a) Ta có: AO = $\frac{AC}{2}$ = $\frac{\sqrt{AB^{2} + BC^{2}}}{2}$ = $\frac{\sqrt{5}a}{2}$
cos$\widehat{BAC}$ = $\frac{AB}{AC}$ = $\frac{2a}{\sqrt{5}a}$ = $\frac{2\sqrt{5}}{5}$
Ta có: $\vec{AB}$. $\vec{AO}$ = |$\vec{AB}$|. |$\vec{AO}$|. cos($\vec{AB}$, $\vec{AO}$) = AB. AO. cos$\widehat{BAC}$ = 2a. $\frac{\sqrt{5}a}{2}$. $\frac{2\sqrt{5}}{5}$ = 2$a^{2}$
b) Vì $\vec{AB}$ $\perp$ $\vec{AD}$ $\Rightarrow$ $\vec{AB}$. $\vec{AD}$ = 0