# Geodesic "perpendicularity"

If we have two points A and B on the surface of a sphere, a geodesic between them, and another point C on the same sphere surface, but not on the geodesic, is there any concept of a "perpendicular" geodesic to AB that passes through C ?

Because I'm not able to describe the problem mathematically (because I don't know what is the exact concept I'm searching for, and I don't have the proper mathematical vocabulary), I'm going to describe the practical problem for which I need this.

A and B are two points on the surface of the Earth with a geodesic between them, C is another point on the surface of the Earth, which does not pass through AB, and I need to calculate the coordinates of D, on the AB geodesic, so that the geodesic distance between C and D is minimized. It is sort of a "shortest distance from point to line" problem applied to geodesics. In 2D geometry D would be the perpendicular foot from C to AB.

• Ignoring the equatorial bulge and considering the Earth to be a perfect sphere, the geodesics are arcs of great circles. So in this restricted case what your intuition tells you is true. By rotation we can consider $A,B$ to be points on an "equator", and thus the shortest line from $C$ to the equator is a line of longitude. This is not trivial for a navigator to achieve without "rotating the Earth", but I can explain the difficulty if that sort of answer would be helpful. Commented Oct 2, 2014 at 18:06
• This calculation is going to be implemented in a computer program, and probably "rotating the Earth" in memory could be an option (speed isn't a problem). Also, this is going to be used mainly on small surfaces (under 1 km) so I think approximating the Earth to a perfect sphere would not distort the results very much. If you wish to give more details about your solution and you think it is feasible to implement in a computer program, it would be greatly appreciated. Thank you Commented Oct 2, 2014 at 18:19

Consider the plane tangent to the sphere at a point where two geodesics intersect. Within this plane, there are two lines, each tangent to one of the geodesics. If these two lines are perpendicular, then the two corresponding geodesics are "perpendicular."

As for your problem, if there isn't a restriction against this, then you could stereographically project (link to Wikipedia) from the sphere onto the plane. Then, you can work in the plane and, when finished, map back into the sphere. This is a common technique in differential geometry.

• Thank you for your answer, looks like an option, I will check that out. Commented Oct 2, 2014 at 18:23
• But in order to perform this stereographic projection, you have to know where the two geodesics intersect, don't you? Which is exactly what you are trying to calculate. Commented Oct 2, 2014 at 18:36
• No. Stereographic projection is applied to the whole sphere, minus a point. That particular point, if your protest is what I think it is, is absolutely irrelevant. If the geodesic happens to intersect this point, just apply a rotation. Commented Oct 2, 2014 at 18:40
• Thank you for the clarification. But now what is the operation that you have to perform in the plane that corresponds to finding the intersection of the two geodesics? Commented Oct 2, 2014 at 20:10
• If the geodesics intersect at a point $p$ on the sphere, then the images of the geodesics under stereographic projection $\sigma$ intersect at $\sigma(p)$. Commented Oct 2, 2014 at 20:24

Obtain the vector AB (that provides is normal vector "N" of the plane, thru C, that will intersect with AB).

Obtain the plane OCN, passing thru the Origin and C, with the normal provided by AB.

Obtain the intersection point between plane OCN and geodesic AB (on surface of sphere), call that D.

The geodesic from C to D will be the shortest geodesic (same as arc of intersection between the plane and the sphere, from point C to point D).