He states after recalling, $\textbf{B}=\frac{\mu_{0}e\textbf{v}\times\textbf{r}}{4\pi r^{3}}$, that $\frac{\textbf{r}}{r^{3}} = - grad(\frac{1}{r})$. First of all I am unsure how we comes to this conclusion. When he proceeds to find the divergence he discards the scalar quantities and writes it as:
div$(\textbf{v} \times \frac{\textbf{r}}{r^{3}}) = \textbf{v}\times \textrm{curl} \;(\textrm{grad}\frac{1}{r}) = 0$
I understand why it is equal to $0$ due to vector identities but I am unsure how he developed this equation please may someone help clear this up.
Observe that if $\;r=(x,y)\;$ , then $\;\left\|r\right\|=\sqrt{x^2+y^2}\;$ , so
$$\nabla\left(-\frac1{\left\|r\right\|}\right)=\left(\frac x{(x^2+y^2)^{3/2}}\,,\,\,\frac y{(x^2+y^2)^{3/2}}\right)=\frac{(x,y)}{\left(\sqrt{x^2+y^2}\right)^3}=\frac r{\left\|r\right\|^3}$$
Most probably (haven't read that book) he uses what many physicists use: the notation r for a vector and then r for its norm...