$Problem:$ If a vector function $V=V(x,y,z)$ is not irrotational, show that if there exists a scalar function $g=g(x,y,z)$ such that $gV$ is irrotational, then $$V\cdot (\nabla \times V )=0$$
Remember, $V$ Irrotational $\iff \nabla\times V=0$
My attempt:
Since $\nabla\times V\not=0$, $\nabla\times V$ is a vector that is orthogonal to $V$, and so $V\cdot \nabla\times V=0$ by definition of the dot product.
I am totally unconvinced this can be the answer as nowhere did I use the fact that there exists a scalar function $g$ in my solution.
Any help is greatly appreciated!
Since $g\vec{V}$ is irrotational, $\nabla\times(g\vec{V}) = 0$. Therefore, $\nabla g \times \vec{V} + g\nabla\times\vec{V} = 0$ or $\vec{V} \times \nabla g = g\nabla\times\vec{V}$. Now take the dot product with $\vec{V}$ so that \begin{equation} \vec{V}\cdot(\vec{V} \times \nabla g) = g\vec{V} \cdot (\nabla\times\vec{V}). \end{equation} The left hand side is zero. Since $g$ is not identically zero, it follows that $\vec{V} \cdot (\nabla\times\vec{V}) = 0$.