\(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB\cdot AC\cdot cos60=\dfrac{1}{2}a^2\)
Kẻ vecto AE=vecto BC
=>AE//BC
=>góc EAB=120 độ
\(\overrightarrow{AB}\cdot\overrightarrow{BC}=\overrightarrow{AB}\cdot\overrightarrow{AE}=AB\cdot AE\cdot cos\left(\overrightarrow{AB},\overrightarrow{AE}\right)=-\dfrac{a^2}{2}\)