Câu 10:

Ta có:

$\overrightarrow{AD}.\overrightarrow{AC} = |\overrightarrow{AD}| \cdot |\overrightarrow{AC}| \cdot cos(\overrightarrow{AD}, \overrightarrow{AC})$

Vì `ABCD` là hình chữ nhật nên $\overrightarrow{AD} \perp \overrightarrow{AB}$

$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$

$\overrightarrow{AD}.\overrightarrow{AC} = ;0>\overrightarrow{AD}.(\overrightarrow{AB} + \overrightarrow{AD})$

$= \overrightarrow{AD}.\overrightarrow{AB} + \overrightarrow{AD}.\overrightarrow{AD}$

Vì $\overrightarrow{AD} \perp \overrightarrow{AB}$ nên $\overrightarrow{AD}.\overrightarrow{AB} = 0$

$\overrightarrow{AD}.\overrightarrow{AC} = 0 + |\overrightarrow{AD}|^2 = (a\sqrt{3})^2 = 3a^2$

`-> bbD`