Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 11 / b / iii

The plane equation $x+z=0$ has normal vector $(1,0,1{)}^T=\sqrt{2}\hat{\mathbf{n}}$. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis $\mathbf{e}=(1,0,-1{)}^T/\sqrt{2}$ and $\mathbf{f}=(0,1,0{)}^T$ of the plane, write $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$. The cylinder equation becomes $s^2+t^2=4$. Equivalently, the intersection has the parametric equation