TITLE: Parametrization for intersection of sphere and plane
QUESTION [6 upvotes]: Given is the sphere $x^2 + y^2 + z^2 = 4$ and the plane $x + y = 2$ in $\mathbb R^3 $.
How can I find a parametrization for the intersection of the two?

REPLY [4 votes]: Let $\mathbf c = (1,1,0)$, and define two orthogonal unit vectors by $\mathbf u = (1,-1,0)/\sqrt 2$, and $\mathbf v = (0,0,1).$ Then Fly By Night's equations can be written as:
$$\gamma(\theta) = 
\mathbf c + (\sqrt 2 \cos\theta)\mathbf u + (\sqrt 2 \sin\theta)\mathbf v $$
This makes it clear that the curve is a circle with center at $\mathbf c$ and radius $\sqrt 2$.
You only asked about the parametrisation of the curve, not its shape, so this is slightly off topic, I suppose, but it might still be of interest.