Can this vector equation be proven?

[R3DH34RT]

delta x (a x b) = (b . delta) a - b (delta . a) + a (delta . b) - (a . delta) b
all terms are in vectors, so delta x means curl.
Can anybody prove that?
Because I have tried to prove it, but it keeps failing.
Please help me to figure this out...
Thanks a lot...

 

[AlphaNumeric2]

Do it in component form :

\big( \nabla \times A \big)_{i} = \epsilon_{ijk} \partial_{j} A_{k} where A_{k} = \epsilon_{klm} a_{l}b_{m}, so

\big( \nabla \times ( a \times b) \big)_{i} = \epsilon_{ijk} \partial_{j} (\epsilon_{klm} a_{l}b_{m})

Remember the identity \epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} and just do some clearing up to get your answer.

 

[R3DH34RT]

But, with that, there will be only 2 terms left, instead of 4...
Is that true?
And what will happen with the other 2 terms?
I'm confused.
Thanks a lot...

 

[AlphaNumeric2]

Use the Leibnitz rule on \partial_{j}(a_{l}b_{m}), that'll double the number of terms.

 

[R3DH34RT]

I've never heard about that. Can you please explain a bit?
Thanks...

 

[CompuChip]

With ordinary functions it's also called the product rule:
\frac{d}{dx}(f g) = g \frac{df}{dx} + f \frac{dg}{dx}.
In this case,
\partial_j(a_l b_m) = a_l \partial_j b_m + b_m \partial_j a_l

 

[AlphaNumeric2]

I've never heard about that. Can you please explain a bit?

If you're being asked to do vector calculus identities, you must have been taught the Liebnitz rule?! It's one of the most important properties of a derivative.

Are you learning just from books? Because if you're just picking up a random book on calculus and geometry, you might be missing the essential requirements by skipping over the prerequesite books.

There's no point doing calculus in n dimensions if you can't do it in 1.

 

[R3DH34RT]

Oh yes, I forgot the name, but I remember that equation...
Thanks... :)