# 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.

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 $$\vec f\cdot(\nabla\times\vec f) = 0$$

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

$$f \cdot \nabla \times f = 0. \tag 6$$