Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 6

The sphere consists of points whose displacement from its center has fixed norm. Its vector and Cartesian equations areEquivalently, $\mathbf{r}=(1,2,3)+4\mathbf{n}$ with $|\mathbf{n}|=1$.

The coefficient vector $\mathbf{n}=(1,2,2)$ is a normal vector, so a vector equation of the plane isA parametric version is $\mathbf{r}=(20,0,0)+s(-2,1,0)+t(-2,0,1)$, since both independent direction vectors are perpendicular to $\mathbf{n}$.

The distance from a point to a plane is the absolute residual of the plane equation divided by its normal's length. Here $\mathbf{n}\cdot \mathbf{c}=11$ and $|\mathbf{n}|=3$, so

The center of the sphere-plane intersection is the perpendicular projection of the sphere's center onto the plane:The distance from $\mathbf{c}$ to the plane is $3<4$, so the intersection is a genuine circle. A right triangle with hypotenuse equal to the sphere radius givesIts plane is the given plane; displacements within it are perpendicular to $\mathbf{n}$.