Cho tam giác $ABC$ có $\widehat{A}=90^\circ$, $BC=\frac{2a}{\sqrt{3}}$, $AC=a$ (với $a>0$). Tính $\overrightarrow{AB} \cdot \left(\overrightarrow{AC}-2\overrightarrow{BC}\right)$ với $a=\sqrt{3}$

Tam giác ABC vuông tại A nên $AB=\sqrt{BC^2-AC^2} = \sqrt{\frac{4a^2}{3}-a^2}=\frac{a}{\sqrt{3}}$. $\overrightarrow{AB} \cdot \left(\overrightarrow{AC}-2\overrightarrow{BC}\right) = \overrightarrow{AB} \cdot \left(\overrightarrow{AC}+\overrightarrow{CB}-\overrightarrow{BC}\right) = \overrightarrow{AB}^2-\overrightarrow{AB} \cdot \overrightarrow{BC} = AB^2-AB \cdot BC \cdot \cos\left(\overrightarrow{AB}, \overrightarrow{BC}\right)$ $=AB^2-AB \cdot BC \cdot \left[-\cos\left(\overrightarrow{BA}, \overrightarrow{BC}\right)\right]=\frac{a^2}{3}+\frac{a}{\sqrt{3}} \cdot \frac{2a}{\sqrt{3}} \cdot \cos \widehat{ABC}$ $=\frac{a^2}{3}+\frac{a}{\sqrt{3}} \cdot \frac{2a}{\sqrt{3}} \cdot \frac{AB}{BC}=\frac{a^2}{3}+\frac{a}{\sqrt{3}} \cdot \frac{2a}{\sqrt{3}} \cdot \frac{1}{2}=\frac{2a^2}{3}=\frac{2 \cdot (\sqrt{3})^2}{3}=2$.