Let Φ be a scalar function. Then ∇Φ is

The correct option is: always perpendicular to surface of constant Φ.

To understand why this is the case, let us analyze the geometric interpretation of the gradient of a scalar function.
Let $\Phi (x,y,z)$ be a scalar field. A surface of constant $\Phi$ (also known as a level surface or an equipotential surface) is defined by the equation:

$\Phi (x,y,z)=C$

where $C$ is a constant.

Now, let us consider any point on this surface and a small displacement vector $d\mathbf{r}$ that lies entirely along the surface of constant $\Phi$:

$d\mathbf{r}=dx\hat{\mathbf{i}}+dy\hat{\mathbf{j}}+dz\hat{\mathbf{k}}$

Since we are moving along a surface where $\Phi$ is constant, the change in the value of the scalar function, $d\Phi$, must be zero:

$d\Phi =0$

Using the definition of the total differential, the change $d\Phi$ is given by:

$d\Phi =\frac{\partial \Phi }{\partial x}dx+\frac{\partial \Phi }{\partial y}dy+\frac{\partial \Phi }{\partial z}dz$

This total differential can be written mathematically as the dot product of the gradient vector $\nabla \Phi$ and the displacement vector $d\mathbf{r}$:

$d\Phi =\nabla \Phi \cdot d\mathbf{r}$

Since $d\Phi =0$ for any displacement along the surface, we get:

$\nabla \Phi \cdot d\mathbf{r}=0$

By vector algebra, if the dot product of two non-zero vectors is zero, the vectors are perpendicular to each other. Because the displacement vector $d\mathbf{r}$ is tangent to the surface of constant $\Phi$, the gradient vector $\nabla \Phi$ must be orthogonal (perpendicular) to the tangent plane of the surface.
Therefore, the gradient $\nabla \Phi$ is always perpendicular to the surface of constant $\Phi$ at any point.