<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\( \overrightarrow{Q_1Q_2} = \frac{\boxed{15-1}}{\boxed{}} \overrightarrow{P_1Q_1} + \frac{\boxed{15-2}}{\boxed{15-3}} \overrightarrow{P_2Q_2} \)。
","首先，根據向量的分解與運算：
由\( \overrightarrow{P_1R} = 4\overrightarrow{P_1Q_1} \)，得\( \overrightarrow{R} = \overrightarrow{P_1} + 4\overrightarrow{P_1Q_1} = \overrightarrow{P_1} + 4(\overrightarrow{Q_1} - \overrightarrow{P_1}) = -3\overrightarrow{P_1} + 4\overrightarrow{Q_1} \)。
由\( \overrightarrow{P_2R} = 7\overrightarrow{P_2Q_2} \)，得\( \overrightarrow{R} = \overrightarrow{P_2} + 7\overrightarrow{P_2Q_2} = \overrightarrow{P_2} + 7(\overrightarrow{Q_2} - \overrightarrow{P_2}) = -6\overrightarrow{P_2} + 7\overrightarrow{Q_2} \)。
因此，\( -3\overrightarrow{P_1} + 4\overrightarrow{Q_1} = -6\overrightarrow{P_2} + 7\overrightarrow{Q_2} \)，整理得：
\( 4\overrightarrow{Q_1} - 7\overrightarrow{Q_2} = 3\overrightarrow{P_1} - 6\overrightarrow{P_2} \)，
\( 4\overrightarrow{Q_1} - 7\overrightarrow{Q_2} = 3(\overrightarrow{P_1} - 2\overrightarrow{P_2}) \)。
另，\( \overrightarrow{Q_1Q_2} = \overrightarrow{Q_2} - \overrightarrow{Q_1} \)，嘗試用\( \overrightarrow{P_1Q_1} \)和\( \overrightarrow{P_2Q_2} \)表示：
由\( \overrightarrow{P_1Q_1} = \overrightarrow{Q_1} - \overrightarrow{P_1} \)，得\( \overrightarrow{Q_1} = \overrightarrow{P_1} + \overrightarrow{P_1Q_1} \)；
由\( \overrightarrow{P_2Q_2} = \overrightarrow{Q_2} - \overrightarrow{P_2} \)，得\( \overrightarrow{Q_2} = \overrightarrow{P_2} + \overrightarrow{P_2Q_2} \)。
將\( \overrightarrow{Q_1}、\overrightarrow{Q_2} \)代入\( \overrightarrow{Q_1Q_2} = \overrightarrow{Q_2} - \overrightarrow{Q_1} \)：
\( \overrightarrow{Q_1Q_2} = (\overrightarrow{P_2} + \overrightarrow{P_2Q_2}) - (\overrightarrow{P_1} + \overrightarrow{P_1Q_1}) = (\overrightarrow{P_2} - \overrightarrow{P_1}) + \overrightarrow{P_2Q_2} - \overrightarrow{P_1Q_1} \)。
再結合\( \overrightarrow{R} \)的兩種表示相等，即\( -3\overrightarrow{P_1} + 4\overrightarrow{Q_1} = -6\overrightarrow{P_2} + 7\overrightarrow{Q_2} \)，代入\( \overrightarrow{Q_1}、\overrightarrow{Q_2} \)的表達式：
\( -3\overrightarrow{P_1} + 4(\overrightarrow{P_1} + \overrightarrow{P_1Q_1}) = -6\overrightarrow{P_2} + 7(\overrightarrow{P_2} + \overrightarrow{P_2Q_2}) \)，
\( \overrightarrow{P_1} + 4\overrightarrow{P_1Q_1} = \overrightarrow{P_2} + 7\overrightarrow{P_2Q_2} \)，
\( \overrightarrow{P_2} - \overrightarrow{P_1} = 4\overrightarrow{P_1Q_1} - 7\overrightarrow{P_2Q_2} \)。
將其代入\( \overrightarrow{Q_1Q_2} \)的表達式：
\( \overrightarrow{Q_1Q_2} = (4\overrightarrow{P_1Q_1} - 7\overrightarrow{P_2Q_2}) + \overrightarrow{P_2Q_2} - \overrightarrow{P_1Q_1} = 3\overrightarrow{P_1Q_1} - 6\overrightarrow{P_2Q_2} = \frac{3}{1}\overrightarrow{P_1Q_1} + \frac{-6}{1}\overrightarrow{P_2Q_2} \)。
因此，\( \overrightarrow{Q_1Q_2} = \frac{\boxed{3}}{\boxed{1}} \overrightarrow{P_1Q_1} + \frac{\boxed{-6}}{\boxed{1}} \overrightarrow{P_2Q_2} \)（若題中15-2、15-3為分子分母結構，也可整理為\( \overrightarrow{Q_1Q_2} = \frac{\boxed{3}}{\boxed{1}} \overrightarrow{P_1Q_1} + \frac{\boxed{-6}}{\boxed{1}} \overrightarrow{P_2Q_2} \)，具體依題目格式調整）。 報錯
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