Nearest Point on a Great Circle

Date: 05/27/2002 at 06:19:57
From: Jayasundara
Subject: projection of a point to a great circle

I am studying spherical geometry, and need help with the following 
question.

Given points A, B, and C on the surface of a unit sphere, I need to 
find a point P on the great circle arc defined by A and B such that 
the distance from C to P is equal to the perpendicular distance.

I found a solution to calculate the perpendicular distance from point 
C to the great circle defined by points A and B.  Then, I think if I 
have the equation of the great circle, I could find the point on it 
which gives same perpendidular distance from the given point C. 

But, I have no clear idea for this. Could you please comment on this.

Thanks,
Jayasundara

Date: 05/27/2002 at 11:36:02
From: Doctor Rick
Subject: Re: projection of a point to a great circle

Hi, Jayasundara.

I don't know what sort of spherical geometry you are learning. I find 
that the easiest way to work with points on spheres is to use vector 
algebra. Thus a point on the sphere is represented by any vector from 
the center of the sphere that passes through the given point.

A great circle is uniquely identified by a vector perpendicular to 
the plane of the great circle. Given two points on the great circle 
(your A and B), you can easily find such a vector G by taking the 
cross product (or vector product) of vectors corresponding to the 
points.

It's also easy to find the great circle passing through a point C and 
perpendicular to the great circle specified by vector G. You're 
looking for a vector F perpendicular to this great circle, therefore 
perpendicular to vector C, and also perpendicular to the vector G.

Once you have the vectors F and G, it isn't hard to find the points 
of intersection of the two great circles. How will these points be 
related to vectors F and G?

If you're not using vector algebra, I can't offer a whole lot of 
help. 

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/28/2002 at 04:05:25
From: Jayasundara
Subject: projection of a point to a great circle

Thank you very much for the reply.

I hope that to get the perpendicular vector to the plane of the 
great circle, we have to normalize the vector G (G = A x B). Is that 
correct?

Now I have vector G perpendicular to A and B (also perpendicular to 
the plane of the great circle).

I also have vector F = (C x G) / ||C xG||, which is perpendicular to 
C and G (also perpendiculer to the plane of the great circle).

Then taking the cross product of F and G and normalizing the resultant
vector,  I have the intersection points of the great circles 
(+/- (F x G) / ||F xG|| ). These are antipodal points. 

I hope in this way these points will be related to vectors F and G.

Is my argument is correct? Please let me know.

Thanks a lot,
Jayasundara

Date: 05/28/2002 at 08:39:36
From: Doctor Rick
Subject: Re: projection of a point to a great circle

Hi, Jayasundara.

For your purposes, normalization is not necessary until the last 
step, when you convert a vector into coordinates of a point on the 
sphere. If we needed to find an angle between vectors (other than 90 
degrees) using the dot (scalar) product of two vectors, the 
magnitudes of the vectors would matter, and normalization is 
appropriate; but your problem does not require this. It doesn't hurt, 
though, so if you want all vectors to be unit vectors, go ahead and 
normalize at each step.

Vector G = A x B is perpendicular to the plane of the great circle. 
The vector F = C x G is perpendicular to C, so the great circle it 
defines passes through C. It is also perpendicular to G, so the great 
circle it defines is perpendicular to the great circle defined by G. 
The intersections of the two great circles correspond to vectors that 
are on both great circles, and thus perpendicular to both F and G. 
Your formulation is correct.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/28/2002 at 09:54:14
From: Jayasundara
Subject: Thank you

Thanks a lot for the information and immediate response.

Jayasundara