Multivariable calculus del notation question

Question

I've been asked to prove that $\frac{1}{2}\nabla(F\cdot F)=F\times(\nabla \times F)+(F\cdot\nabla)F$ where $F$ is the vector field $\langle P(x,y,z),Q(x,y,z),R(x,y,z) \rangle$ and $\nabla=\langle \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\rangle$, but I keep disproving it! Please can someone help me with this. (here are my workings so far):

Answer 1

Hint:

It is a consequence of the general rue: $$ \nabla(\vec u \cdot \vec v)=\vec u \times (\nabla \times \vec v)+\vec v \times (\nabla \times \vec u)+(\vec u \cdot \nabla)\vec v + (\vec v \cdot \nabla)\vec u $$ for $\vec u= \vec v$

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You have a mistake in the calculus of $(\vec F \cdot \nabla)\vec F$.

Note that $ \vec F \cdot \nabla$ is a ''scalar'' operator: $$ \vec F \cdot \nabla=P\frac{\partial}{\partial x}+Q\frac{\partial}{\partial y}+R\frac{\partial}{\partial z} $$ So, when you multiply it by $\vec F$ you have to multiply all the components by this scalar:

$$ (\vec F \cdot \nabla)\vec F= \left(PP_x+QP_y+RP_z\,,\,PQ_x+Qq_y+RQ_z\,,\,PR_x+QR_y+RR_z \right) $$

Correct this and you can find the result.