Find the shortest distance from a point to curved surface 
Find the shortest distance from a point $(0,0,0)$ to curved surface $x^2+2y^2-z^2=5.$  


What I have done is,
Let a point in the curved surface be $(a,b,c)$, and $\begin{bmatrix} {a - 0}  \\
{b - 0}\\
{c - 0}\\
\end{bmatrix}$ be vector v.
A vector w perpendicular to curved suface at (a,b,c) would be 
$ \begin{bmatrix} {2a}  \\
{4b}\\
{-2c} \\
\end{bmatrix} $.
Distance is $\frac{10}{2\sqrt{a^2+4b^2+c^2}}$ .
Then $\begin{bmatrix} {a - 0}  \\
{b - 0}\\
{c - 0}\\
\end{bmatrix} \cdot \begin{bmatrix} {2a}  \\
{4b}\\
{-2c} \\
\end{bmatrix} = 2a^2+4b^2-2c^2 = \sqrt{a^2+b^2+c^2}\cdot 2\sqrt{a^2+4b^2+c^2}\cdot \cos{0} =10$.
and $\begin{bmatrix} {a - 0}  \\
{b - 0}\\
{c - 0}\\
\end{bmatrix} \times \begin{bmatrix} {2a}  \\
{4b}\\
{-2c} \\
\end{bmatrix} = \begin{bmatrix} {-2bc-4bc}  \\
{2ac+2ac}\\
{4ab-2ab} \\
\end{bmatrix} = \begin{bmatrix} {-6bc}  \\
{4ac}\\
{2ab} \\
\end{bmatrix}= \sqrt{()} \cdot \sqrt{()} \cdot \sin{0}$ = 0
 $\quad \to$ For this to be true, two of a,b, and c have to be zero.
So the answer is $\frac{10}{2\sqrt{10}}=\frac{\sqrt{10}}{2}$.   
Is this correct? and is there a better way? 

No, this is not correct. The vector $w$ is indeed orthogonal to the surface. Therefore, what you want to know is when $v$ and $w$ are collinear. That is, when is there a number $\lambda$ such that $v=\lambda w$. What this means is that you must solve the system$$\left\{\begin{array}{l}a=\lambda a\\b=2\lambda b\\c=-2\lambda c\\a^2+2b^2-c^2=5.\end{array}\right.$$This system has $4$ solutions: $\pm\left(0,\sqrt{\frac52},0\right)$ and $\pm\left(\sqrt5,0,0\right)$. Therefore, the minimal distance is $\sqrt{\frac52}$.
