a: Xét ΔBAC vuông tại A có cos B=AB/BC
=>a/BC=1/2
=>BC=2a
=>AC=a căn 3
\(\overrightarrow{AC}\cdot\overrightarrow{CB}=-\overrightarrow{CA}\cdot\overrightarrow{CB}=-CA\cdot CB\cdot cosC=-a\sqrt{3}\cdot2a\cdot\dfrac{\sqrt{3}}{2}\)
\(=-3a^2\)
b: \(\overrightarrow{AB}\cdot\overrightarrow{IC}=\overrightarrow{AB}\cdot\overrightarrow{AC}-\overrightarrow{AB}\cdot\overrightarrow{AI}\)
\(=0-AB\cdot AI\cdot cos60=-a\cdot a\cdot\dfrac{1}{2}=-\dfrac{1}{2}a^2\)
\(\overrightarrow{AB}\cdot\overrightarrow{IM}=\overrightarrow{AB}\cdot\overrightarrow{AM}-\overrightarrow{AB}\cdot\overrightarrow{AI}\)
\(=0-AB\cdot AI\cdot cosBAI=-a\cdot a\cdot\dfrac{1}{2}=-\dfrac{1}{2}a^2\)