Projecting a line onto a sphere?

I know how to project a point onto a sphere, but what about a line? I have a line in parametric form and know the equation of my sphere. My initial guess is to project the point which I know lies on my line onto the surface of the sphere, but then I’m unsure as to whether the direction vector is altered. Is it the case that I simply need to find a new point lieing on the line that has been projected whilst retaining the same direction vector?

I searched the internet, Dr. Math, my geometry course notes and my textbooks for an answer, but nothing!

Many thanks in advance for any insight.

This isn’t really my area, but it seems intuitively reasonable that you can project a line onto a sphere by projecting two points on that line onto the sphere.

Is a line mapped onto a sphere necessarily a great circle?

Seems like it would be half of one. Draw the cylinder of tangents to the sphere and orthogonal to the line. Project the two points of intersection of the line and cylinder onto the sphere, and you get half a great circle.

Thanks, I actually realised this on my way to a lecture just a while ago :smack:

If it is a normal projection then it will be a segment the arc of a great circle. If it is not a normal projection, it will be a segment of arc of a small circle. Note the if it is not a normal “great circle” projection, projecting two points and drawing the minimum length arc between them is not a correct projection, but this will work for a strictly normal projection.

Stranger

Sorry, I’ve thought about it a bit more and I don’t think this is the solution. Won’t projecting two points that lie on the line onto the sphere then finding the equation of the line that passes through these points have a line that cuts through the circle at those two points and not follow the circle’s curvature around?

Also it seems to me that you could have an infinite number of lines projected onto the surface which could have those two endpoints. You’d have to specify more than just the two points.

If it’s a “normal” projection, you’d just need to project the endpoints and then find the equation of a circle with the same center as the sphere and which passes through the endpoints.

If it’s not a “normal” projection, I think it would be more complicated.

Who said anything about that? The thing to do is to find the circle passing through those two points that has the same radius and center as the sphere.

True, but that’s only an issue if the two points are antipodal.

Assuming by “project” you mean the standard retract from pierced Euclidean n-space to the (n-1)-sphere, then you just do it for each point. I don’t see what’s so complicated about it. The projection of the line is the locus of the projection of each point.

I think the easiest way to conceptualize the line you want is to first make the plane containing the original line and the center of the sphere. The line on the sphere you want is on the intersection of the plane with the sphere. The boundary points of the line on the sphere are where lines between the sphere center and the ends of the original line intersect the sphere.

If you have the vector of the line and the vector of the projection, you can form a plane and use parametric equations for the intersection of a sphere with a plane.

Yeah, I eventually did the plane thing. Thanks all.