I will try to express everything in terms of the unit vectors p , q and r , which can be considered as points on the unit sphere & Sigma; centered at the beginning of 0 . You can convert this to terrestrial values by increasing the radius of the earth. There is background material here .
We want to find a large distance d from p to a large circle C passing through q and r strong>. C is the intersection of the plane P and the sphere & Sigma; , where P is the plane passing through q , r and the beginning 0 . d is just the angle & theta; (expressed in radians) between p and P. The normal vector for P is the normalized transverse product q x r / sin? phi; where? phis; is the angle between q and r .
As a result, we get
& theta; = arcsin ( p & sdot; ( q * r ) / sin & phi;)
As I said, everything here expands with the radius R of the earth. Thus, there are three points: * R *** p **, * R *** q **, * R *** r **, and the distance is R the.
However, if all you want is to find combos with points / lines with the shortest distance, you can omit the multiplication by R. In fact, you can omit arcsin () and just look at the relative sizes of p & SDOT; ( d & t ; t ) / sin & phi ;.
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