Is the curl of a div. free vector field perpendicular to the field?

[Wuberdall]

Hi PF-members.

My intuition tells me that: Given a divergence free vector field $\mathbf{F}$, then the curl of the field will be perpendicular to field.
But I'm having a hard time proving this to my self.

I'know that : $\nabla\cdot\mathbf{F} = 0 \hspace{3mm} \Rightarrow \hspace{3mm} \exists\mathbf{A}: \mathbf{F} = \nabla\times\mathbf{A}$

Therefore : $\mathbf{F}\cdot(\nabla\times\mathbf{F}) = 0 \hspace{3mm} \Rightarrow \hspace{3mm} [\nabla\times\mathbf{A}]\cdot[\nabla\times(\nabla\times\mathbf{A})] = 0$

But I can't prove that this actually equals zero... Please help!

 

[DeIdeal]

Off the top of my head (as in, there are probably less artificial examples):

Let $\mathbf{F} = (x-z,y-z,-2z)$. As $\nabla \cdot \mathbf{F} = 1+1-2=0$, the field is divergence-free. However, $\nabla \times \mathbf{F} = (1,-1,0)$, so, clearly, $\mathbf{F}\cdot (\nabla \times \mathbf{F})\not{\equiv}0$ and by contradiction, your intuition seems to be wrong.

 

[Wuberdall]

Gracias.