Use Lagrange multipliers to find the minimum distance to the cone $z^2 - x^2 - y^2 = 0$ that are closest to the point $(1,3,1)$

Question

I've got the problem started but I'm stuck on getting the critical points. So far I got the 4 equations

1) $2(x-1)=-2\lambda x$

2) $2(y-3)=-2\lambda y$

3) $2(z-1)=2\lambda z$

4) $z^2 - x^2 - y^2 =0$

I know i have to use these equations to solve for each unknown but I'm stumped on how to so.

Answer 1

Multiply the fourth equation by $ \lambda^2$ and sub ... gives \begin{eqnarray*} (z-1)^2 -(x-1)^2-(y-3)^2=0 \\ 2x+6y-2z=9. \end{eqnarray*} Now rearrange the first $3$ equations \begin{eqnarray*} x= \frac{1}{1+\lambda} \\ y= \frac{3}{1+\lambda} \\ z= \frac{1}{1-\lambda}. \\ \end{eqnarray*} sub these into $x+3y-z=4$ ... should be a doodle from here ?

Answer 2

By C-S $$\sqrt{(x-1)^2+(y-3)^2+(z-1)^2}=\sqrt{2z^2-2z+11-2(x+3y)}\geq$$ $$\geq\sqrt{2z^2-2z+11-2\sqrt{(1+9)(x^2+y^2)}}=\sqrt{2z^2-2z+11-2\sqrt{10}|z|}.$$ Now, let $z\geq0.$

Thus, $$\sqrt{2z^2-2z+11-2\sqrt{10}|z|}=\sqrt{2z^2-2(1+\sqrt{10})z+11}=$$ $$=\sqrt{2\left(z-\frac{1+\sqrt{10}}{2}\right)^2+11-\frac{(1+\sqrt{10})^2}{2}}\geq\sqrt{\frac{11}{2}-\sqrt{10}}.$$ The equality occurs for $z=\frac{1+\sqrt{10}}{2}$ and $(1,3)||(x,y).$

For $z<0$ by the same way we obtain $$\sqrt{(x-1)^2+(y-3)^2+(z-1)^2}\geq\sqrt{\frac{11}{2}+\sqrt{10}}>\sqrt{\frac{11}{2}-\sqrt{10}},$$ which says that $\sqrt{\frac{11}{2}-\sqrt{10}}=\frac{\sqrt{10}-1}{\sqrt2}=\sqrt5-\frac{1}{\sqrt2}$ is an answer.

Answer 3

Donald Splutterwit already posted the correct answer, this is only additional. Notice that, by using cylindrical coordinates, we can reduce the problem statement to be:

Find the minimum distance between point $A(r,z)=(\sqrt{10},1)$ to the lines $r\pm z=0$.

$min\left(\frac{\left|\sqrt{10}+1\right|}{\sqrt{2}}, \frac{\left|\sqrt{10}-1\right|}{\sqrt{2}}\right)=\sqrt{5}-\frac{1}{\sqrt{2}}$