Why the vector $(\nabla f(a), -1)$ is perpendicular to the tangent plane at the point $(a, f(a))$

Question

I can't see very clearly why the following scalar product is true:

$\langle (\nabla f(a), -1), v \rangle = 0$

where $ f : \mathbb{R}^2 \to \mathbb{R} $ and $ v \in \mathbb{R}^3 $ is a vector collinear to the tangent plane at the point $(a, f(a))$.

I already know that the gradient is perpendicular to the level curves, but I don't see how to use this information.

I need a conceptual explanation (a proof by computation is well-appreciated too).

Thank you in advance.

Answer 1

Consider the function (on $\Bbb R^3$) defined by $$ G(x, y, z) = z - f(x, y) $$

How is the level surface $G = 0$ related to the graph of $f$?

What does the gradient-perpendicular-to-level-sets theorem tell you about this level surface?