Gradient of an implicit surface at points not on the surface (clarification)

Question

Given an implicit surface $f(x,y,z)=0$ and it's gradient $\nabla f(x,y,z)$

We know that if $(x_0,y_0,z_0)$ is a point on the surface then $\nabla f(x_0,y_0,z_0)$ is the normal to the tangent plane at $(x_0,y_0,z_0)$.

What happens if $(x_0,y_0,z_0)$ is not on the surface, what does the gradient return?

Edit:

I am concerned about what the output of $\nabla f(x,y,z)$ with respect to the original shape only.

Answer 1

If $f(x_0,y_0,z_0)=c$ the gradient returns the normal to the surface given by the implicit equation $$f(x,y,z)=c \;.$$

Answer 2

There are many different defining functions $f$ with the same level surface $\{f=0\}.$ It's a very nice fact that when $p$ is a point on the surface, the direction of $\nabla f(p)$ is independent of the particular defining function we choose. This is no longer true for points outside the surface: if $\Sigma \subset \mathbb R^3$ is a surface and $p \in \mathbb R^3 \setminus \Sigma,$ then for any vector $v$ you can find a smooth function $f : \mathbb R^3 \to \mathbb R$ with $\Sigma = \{f = 0\}$ and $\nabla f(p) =v.$

Thus if $p=(x_0,y_0,z_0)$ is not on the surface, there is no essential relationship between $\nabla f(x_0,y_0,z_0)$ (away from the surface) and the shape of the surface itself.