<選填>已知\( P_1、P_2、Q_1、Q_2、R \)為平面上相異五點，其中\( P_1、P_2、R \)三點不共線，且滿足\( \overrightarrow{P_1R} = 4\overrightarrow{P_1Q_1} \)，\( \overrightarrow{P_2R} = 7\overrightarrow{P_2Q_2} \)，則\( \overrightarrow{Q_1Q_2} =\) __________\(\overrightarrow{P_1Q_1}\) + __________\( \overrightarrow{P_2Q_2} \)。
答案
$\begin{align*}
&\because \overrightarrow{P_1R}=4\overrightarrow{P_1Q_1},\ \overrightarrow{P_2R}=7\overrightarrow{P_2Q_2} \\
&\implies \overrightarrow{P_1Q_1}:\overrightarrow{Q_1R}=1:3,\ \overrightarrow{P_2Q_2}:\overrightarrow{Q_2R}=1:6 \\
&\implies \overrightarrow{Q_1Q_2}=\overrightarrow{Q_1R}+\overrightarrow{RQ_2}=3\overrightarrow{P_1Q_1}-6\overrightarrow{P_2Q_2}
\end{align*}$
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