Past exam of the options II course of the University of Oxford / 2009 / P / P4 / c / Solution

Let $r=\Vert x\Vert >0$. Choose $c=-r$ if $x_1\ge 0$ and $c=r$ if $x_1<0$, and set $v=x-c{e}_1$. This opposite-sign choice ensures $v\ne 0$. Because $c^2=r^2$,ThereforeThe denominator is positive, including when $x$ already lies on the first coordinate axis.