Suppose $\textbf{A}$ is radial. Prove that $\operatorname{rot}\textbf{A}$ is a vector orthogonal to $\textbf{A}$. First show that $\varepsilon_{ijk}k_jk_k=0$.
I have no problem showing the second part but I don't know how to use it in the proof.
I want to prove the orthogonality with scalar product so I should obtain 0 from the equation. So far I have:
$$\boldsymbol{\nabla}\times\textbf{A}\ \cdot \ \textbf{A} = \hat{e_i}a_i(\varepsilon_{jkl}\hat{e_j}\frac{\partial}{\partial x_k}a_l)=\delta_{ij}\varepsilon_{jkl}\frac{\partial}{\partial x_k}a_la_i$$
I don't know what to do after this so that I use the shown property.
Best Answer
Write $A_i=fx_i$ so$$A_j\varepsilon_{jkl}\partial_kA_l=fx_j\varepsilon_{jkl}\partial_k(fx_l)=f\underbrace{x_jx_l\varepsilon_{jkl}}_{0}\partial_kf+f^2x_j\underbrace{\varepsilon_{jkl}\partial_kx_l}_{\varepsilon_{jkl}\delta_{kl}=0}.$$