Cho Δ ABC . Gọi M , N lần lượt là trung điểm của AB , AC . Khi đó: a) 2 vecto CM = vecto CB + vecto CA

a) Đúng

b) Đúng

c) Sai

d) Đúng

 

Ta có: \( - \frac{2}{3}\overrightarrow {CM}  - \frac{4}{3}\overrightarrow {BN}  =  - \frac{1}{3}(\overrightarrow {CA}  + \overrightarrow {CB} ) - \frac{2}{3}(\overrightarrow {BA}  + \overrightarrow {BC} )\)

\( = \frac{1}{3}\overrightarrow {AC}  + \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {BC}  - \frac{2}{3}\overrightarrow {BC}  = \frac{1}{3}\overrightarrow {AC}  + \frac{2}{3}\overrightarrow {AB}  - \frac{1}{3}\overrightarrow {BC} \)\(\)

\( = \frac{1}{3}(\overrightarrow {AC}  - \overrightarrow {BC} ) + \frac{2}{3}\overrightarrow {AB}  = \frac{1}{3}\overrightarrow {AB}  + \frac{2}{3}\overrightarrow {AB}  = \overrightarrow {AB} {\rm{. }}\)

Ta có: \( - \frac{4}{3}\overrightarrow {CM}  - \frac{2}{3}\overrightarrow {BN}  =  - \frac{2}{3}(\overrightarrow {CA}  + \overrightarrow {CB} ) - \frac{1}{3}(\overrightarrow {BA}  + \overrightarrow {BC} ) = \frac{2}{3}\overrightarrow {AC}  + \frac{2}{3}\overrightarrow {BC}  + \frac{1}{3}\overrightarrow {AB}  - \frac{1}{3}\overrightarrow {BC} \) \( = \frac{1}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {BC}  + \frac{2}{3}\overrightarrow {AC}  = \frac{1}{3}(\overrightarrow {AB}  + \overrightarrow {BC} ) + \frac{2}{3}\overrightarrow {AC}  = \frac{1}{3}\overrightarrow {AC}  + \frac{2}{3}\overrightarrow {AC}  = \overrightarrow {AC} \)

Ta có:

\(\begin{array}{l}\overrightarrow {AC}  - \overrightarrow {AB}  =  - \frac{2}{3}\overrightarrow {CM}  + \frac{2}{3}\overrightarrow {BN}  \Leftrightarrow \overrightarrow {BC}  = \frac{2}{3}\overrightarrow {BN}  - \frac{2}{3}\overrightarrow {CM}  \Leftrightarrow 2\overrightarrow {MN}  = \frac{2}{3}\overrightarrow {BN}  - \frac{2}{3}\overrightarrow {CM} .\\ \Leftrightarrow \overrightarrow {MN}  = \frac{1}{3}\overrightarrow {BN}  - \frac{1}{3}\overrightarrow {CM} .\end{array}\)