a) Do P nằm trên tia đối BA nên B nằm giữa AP.
Mặt khác, lại có
$\dfrac{AB}{AP} = \dfrac{2BP}{AB + BP} = \dfrac{2BP}{2BP + BP} = \dfrac{2}{3}$
Vậy ta có
$\vec{AB} = \dfrac{AB}{AP} \vec{AP} = \dfrac{2}{3} \vec{AP}$.
Ta có
$\dfrac{AC}{AN} = \dfrac{AN + NC}{AN} = \dfrac{3NC + NC}{3NC} = \dfrac{4}{3}$
Vậy ta có
$\vec{AC} = \dfrac{AC}{AN} \vec{AN} = \dfrac{4}{3} \vec{AN}$
b) Ta có
$\vec{AM} = \vec{AB} + \vec{BM}$
$= \dfrac{2}{3} \vec{AP} + \dfrac{1}{2} \vec{BC}$
$= \dfrac{2}{3} \vec{AP} + \dfrac{1}{2} (\vec{AC} - \vec{AB})$
$= \dfrac{2}{3} \vec{AP} + \dfrac{1}{2} (\dfrac{4}{3} \vec{AN} - \dfrac{2}{3} \vec{AP})$
$= \dfrac{2}{3} \vec{AP} + \dfrac{2}{3} \vec{AN} - \dfrac{1}{3} \vec{AP}$
$= \dfrac{1}{3} \vec{AP} + \dfrac{2}{3} \vec{AN}$