In multivariable calculus, a common mnemonic to remember $\text{div}(\text{curl}(\mathbf{F}))=0$ for vector fields $\mathbf{F}$ on $\mathbb{R}^3$ is treating the gradient operator $\nabla=\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}\rangle$ as a vector, writing
$$\text{div}(\text{curl}(\mathbf{F}))=\nabla\cdot\nabla\times \mathbf{F},$$
and using the fact that for any vectors $\textbf{u},\textbf{v}\in\mathbb{R}^3$, there holds $\textbf{u}\cdot \textbf{u}\times\textbf{v}=0.$
Is there a simple classification of vector fields $\mathbf{F}$ on $\mathbb{R}^3$ that satisfy $\mathbf{F}\cdot \nabla\times\mathbf{F}=0$?
A trivial example of such a vector field is any $\mathbf{F}(x,y)=\langle P(x,y), Q(x,y), 0\rangle$. This leads to a less general question: are there vector fields on $\mathbb{R}^3$ not contained in a plane for which the above equation holds? Does this equation have any special physical meaning?
Thank you.
It's helpful to translate this question into the notation of differential forms. If the vector field $\mathbf F$ corresponds to the $1$-form $F_1\,dx + F_2\,dy + F_3\,dz$, then $\text{curl}\,\mathbf F$ corresponds to the $2$-form $d\omega$ and the dot product $\mathbf F\cdot \text{curl}\,\mathbf F$ is the coefficient of the $3$-form $\omega\wedge d\omega$. So we're asking when $\omega\wedge d\omega = 0$.
This is precisely the integrability condition for the two-planes given by $\omega=0$ (i.e., the two-planes orthogonal to $\mathbf F$) to have everywhere integrable submanifolds. That is, $\Bbb R^3$ is foliated by surfaces with normal vector $\mathbf F$. [Of course, I'm assuming $\mathbf F$ is nowhere-zero.] More analytically, you can rescale $\mathbf F$ by a smooth (nowhere-zero) function to make it conservative.