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\[\vec \nabla \times \vec \nabla \times \vec f=\vec \nabla(\vec \nabla . \vec f)-\nabla^2 \vec f\] Can we use vector triple product to prove this identity??Expanding it is long and unecessary :/
are you familiar with the Ricci's symbol
Nope

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you can apply this identity: \[\Large {\mathbf{a}} \times \left( {{\mathbf{b}} \times {\mathbf{c}}} \right) = \left( {{\mathbf{a}} \cdot {\mathbf{c}}} \right){\mathbf{b}} - \left( {{\mathbf{a}} \cdot {\mathbf{b}}} \right){\mathbf{c}}\]
that's what I am asking if it's ok to use that
It looks like a quick and cheat method lol
yes! you have to memorize that identity :)
alright sweet :D
:)
do it!! you're just pattern matching after all and \(\nabla\) is functionally a vector....
yeah :P

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