Past exam of the mathematical methods I course of the University of Oxford / 2014 / 10 / a

A direction of the intersection line must be orthogonal to both normal vectors. Their cross product isThus a convenient parallel vector is $\boxed{u=(0,1,2)}$.

Eliminating $2y-z$ from the two plane equations gives $x=2$, followed by $2y-z=2$. Taking $y=1$ gives the point $\boxed{p=(2,1,0)}$. The entire intersection line is $\mathbf{r}=p+tu$, where $u=(0,1,2)$.

The orthogonal projection of the origin onto the line occurs at $t_{\ast }=-p\cdot u/|u{|}^2=-1/5$, giving closest point $(2,4/5,-2/5)$. Hence the distance from a point to a line in three dimensions is