Vector calculus identities

Question

Let $f$ be scalar potential for the vector field $\underline u $ (i.e $\underline u = -\underline \nabla f$). Prove that the vector field $$ \underline r \wedge \underline u $$ has magnetic potential and find it.

Now, I have the solution to the exercise (which I almost completly don't understand) and its states the following:

$$ \underline \nabla \cdot (\underline r \wedge \underline u)=\underbrace {(\underline \nabla \wedge \underline r )}_\underline 0\cdot \underline u -\underline r \cdot \underbrace {(\underline \nabla \wedge \underline u )}_{\underline \nabla \wedge (-\underline \nabla f)=\underline 0}=0$$ and that is why there exit a magnetic potential. Which means, there exist $\underline A $ such that $\underline \nabla \wedge \underline A =\underline r \wedge \underline u $.

Now: $$\underline \nabla \wedge (f \underline r)=(\underline \nabla f)\wedge \underline r +\underbrace {f (\underline \nabla \wedge \underline r)}_{\underline 0}=-\underline u \wedge \underline r =\underline r \wedge \underline u $$

So that $f\underline r$ is answer.

I don't understand why the following are true:

• ${(\underline \nabla \wedge \underline r )\cdot \underline u}=0$
• $\underline r \cdot (\underline \nabla \wedge \underline u ) =-\underline \nabla \wedge (-\underline \nabla f)=\underline 0$

and why

$f (\underline \nabla \wedge \underline r)=\underline 0$

any help?

Answer 1

Hints:

1) What is ${(\underline \nabla \wedge \underline r )}$? (it's $\underline 0$)

2) You've misread/miswritten something. $\underline \nabla \wedge \underline u=\underline \nabla \wedge (- \underline \nabla f)=\underline 0$

3) Again, what's ${(\underline \nabla \wedge \underline r )}$?