TITLE: Divergence of vector potential
QUESTION [0 upvotes]: I was given the vector potential $$\vec A (\vec r) = - \vec a \times \nabla \frac{1}{r}$$ with a constant vector $\vec a$. Now, I found the $\vec B$ field which is I think $- \vec a \frac{2}{r^3}$, but the divergence of $\vec A$ seems to vanish. What I tried was this:
$$\nabla \cdot \vec A = - \nabla \cdot (\vec a \times \frac{\vec r}{r^3}) = \frac{\vec r}{r^3}\cdot (\nabla \times \vec a) - \vec a \cdot(\nabla \times \frac{\vec r}{r^3}) = - \vec a \cdot(\frac{1}{r^3} \nabla \times \vec r + \nabla \frac{1}{r^3} \times \vec r) = \vec a (\frac{3 \vec r}{r^5} \times \vec r) = 0,$$ since the last term contains two parallel vectors.
Where is my mistake? I used $$\nabla \frac{1}{r} = \frac{\vec r}{r^3},$$ can I only use that in certain cases, for example not close to zero? 
The follow-up question asks about what current density is produced by this vector potential. Using $$-\nabla^2 \vec A = \mu_0 \vec j$$ and plugging in the zero result from above, wouldn't I end up with no current density?
Any help would be appreciated.

REPLY [1 votes]: $\nabla^2\vec A = (\nabla \cdot \nabla)\vec A$, not $\nabla(\nabla \cdot \vec A)$. The former is usually written $\nabla^2$, whereas the latter is usually written out explicitly.
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