Đáp án B

Giải thích:

Có $\overrightarrow{MA} = \overrightarrow{MB} + \overrightarrow{BA} = -\overrightarrow{AB} - \frac{2}{3}\overrightarrow{BC} = - \frac{1}{3}\overrightarrow{AB} - \frac{2}{3}\overrightarrow{AC}$

$\overrightarrow{MC} = \frac{1}{3}\overrightarrow{CB} = \frac{1}{3}(\overrightarrow{AB} - \overrightarrow{AC}) = \frac{1}{3}\overrightarrow{AB} - \frac{1}{3}\overrightarrow{AC}$

Có $\overrightarrow{MA} . \overrightarrow{MC} = (- \frac{1}{3}\overrightarrow{AB} - \frac{2}{3}\overrightarrow{AC}) . (\frac{1}{3}\overrightarrow{AB} - \frac{1}{3}\overrightarrow{AC})$

$ = -\frac{1}{9}\overrightarrow{AB}^{2} + \frac{2}{9}\overrightarrow{AC}^{2} - \frac{1}{9}\overrightarrow{AB} . \overrightarrow{AC}$

$ = -\frac{1}{9}\overrightarrow{AB}^{2} + \frac{2}{9}\overrightarrow{AC}^{2} - \frac{1}{9}\overrightarrow{AB} . \overrightarrow{AC} . cos\widehat{BAC}$

$ = -\frac{1}{9}9a^{2} + \frac{2}{9}9a^{2} - \frac{1}{9}.9a^{2}. cos60^{o}$

$ = -\frac{1}{9}9a^{2} + \frac{2}{9}9a^{2} - \frac{1}{9}.9a^{2}. cos60^{o}$

$ = -\frac{a^{2}}{2}$