I'd like to cut the $3$-sphere of radius $1$ with the plane $x+y+z=k$. Of course the cross-section is a circle: the problem is that I'd like to have the equation of the circle expressed by two "coordinates" on the plane.
An orthogonal basis for the plane (better, of the linear subspace associated to it) is $\{u_1,u_2\}$ where $u_1=1/\sqrt{2}(1,-1,0)$ and $u_2=\sqrt{2/3}(1/2,1/2,-1)$ (obtained by Gram-Schmidt).
Having a point $P$ on the intersection between the sphere and the plane, I could get the coordinates $X,Y$ on the plane simply by inner product with the basis vectors: $X=\langle P,u_1\rangle$ and $Y=\langle P,u_2\rangle$.
The calculations appear to be horrible: is there a neat way to do it?
The problem is not really limited to a sphere: i'd be interested in the general case of a generic surface cut by the same particular plane.
You have two equations: $x^2+y^2+z^2=1$ and $x+y+z=k$. The coordinates here relate to your new coordinates as follows:
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix} \sqrt{1/2} & \sqrt{1/6} & \sqrt{1/3} \\ -\sqrt{1/2} & \sqrt{1/6} & \sqrt{1/3} \\ 0 & -\sqrt{2/3} & \sqrt{1/3} \end{pmatrix}\begin{pmatrix}X\\Y\\Z\end{pmatrix}$$
The first two columns are your $u_1$ and $u_2$, the last is the normal vector of the plane. Now use this to express your equations in the new coordinates. Since the matrix describes just a rotation (or since your new basis is still orthonormal), the sphere still has $X^2+Y^2+Z^2=1$. For the plane, you multiply the row vector $(1,1,1)$ (which is the normal vector of the plane) from the left to the matrix, or in other words, you add the rows and obtain $\sqrt3Z=k$. Now you can combine these two equations and eliminate $Z$ from them. One approach would be computing the resultant using the Sylvester matrix:
\begin{align*} 1Z^2 + 0Z + (X^2+Y^2-1) &= 0 \\ \sqrt3Z - k &= 0 \\ 0=\begin{vmatrix} 1 & 0 & (X^2+Y^2-1) \\ \sqrt3 & -k & 0 \\ 0 & \sqrt3 & -k \end{vmatrix}&=k^2+3(X^2+Y^2-1) \end{align*}
So that's your implicit equation for the intersection figure. You could look at this and see that it's a circle, and then parametrize it using trigonometric functions of a single parameter if you like.