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Given that I have a vector in R3 (7t, 10t - 2t^2, 5t) | (These numbers are arbitrary for the sake of the process)

A sphere centered at the point ( 15, 25, 10) with a radius of 20

There is a possibility of 1 to 5 intersection points. (From what I can tell)

Is it possible to find all these points with conventional algebra?

There are countless linear line-sphere intersection equations online but I don't think those would apply in this case.

user1828526
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The sphere is $(x-15)^2+(y-25)^2+(z-10)^2 = 20^2$. Substitute in the vector you have for x, y and z and you get a 4th order polynomial in t to solve. In 3D the curve traced out by your vector is intersecting the sphere at 4 points.

Paul
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  • Also this post http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations was really helpful in solving 4th order equations. – user1828526 Apr 02 '14 at 09:43