Calculus normal lines to a surface parallel to a line For the hyperboloid $x^2 -y^2 +2z^2 = 1$, at what point(s) is the normal line to the surface parallel to the line through the points $(3, -1, 0)$ and $(5,3,6)$?
I tried finding the gradient vector for the hyperboloid and setting it equal to the vector for the line but I got the points where the tangent is parallel, not the normal. Any ideas?
 A: Write
$$
u = x^{\,2}  - y^{\,2}  + 2z^{\,2}  = 1
$$
and take the differential of each of the three terms
$$
du = 2xdx - 2ydy + 4zdz = 0
$$
or
$$
0 = xdx - ydy + 2zdz
$$
Now, $\left( {dx,\;dy,\;dz} \right)$ is a vector on the surface and 
${\bf n} = \left( {x,\; - y,\;2z} \right)$ is therefore a vector normal to the above, for whichever value of $dx,dy,dz$,
thus it is normal to the surface.
But,

it shall be understood that the point $(x,y,z)$ is bound to stay on the surface.

So
$$
\left\{ \matrix{
  x^{\,2}  - y^{\,2}  + 2z^{\,2}  = 1 \hfill \cr 
  {\bf n} = \left( {x,\; - y,\;2z} \right) \hfill \cr}  \right.
$$
which means that you have only two degrees of freedom, i.e. you have only $2$ independent parameters.   
You can choose them to be $x,z$, then
$$
\left\{ \matrix{
  x,z \in R\;:\quad 1 \le x^{\,2}  + 2z^{\,2}  \hfill \cr 
  y =  \pm \sqrt {x^{\,2}  + 2z^{\,2}  - 1}  \hfill \cr 
  {\bf n} = \left( {x,\; \mp \sqrt {x^{\,2}  + 2z^{\,2}  - 1} ,\;2z} \right) \hfill \cr}  \right.
$$
and the rest should be easy for you to do.
A: If $f(x,y,z) = x^2-y^2+2z^2$, then the gradient vector $\nabla f = 2(x,-y,2z)$ is normal to level surfaces of $f$. So you were exactly on the right track. We want points of the level surface $f(x,y,z)=1$ where $\nabla f$ is parallel to (not necessarily equal to) $(2,4,6) = 2(1,2,3)$. 
So I get $(x,-y,2z) = \lambda (1,2,3)$ for some scalar $\lambda$, which means that $$x=-\frac y2 = \frac {2z}3.\tag{$\star$}$$
I find that $x=\pm \sqrt{2/3}$, and $y$ and $z$ follow from ($\star$). Is this what you got?
A: Most likely where you went wrong was in setting the gradient equal to the direction vector of the line (although I don’t really see how this would lead you to points at which the tangent is parallel to the line). This overconstrained the problem: there’s no particular reason to expect equality, particularly since you can choose an arbitrary length for the line’s direction vector. The correct way to express this condition is that the normal is equal to a nonzero scalar multiple of a direction vector of the line. I.e., if we set $\mathbf v=(5,3,6)-(3,-1,0)$, then we want $\nabla f=\lambda\mathbf v$ for some unknown scalar $\lambda\ne0$. 
Since you’re working in $\mathbb R^3$, you can take advantage of the cross product to avoid introducing the additional unknown $\lambda$: two nonzero vectors are parallel iff their cross product vanishes, so the parallelism condition in this problem can be expressed as $\mathbf v\times\nabla f=0$. This generates three equations, only two of which are independent. Together with the equation of the surface, this is enough to find the solution.
