在 $\triangle \mathrm{ABC}$ 中, $\mathrm{AB}=3, \mathrm{AC}=2, \mathrm{BC}=\sqrt{10}$, 则 $\overrightarrow{B A} \cdot \overrightarrow{A C}$ 等于 $(\quad)$
解析 由余弦定理得

$\cos A=\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C}=\frac{9+4-10}{12}=\frac{1}{4}$.
$\therefore \overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cdot \cos A=3 \times 2 \times \frac{1}{4}=\frac{3}{2}$.

$\therefore \overrightarrow{B A} \cdot \overrightarrow{A C}=-\overrightarrow{A B} \cdot \overrightarrow{A C}=-\frac{3}{2}$.