110指考數學甲選填第11題詳解

1. 為簡化計算，設 $D$ 為原點，令 $\overset{\large\rightharpoonup}{DC}=\overset{\large\rightharpoonup}{d}$、$\overset{\large\rightharpoonup}{DA}=\overset{\large\rightharpoonup}{a}$。 2. 因為 $\overline{AB}\parallel\overline{DC}$ 且 $\overline{AB}=\frac{2}{5}\overline{DC}$，所以 $\overset{\large\rightharpoonup}{AB}=\frac{2}{5}\overset{\large\rightharpoonup}{d}$，因此 $\overset{\large\rightharpoonup}{DB}=\overset{\large\rightharpoonup}{DA}+\overset{\large\rightharpoonup}{AB}=\overset{\large\rightharpoonup}{a}+\frac{2}{5}\overset{\large\rightharpoonup}{d}$。 3. 因為 $\overline{AE}:\overline{EC}=3:2$，所以 $\overset{\large\rightharpoonup}{AE}=\frac{3}{5}\overset{\large\rightharpoonup}{AC}$，即 $\overset{\large\rightharpoonup}{OE}=\overset{\large\rightharpoonup}{OA}+\frac{3}{5}\overset{\large\rightharpoonup}{AC}=\overset{\large\rightharpoonup}{a}+\frac{3}{5}(\overset{\large\rightharpoonup}{d}-\overset{\large\rightharpoonup}{a})=\frac{2}{5}\overset{\large\rightharpoonup}{a}+\frac{3}{5}\overset{\large\rightharpoonup}{d}$。 4. 因為 $\overline{BF}:\overline{FD}=2:3$，所以 $\overset{\large\rightharpoonup}{BF}=\frac{2}{5}\overset{\large\rightharpoonup}{BD}$，即 $\overset{\large\rightharpoonup}{OF}=\overset{\large\rightharpoonup}{OB}+\frac{2}{5}(\overset{\large\rightharpoonup}{OD}-\overset{\large\rightharpoonup}{OB})=\left(\overset{\large\rightharpoonup}{a}+\frac{2}{5}\overset{\large\rightharpoonup}{d}\right)+\frac{2}{5}\left(-\overset{\large\rightharpoonup}{a}-\frac{2}{5}\overset{\large\rightharpoonup}{d}\right)=\frac{3}{5}\overset{\large\rightharpoonup}{a}+\frac{6}{25}\overset{\large\rightharpoonup}{d}$。 5. 因此 $$\overset{\large\rightharpoonup}{FE}=\overset{\large\rightharpoonup}{OE}-\overset{\large\rightharpoonup}{OF}=\left(\frac{2}{5}\overset{\large\rightharpoonup}{a}+\frac{3}{5}\overset{\large\rightharpoonup}{d}\right)-\left(\frac{3}{5}\overset{\large\rightharpoonup}{a}+\frac{6}{25}\overset{\large\rightharpoonup}{d}\right)=-\frac{1}{5}\overset{\large\rightharpoonup}{a}+\frac{9}{25}\overset{\large\rightharpoonup}{d}$$ 6. 注意到 $\overset{\large\rightharpoonup}{AC}=\overset{\large\rightharpoonup}{d}-\overset{\large\rightharpoonup}{a}$、$\overset{\large\rightharpoonup}{AD}=-\overset{\large\rightharpoonup}{a}$。設 $$\overset{\large\rightharpoonup}{FE}=\alpha\overset{\large\rightharpoonup}{AC}+\beta\overset{\large\rightharpoonup}{AD}=\alpha(\overset{\large\rightharpoonup}{d}-\overset{\large\rightharpoonup}{a})+\beta(-\overset{\large\rightharpoonup}{a})=-\alpha\overset{\large\rightharpoonup}{a}-\beta\overset{\large\rightharpoonup}{a}+\alpha\overset{\large\rightharpoonup}{d}$$ 7. 比較係數：$\alpha=\frac{9}{25}$，$-\alpha-\beta=-\frac{1}{5}$，所以 $\beta=\frac{1}{5}-\frac{9}{25}=\frac{5-9}{25}=-\frac{4}{25}$。 故 $\alpha=\frac{9}{25}$、$\beta=-\frac{4}{25}$。