# Homework Help: Vector triple product causing a contradiction in this proof

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1. Jul 6, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

Prove the following identity

$$\nabla (\vec{F}\cdot \vec{G}) = (\vec{F}\cdot \nabla)\vec{G} + (\vec{G}\cdot \nabla)\vec{F} + \vec{F} \times (\nabla \times \vec{G}) + \vec{G}\times (\nabla \times \vec{F})$$

2. Relevant equations

vector triple product

$$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a}\cdot \vec{c}) - \vec{c}(\vec{a}\cdot \vec{b})$$

3. The attempt at a solution

The first thing I wanted to do was investigate what expanding according to the vector triple product would do to the original statement I am trying to prove. This happens:

$$\nabla (\vec{F}\cdot \vec{G}) = (\vec{F}\cdot \nabla)\vec{G} + (\vec{G}\cdot \nabla)\vec{F} + \nabla (\vec{F}\cdot \vec{G}) - (\vec{F}\cdot \nabla)\vec{G} + \nabla (\vec{F}\cdot \vec{G}) - (\vec{G}\cdot \nabla)\vec{F} = 2\nabla (\vec{F}\cdot \vec{G})$$

What's happening here? Is it not valid to use the vector differential operator in an expansion of the vector triple product? Why not?

2. Jul 6, 2015

### Dr. Courtney

It may be because the vector operator does not commute in the dot product like ordinary vectors.