TITLE: Why gradient vector is perpendicular to the plane
QUESTION [29 upvotes]: I know what gradient vector or $\nabla F$ is and I know how to prove that it is orthogonal to the surface (using calculation - not intuitive).
In a particular case, in which we have a three variable function, I want to know why the gradient vector is perpendicular as mentioned. I mean, not in theoretical terms, but intuitive.

REPLY [2 votes]: I am using 3 dimensions in my proof, but it easily extends into higher dimensions.
Proof
First, we need to define the tangent plane to a specific point $(x, y, z)$. Suppose our surface satisfies the equation
$$f(x) + f(y) + f(z) = k$$
for a constant c. Note that
$$f(x+\delta_1) + f(y+\delta_2)+f(z+\delta_3)$$
approaches $c$ for small enough deltas. But,
$$f(x+\delta_1) = f(x) + \delta_1 f'(x)$$
Therefore, the tangent plane is defined as
$$[f(x) + \delta_1 f'(x)] + [g(y) + \delta_2 g'(y)] + [h(z) + \delta_3 h'(z)] = k$$
Or,
$$f'(x)a + g'(y)b + h'(z)c = 0$$
I have replaced the $\delta$s with $(a, b, c)$ to show that the plane extends to infinity (as before, the $\delta$s tended towards $0$).
It is well known that the normal vector to this plane is
$$\langle f'(x),\ g'(y),\ h'(z)\rangle$$
Proof
Suppose $(a, b, c)$ satisfies
$$k_1a + k_2b + k_3c = 0.$$
Note that
$$\langle k_1, k_2, k_3\rangle \cdot \langle a, b, c\rangle = k_1a + k_2b + k_3c = 0$$
Hence they are orthogonal.
$\blacksquare$