If $u = \frac{1}{\sqrt{2}}(0,1,1)$, $v=\frac{1}{\sqrt{6}}(1,2,1)$, find $r>0$ such that $C(u,r)$ is tangent to $L_v$.
My attempt: I tried to relate $x \cdot u = \cos r$, $x\cdot v = 0$, and $\Vert x \Vert=1$, but, obviously, one more equation is needed, since there are four unknowns here. On the other hand, only $r$ needs to be found, and there are only two vectors $x$ satisfying the needed property. If these vectors $x$ could be found, $r$ could be found as well, but I'm missing something as I don't see how to find these vectors $x$. Some hints would be appreciated.