I just started learning this and I don't understand much so how can I prove this? $$\vec\nabla\times(\vec\nabla\times\vec A )=-\vec\nabla^2\vec A +\vec\nabla (\vec\nabla\cdot\vec A )$$

  • 1
    $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ Feb 7 at 14:57
  • $\begingroup$ What have you tried? Have you attempted a brute force approach by carrying out the derivatives? $\endgroup$
    – Mark Viola
    Feb 7 at 15:05
  • $\begingroup$ Possible duplicate: math.stackexchange.com/a/1108604/583883 $\endgroup$
    – student91
    Feb 7 at 15:06
  • $\begingroup$ You can do it with Levi-Civita symbol $\endgroup$
    – Bemciu
    May 2 at 17:39

1 Answer 1


$\def\n{\nabla}$The vector aspect can be addressed using the triple product (aka bac-cab) rule $a\times(b\times c) = b\,(a\cdot c)- c\,(a\cdot b)$, but the derivative aspect requires keeping the $\n$ operator on the LHS of each expression
$\eqalign{\n\times(\n\times c) &= \n\big(\n\cdot c\big)- \big(\n\cdot \n\big)\,c \\ &= \n\big(\n\cdot c\big)- \n^2c \\ }$


Not the answer you're looking for? Browse other questions tagged .