tangent plane to a level set I have the function $$f(x,y,z) = x^2 +y^2 -z^2$$ and I am asked to determine for each level set if I have a point that the tangent plane to the level set is parallel  to the plane $ z=0$ and another point that is perpendicular to $z=0$
I am not sure how I should handle this qeustion, what does $z=0$ meaning? I think that I need to find the normal vector of the plane $z=0$ and than check if it can be parallel or perpendicular to the gradient of $f(x,y,z)$ . is what I said is true? if yes how I should do it? and if no, I would like to get an explenation.
thank you kindly.
 A: z=0 means x-y plane so its normal vector is (0,0,1), and what you said is true.Just find that function's gradiant and check whether its perpendicular or parallel to (0,0,1).
A: I think you want to determine whether or not there exist points on the level set of $f$ such that their tangent planes are parallel (resp. perpendicular) to the plane determined by $z=0$.
Your idea is correct! The unit normal to the $z=0$ plane is $n=(0,0,1)$. So, you want to see if you can find solutions to the equations 
$$ \nabla f\cdot n=0$$
$$ \lVert \nabla f\cdot n\rVert=\lVert \nabla f\rVert.$$
The first equation describes exactly $\nabla f\perp n$, while the second equation is equivalent to $\nabla f$ and $n$ pointing in the same direction by $\lVert \nabla f\rVert =\lVert \nabla f\rVert \cdot \lVert n\rVert\cos \theta$, where we note $\lVert n\rVert=1$ and $\cos\theta$ is the angle between the two vectors. 
Another tip: in this case, you should try to draw the level surface of $f(x,y,z)$. This will give you a guess as to what the answer should be in this case. 
