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

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

For $r=0$, the norm equation requires $x=c$. Thus the locus isThis includes $a=0$, when the first condition imposes no restriction.

First assume $a\ne 0$. The first equation is a plane, and the second is a sphere centred at $c$. Define its signed normal separation and projected centre byThe intersection of a plane and a sphere isThe circle lies in the given plane. Its radius follows from the Pythagorean theorem, splitting $x-c=(x-q)+(q-c)$ into orthogonal components. If $a=0$, the first equation is the identity $0=0$, so the locus is the entire sphere, rather than a planar section.

A norm is nonnegative, so it cannot equal a negative $r$. The locus is therefore $\boxed{\varnothing }$, irrespective of $a,b,c$.