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.
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.