(b) Vector from of Coulomb’s law states that $\mathbf{F}_{AB}\ $and $\mathbf{F}_{BA}$ are equal and opposite.
Since, ${\widehat{\mathbf{r}}}_{AB\ }and\ {\widehat{\mathbf{r}}}_{BA}$ are unit vectors pointing in opposite direction, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ {\ \widehat{\mathbf{r}}}_{AB\ } = \ - {\widehat{\mathbf{r}}}_{BA}$
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Therefore, equations of force is
Given by
$\mathbf{F}_{AB} = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}({\widehat{\mathbf{r}}}_{AB})$
$\ \ = - \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}{\widehat{\mathbf{r}}}_{BA} = - \mathbf{F}_{BA}$