Cho tam giác ABC có AB = 2, AC = 3, $\widehat{\text{BAC}}\text{ = 60°}$. Gọi M là trung điểm của đoạn thẳng BC. Điểm D thỏa mãn $\overrightarrow{\text{AD}}\text{ = }\frac{\text{7}}{\text{12}}\overrightarrow{\text{AC}}$

a) Tính $\overrightarrow{\text{AB}}\text{ . }\overrightarrow{\text{AC}}$

b) Biểu diễn $\overrightarrow{\text{AM}}\text{, }\overrightarrow{\text{BD}}$ theo $\overrightarrow{\text{AB}}\text{ , }\overrightarrow{\text{AC}}$

c) Chứng minh $AM\perp BD$

a) Ta có $\overrightarrow{\text{AB}}\text{ . }\overrightarrow{\text{AC}}\text{ = }\left|\overrightarrow{\text{AB}}\right|\text{ . }\left|\overrightarrow{\text{AC}}\right|\text{ . cos}\left(\overrightarrow{\text{AB}}\text{, }\overrightarrow{\text{AC}}\right){\text{ = 2 . 3 . cos 60}}^{\text{o}}\text{ = 3.}$

Vậy $\overrightarrow{\text{AB}}\text{ . }\overrightarrow{\text{AC}}=3$

b) Do M là trung điểm của BC nên $\overrightarrow{\text{AM}}\text{ = }\frac{\text{1}}{\text{2}}\overrightarrow{\text{AB}}\text{ + }\frac{\text{1}}{\text{2}}\overrightarrow{\text{AC}}$

Ta có $\overrightarrow{\text{BD}}\text{ = }\overrightarrow{\text{AD}}\text{ - }\overrightarrow{\text{AB}}\text{ = }\frac{\text{7}}{\text{12}}\overrightarrow{\text{AC}}\text{ - }\overrightarrow{\text{AB}}$

c) Ta có: $\overrightarrow{\text{AM}}\text{ . }\overrightarrow{\text{BD}}\text{ = }\left(\frac{\text{1}}{\text{2}}\overrightarrow{\text{AB}}\text{ + }\frac{\text{1}}{\text{2}}\overrightarrow{\text{AC}}\right)\text{ . }\left(\frac{\text{7}}{\text{12}}\overrightarrow{\text{AC}}\text{ - }\overrightarrow{\text{AB}}\right)$

$\text{= }\frac{\text{7}}{\text{24}}\overrightarrow{\text{AB}}\text{ . }\overrightarrow{\text{AC}}\text{ - }\frac{\text{1}}{\text{2}}{\overrightarrow{\text{AB}}}^{\text{2}}\text{ + }\frac{\text{7}}{\text{24}}{\overrightarrow{\text{AC}}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\overrightarrow{\text{AC}}\text{ . }\overrightarrow{\text{AB}}$

$\text{= }\frac{\text{7}}{\text{24}}\text{ . 3 - }\frac{\text{1}}{\text{2}}{\text{ . 2}}^{\text{2}}\text{ + }\frac{\text{7}}{\text{24}}{\text{ . 3}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\text{ . 3 = 0}$

Do đó $AM\perp BD$.