If $u\vec{f} = \nabla v$ where then prove that $\vec{f}.\text{curl F} = 0$ If $u\vec{f} = \nabla v$ where $u$ and $v$ are scalar fields and $\vec{f}$ is a vector  then prove that $\vec{f}.\text{curl F} = 0$
So, $u\vec{f} = \nabla v$ Taking Curl both sides 
 we get:
$\nabla u \times f + u \nabla \times f =0$ Now if I take dot product with $f$ I get:
$f.\left(\nabla u \times f\right) + u f.\left(\nabla \times f\right) = 0$
After This step I am stuck, Can anyone tell me how should I proceed ?
Thank you.
 A: From
$$\vec f\cdot (\nabla u\times\vec f) + u\vec f\cdot(\nabla\times\vec f) = 0$$
We need only prove that 
$$\vec f\cdot (\nabla u\times\vec f) = 0$$
and the proof is complete.
Since the cross product of any two vectors produces a vector perpendicular to the two, the dot product of any vector $\vec v$ with the cross product of $\vec v$ with and arbitrary vector $\vec u$ is $0$.
$$\vec v\cdot (\vec u\times\vec v) = 0$$
Hence we have 
$$\vec f\cdot (\nabla u\times\vec f) = 0$$
and thus
$$u\vec f\cdot(\nabla\times\vec f) = 0$$
Since $u$ is an arbitrary scalar we have
$$\vec f\cdot(\nabla\times\vec f) = 0$$
A: With
$u\vec f = \nabla v, \tag 1$
taking $\nabla \times$ yields
$\nabla u \times \vec f + u \nabla \times \vec f = \nabla \times (u \vec f) = \nabla \times \nabla v = 0; \tag 2$
we $\cdot$ with $\vec f$:
$\vec f\cdot (\nabla u \times \vec f) + u \vec f \cdot \nabla \times \vec f = 0; \tag 3$
recall that
$\vec f \cdot (\nabla u \times \vec f) = 0, \tag 4$
since the cross product of two vectors is always normal to each of them--cf. this wikipedia entry; in light of this we have from (3)
$u\vec f \cdot \nabla \times \vec f = 0, \tag 5$
which, provided $u \ne 0$, forces
$f \cdot \nabla \times f = 0. \tag 6$
