Vector cross product identity proof

[notReallyHere]

Homework Statement

\bigtriangledown\times\\(v\times w)= v(\bigtriangledown\cdot w) - w(\bigtriangledown\cdot v)+ (v\cdot\bigtriangledown)w - (w\cdot\bigtriangledown) v


I've tried expanding left side and get
[v1(dw2/dy+dw3/dz)-w1(dv2/dy+dv3/dz)]i +
[v2(dw3/dz+dw1/dx)-w2(dv3/dz+dv1/dx)]j +
[v3(dw1/dx+dw2/dy)-w3(dv2/dy+dv1/dx)]k

is this right way to start it? another way to attack it?

 

[meopemuk]

Homework Statement

\bigtriangledown\times\\(v\times w)= v(\bigtriangledown\cdot w) - w(\bigtriangledown\cdot v)+ (v\cdot\bigtriangledown)w - (w\cdot\bigtriangledown) v


I've tried expanding left side and get
[v1(dw2/dy+dw3/dz)-w1(dv2/dy+dv3/dz)]i +
[v2(dw3/dz+dw1/dx)-w2(dv3/dz+dv1/dx)]j +
[v3(dw1/dx+dw2/dy)-w3(dv2/dy+dv1/dx)]k

is this right way to start it? another way to attack it?

You can try the following: First note that \nabla is a differential operator, so its action on the product is the sum of actions on each factor

\bigtriangledown\times\\(v\times w)= \bigtriangledown\times\\(V\times w) + \bigtriangledown\times\\(v\times W)

(I used capital letters to denote the factor on which \nabla acts in each term. The other factor can be treated as a constant). Then use the vector identity

a \times( b \times c) = b(a \cdot c) - c (a \cdot b)