Cho  $\Delta ABC, M  và  N$  xác định bởi:
     $3 \overrightarrow{MA}+4 \overrightarrow{MB}=\overrightarrow{0}; \overrightarrow{NB}-3 \overrightarrow{NC}=\overrightarrow{0} $. Trọng tâm của  $\Delta ABC$  là  $G$.
1. Chứng mỉnh rằng  $M,N,G$  thẳng hàng.
2. Tính  $\overrightarrow{AC} $  theo  $\overrightarrow{AG} $  và  $\overrightarrow{AN} $, và $AC$  cắt  $GN$  tại $P$. Tính  $\frac{PA}{PC}. $


$1.$ Ta có:  
$3 \overrightarrow{MA}+4 \overrightarrow{MB}=\overrightarrow{0}\Rightarrow 3 \overrightarrow{GA}+4 \overrightarrow{GB}= 7 \overrightarrow{GM}   $
$\overrightarrow{NB}-3 \overrightarrow{NC}=\overrightarrow{0}\Rightarrow \overrightarrow{GB}-3 \overrightarrow{GC}=-2 \overrightarrow{GN}   $
$\Rightarrow    \overrightarrow{GB}+3(\overrightarrow{GA}+\overrightarrow{GB}  )=-2 \overrightarrow{GN}  $
$\Rightarrow    3\overrightarrow{GA} +4 \overrightarrow{GB}=-2 \overrightarrow{GN}   $  ( vì $G$ là trọng tâm của $\triangle ABC$)
Vậy:  $7 \overrightarrow{GM}=-2 \overrightarrow{GN}  $. Suy ra  $G, N, M$  thẳng hàng.

$2.$ Gọi  $E$ là trung điểm của  $BC$, ta có:
 $EC=EB=CN                 \Rightarrow     2 \overrightarrow{AC}=\overrightarrow{AE}+\overrightarrow{AN}   $
$\Rightarrow 2 \overrightarrow{AC}=\frac{3}{2} \overrightarrow{AG}+\overrightarrow{AN}       \Rightarrow     \overrightarrow{AC}=\frac{3}{4}\overrightarrow{AG}+\frac{1}{2}\overrightarrow{AN}        (*)$
Gọi  $P'$  thuộc đường thẳng  $GN$  và thỏa mãn:  $\frac{3}{4}\overrightarrow{P'G}+\frac{1}{2}\overrightarrow{P'N}=\overrightarrow{0}     $
Từ  $(*)$  ta có:  $\overrightarrow{AC}=(\frac{3}{4}+\frac{1}{2}  )\overrightarrow{AP'}     \Rightarrow      A,C,P'$  thẳng hàng,
Vậy:   $P'= AC \cap  GN      \Rightarrow     P \equiv P'    \Rightarrow     \overrightarrow{AC}=\frac{5}{4} \overrightarrow{AP}   $
    $\Rightarrow     \overrightarrow{AP}+\overrightarrow{PC}=\frac{5}{4} \overrightarrow{AP}               \Rightarrow     \overrightarrow{PC}=-\frac{1}{4}\overrightarrow{PA}     \Rightarrow      \frac{PA}{PC}=4.  $


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