Có $cos60^{\circ} = \dfrac{BA}{BC}$
$\to \dfrac{1}{2} = \dfrac{BA}{a} \to BA = \dfrac{a}{2}$
$\overrightarrow{CB}.\overrightarrow{BA} = -\overrightarrow{BC}.\overrightarrow{BA} = \left | BC \right |.\left | BA \right |.cos\left ( \overrightarrow{BC}; \overrightarrow{BA} \right ) = -a.\dfrac{a}{2}.cos60^{\circ} = \dfrac{a^{2}}{4}$