Perpendicular vector to a surface

Question

In calculus (I will ask it for $\mathbb{R}^2$), when I have a function $z=f(x,y)$ (differentiable), the gradient vector $\nabla f(x_0,y_0)$ will be perpendicular to the level curve $f(x,y)=z_0$, and then, transforming it to $f(x,y)-z=g(x,y,z)=0$ one is told that the gradient $\nabla g(x_0,y_0,z_0)$ will be perpendicular to the surface (now level surface to $g(x,y,z)=0$), but the notion does not coincide with the observation that a vector with $z$ component equal to $-1$ will be always downward (though it won't necessarily mean it will be perpendicular, as the comment suggested). Is that consistent with all possible surfaces?

Answer 1

First of all, don't use the same letter $f$ for two different purposes. Call it $g$ or something else. Second, $\nabla g(x_0,y_0,z_0)=\begin{pmatrix}\frac{\partial f}{\partial x}(x_0,y_0) & \frac{\partial f}{\partial y}(x_0,y_0) & -1\end{pmatrix}$. If this is a normal to a plane, it doesn't mean that plane is parallel to the $x$-$y$ plane. For example, $(0,1,-1)$ is normal to a ($45$ degree) tilted plane (sketch what the vector looks like and see for yourself). In order for the plane to be parallel to the $x$-$y$ plane, we need $\nabla f(x_0,y_0)=0$, so that $\nabla g(x_0,y_0,z_0)=\begin{pmatrix}0&0&-1\end{pmatrix}$.