TITLE: Paramertrization of intersection between spehere and plane.
QUESTION [0 upvotes]: I have the normal  $n = (a,b,c)$  for a plane through origo,and want to find the paramertrization of the unit circle.
How can I do this? 
I guess I should eliminate one coordinate from the plane and put into the circle, and then do some paramertrization of the other two. But the calculations gets messy.

REPLY [1 votes]: If  $n$ is normal to the plane through the origin, the plane's equation is $ax+by+cz=0.$ The vector $u=(-b,a,0)$ lies in this plane, and another vector $v$ in this plane which is orthogonal to $u$ may be found from the cross product of $n$ with $u$. So one choice for $v$ is $v=(-ac,-bc,a^2+b^2).$ 
Now these vectors $u,v$ should be normalized to have unit length. I'll assume they have been normalized to $u_1,v_1$. You now have two orthogonal unit vectors in the plane $ax+by+cz=0$, so a parametrization of the unit circle in that plane is
$$F(t)=\cos(t)u_1+\sin(t)v_1.$$
There are of course infinitely many choices for two vectors $u_1,v_1$ which are orthogonal unit vectors in the plane $ax+by+cz=0$. Depending on how you wish to use the parametrization, some particular choice of $u_1,v_1$ (other than the somewhat arbitrary one outlined above) might be more convenient.
NOTE: In the above I assumed $a,b$ not both $0$. This guarantees the vectors $u,v$ are nonzero, since the third coordinate of $v$ is $a^2+b^2$. If it happens that $a=b=0$ then the plane is simply the $xy$ plane, whose unit circle parametrizes as $(\cos t,\sin t,0).$
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