can I find out the point $Q \in AB$ which is closest to $P$ (in the geodesic sense)?
 A: There are a couple of online references.  
Mathoverflow
https://mathoverflow.net/questions/101776/altitudes-of-a-triangle
There is a more extensive theory at:
https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins
page 29. And, no I haven't read the whole article :) but it looks powerful.
Setting R=1.
The “$\times$” product below is the normal cross product normalized to 1.
Here is an intuitive method using the picture below with C taking the place of P.
Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $\left[a_{x},a_{y},a_{z}\right],\left[b_{x},b_{y},b_{z}\right],\left[c_{x},c_{y},c_{z}\right]$
 Call the great circle the “equator”.
Define the “North pole” $N$ as the intercept of $A\times B$ with the spherical surface; i.e. normalized.
Now every great circle passing through the North pole is a geodesic path intercepting the equator.
Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.
At this point, we could revert to Spherical coordinates and find $Q$.
Alternately we can stay with Cartesian coordinates.
We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $\text{N,C }$ giving
$Q=\left(A\times B\right)\times\left(\left(A\times B\right)\times C\right)$
This can be rephrased by means of compounded vector triple product.

