# Scalar fields whose gradient lies on a plane?

I'm looking to classify the scalar fields $f : \mathbb{R}^3\to\mathbb{R}$ that satisfy $\nabla f\cdot{\bf A} = 0$ for some ${\bf A}\in\mathbb{R}^3$. Since this is a plane passing through the origin with normal vector $\bf A$, I thought I could write $$\nabla f = \lambda_1{\bf p} + \lambda_2{\bf q}$$ where ${\bf p}\cdot{\bf q} = 0$ and ${\bf p}\times{\bf q} = {\bf A}$, but I wasn't able to figure out any other information about $f$ from this. I also considered the idea that the gradient $\nabla f$ is perpendicular to the level surfaces of $f$, and since $\nabla f$ lies on a plane, maybe this says something about the level surfaces of $f$ (and so about $f$ in turn)? I couldn't take this idea any further however either.

Does anyone have any ideas? Have these types of scalar fields been studied before?

• This would mean that the directional derivative of $f$ in the direction $A$ is zero. In other words, $f$ is constant along lines parallel to $A$. – Nick Mar 14 '18 at 2:33
• If ${\bf A}=(a_1,a_2,a_3)$ with $a_3\neq0$, and $g:\Bbb{R}^2\to\Bbb{R}$ is a smooth enough scalar field, then $$f(x,y,z)=g(x-(a_1z/a_3),y-(a_2z/a_3))$$ has this property. By Nick's observation you get all the functions $f$ from this recipe. If $a_3=0$ then you need to use another non-zero component of ${\bf A}$ in the analogous way. – Jyrki Lahtonen Mar 14 '18 at 8:37