Prove ∇(A x B) = (∇ x A)⋅B - (∇ x B)⋅A where A,B are vectors

[goggles31]

I can prove this relationship by defining A = (A1,A2,A3) and B=(B1,B2,B3) and expanding but I tried another approach and failed.

I read that for any 3 vectors,
a⋅(b x c) = (a x b)⋅c
and thus applying this to the equation, I only get
∇(A x B) = (∇ x A)⋅B
Can anyone explain why this is so?

 

[Charles Link]

That should be the divergence operator $\nabla \cdot (A \times B )$ ,and the divergence operator behaves quite differently from a vector dot product.

 

[fresh_42]

Your mixture of products is wrong. Either use
$\vec{a} \times (\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\cdot\vec{b}-(\vec{a}\cdot\vec{b})\cdot\vec{c}\;$ (Graßmann-identity) or
$\vec{a}\times(\vec{b}\times\vec{c})+\vec{b}\times(\vec{c}\times\vec{a})+\vec{c}\times(\vec{a}\times\vec{b})=0 \;$ (Jacobi-identity).

 

[Igael]

exact, it's not the same operation and then we cannot speak of associativity. ∇ is itself an operator noted as 3 derivators components. See its definition, its components are not numbers

 

[robphy]

Derivative operators satisfy the Leibniz rule.

 

[fresh_42]

Derivative operators satisfy the Leibniz rule.

... which is closely related to the Jacobi-identity or likewise the definition of a derivation.