Del operator crossed with a scalar times a vector proof

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The discussion centers on proving the identity involving the "Del" operator crossed with a scalar times a vector. The identity to be proven is ∇×(φV) = (φ∇)×V - V×(∇φ). Participants express confusion about the initial steps, particularly with the Levi-Civita symbols and the decomposition of the cross product. Clarifications suggest using the product rule for differentiation and reconstructing vectors from their components after applying the rule. The conversation emphasizes the importance of breaking down the proof into manageable steps to achieve clarity.
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"Del" operator crossed with a scalar times a vector proof

Homework Statement


Prove the following identity (we use the summation convention notation)

\bigtriangledown\times(\phi\vec{V})=(\phi \bigtriangledown)\times\vec{V}-\vec{V}\times(\bigtriangledown)\phi

Homework Equations



equation for del, the gradient, curl..

The Attempt at a Solution



im kind of confused on the first step...I broke it down into the following; however, levi civita symbols aren't my cup of tea and I get pretty confused about it...anyway here's what I did:

\bigtriangledown\times(\phi\vec{V})=(\epsilon_{ijk})\partial_i\vec{V}\phi\hat{x}_k

I don't know if this first step is right or if I decomposed the cross product right ?
 
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You should have the j:th component of V in your expression,
\nabla \times (\phi \vec{V}) = \epsilon_{ijk} \partial_i (\phi V_j) \hat{x}_k
 


The form you've written is completely equal to the one seen in math/physics books

\nabla\times \left(\phi\vec{V}\right) = \nabla\phi\times\vec{V} + \phi\nabla\times\vec{V}
 


Thanks for the replies, I'm just not sure what to do after what clamtrox said to do, the whole proofing business if pretty new to me :/
 


You need two more steps. Apply the product rule for differentiation and then once you obtain a sum of two terms, reconstruct vectors from their components.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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