Cho tam giác ABC có BC = a, CA = b, AB = c. Hãy tính vecto AB. vecto AC theo a, b, c

Ta có: $\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.c\text{os}\left(\overrightarrow{AB}.\overrightarrow{AC}\right)=AB.AC.\cos BAC=bc.c\text{osBAC}$

Theo định lí cos, ta có:

$$\begin{gathered} \text{cosBAC=}\frac{b^2+c^2-a^2}{2bc} \\ \overrightarrow{AB}.\overrightarrow{AC}=bc.\frac{{\displaystyle b^2+c^2-a^2}}{{\displaystyle 2bc}}=\frac{{\displaystyle b^2+c^2-a^2}}{{\displaystyle 2}} \end{gathered}$$.

Vậy $\overrightarrow{AB}.\overrightarrow{AC}=\frac{b^2+c^2-a^2}{2}.$