Cho tam giác ABC có AB = 2, AC = 4, góc A = 60

B

- Hướng dẫn giải

\(\begin{array}{l}
\overrightarrow {AB} .\overrightarrow {AC}  = AB.AC.{\rm{cos }}A = 2.4.{\rm{cos }}{60^ \circ }\\
 = 2.4.\frac{1}{2} = 4.\overrightarrow {A{B^2}}  = 4.\overrightarrow {A{C^2}}  = 16\\
\overrightarrow {BN} .\overrightarrow {CM}  = \left( {\overrightarrow {AN}  - \overrightarrow {AB} } \right)\left( {\overrightarrow {AM}  - \overrightarrow {AC} } \right) = \left( {\frac{1}{2}\overrightarrow {AC}  - \overrightarrow {AB} } \right)\left( {\frac{1}{2}\overrightarrow {AB}  - \overrightarrow {AC} } \right)\\
 = \frac{1}{4}\overrightarrow {AC} .\overrightarrow {AB}  - \frac{1}{2}\overrightarrow {A{B^2}}  - \frac{1}{2}\overrightarrow {A{C^2}}  + \overrightarrow {AC} .\overrightarrow {AB}  = \frac{5}{4}\overrightarrow {AC} .\overrightarrow {AB}  - \frac{1}{2}{\overrightarrow {AB} ^2} - \frac{1}{2}\overrightarrow {A{C^2}} \\
 = \frac{5}{4}.4 - \frac{1}{2}.4 - \frac{1}{2}.16 =  - 5
\end{array}\)