a) Ta có:
\(\eqalign{& \overrightarrow {AM} = {2 \over 3}\overrightarrow {AB} \Leftrightarrow \left\{ \matrix{\overrightarrow {AM} \uparrow \uparrow \overrightarrow {AB} \hfill \cr
AM = {2 \over 3}AB \hfill \cr} \right. \cr & \overrightarrow {AN} = - {2 \over 3}\overrightarrow {AC} \Leftrightarrow \left\{ \matrix{\overrightarrow {AN} \uparrow \downarrow \overrightarrow {AC} \hfill \cr AN = {2 \over 3}AC \hfill \cr} \right. \cr} \)
Vẽ được 2 điểm $M,N$ như sau:
b) Ta có:
\(\eqalign{& \overrightarrow {AM} = \alpha \overrightarrow {AB} \cr & \overrightarrow {AN} = \beta \overrightarrow {AC} \cr & \Rightarrow \overrightarrow {AN} - \overrightarrow {AM} = \beta \overrightarrow {AC} - \alpha \overrightarrow {AB} \cr & \Rightarrow \overrightarrow {MN} = \beta \overrightarrow {AC} - \alpha \overrightarrow {AB} \cr }\)
Ta cũng có: \(\overrightarrow {BC} = \overrightarrow {AC} - \overrightarrow {AB}\)
Do đó, để \(MN // BC\) thì $\overrightarrow {MN}=k.\overrightarrow {BC}\Leftrightarrow \beta \overrightarrow {AC} - \alpha \overrightarrow {AB}=k.(\overrightarrow {AC} - \overrightarrow {AB})$
$\Leftrightarrow \beta \overrightarrow {AC} - \alpha \overrightarrow {AB} = k. \overrightarrow {AC} - k. \overrightarrow {AB}$
$\Rightarrow \beta = \alpha = k$
Vậy $MN//BC \Leftrightarrow \beta = \alpha$