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

The dot product with the given unit vector is $\mathbf{r}\cdot \hat{\mathbf{n}}=(x+z)/\sqrt{2}$. Subtracting the corresponding orthogonal projection givesTaking its squared Euclidean norm yields the equivalent Cartesian equation

The removed vector is the orthogonal projection of $\mathbf{r}$ onto the line through the origin in direction $\hat{\mathbf{n}}$. The remaining vector measures the perpendicular displacement from that line. Fixing its Euclidean norm at $2$ therefore describes an infinite circular cylinder of radius $2$, with axis $x=z$, $y=0$.

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

Use the distance from a point to a plane formula with the unit normal $\hat{\mathbf{n}}$. Since the plane passes through the origin, the distance isFor a direct check, the closest point on a plane is $\mathbf{p}-(\mathbf{p}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}=(1,\sqrt{2},-1{)}^T$, whose displacement from $\mathbf{p}$ has Euclidean norm $\sqrt{2}$.