Ive been given a question: Prove that if $f$ is a (smooth) scalar field and $\overrightarrow {G}$ is an irrotational vector field, then $$(∇f × \overrightarrow {G} )f$$ is solenoidal
Ive got the identities in front of me but i dont know how to apply them to this question.
I know I have to use the identity $$∇·(fG)= (∇f)·G+ f∇·G$$
Any help will be appreciated.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\nabla\cdot\bracks{\pars{\nabla f \times \vec{G}}f} \\[5mm] = & \overbrace{\nabla f\cdot \pars{\nabla f \times \vec{G}}}^{\ds{=\ 0}}\ +\ f\,\nabla\cdot\pars{\nabla f \times \vec{G}} = f\bracks{\vec{G}\cdot\underbrace{\pars{\nabla\times\nabla f}}_{\ds{=\ \vec{0}}} - \nabla f \cdot \underbrace{\pars{\nabla \times G}}_{\ds{=\ \vec{0}}}} = \color{#f00}{0} \end{align}