104指考數學乙選填第9題詳解

在長方形 $ABCD$ 中，$\angle CAB = 30^\circ$，且 $\overline{AB} \perp \overline{AD}$。 分解向量 $\overset{\large\rightharpoonup}{AC} = \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD}$。 在直角三角形 $ABC$ 中，$\tan 30^\circ = \frac{\overline{BC}}{\overline{AB}} = \frac{\overline{AD}}{\overline{AB}}$（因為長方形中 $\overline{BC} = \overline{AD}$）。 設 $\overline{AB} = x$，則 $\overline{AD} = x \tan 30^\circ = \frac{x}{\sqrt{3}}$。 $|\overset{\large\rightharpoonup}{AC}| = \sqrt{x^2 + \frac{x^2}{3}} = \frac{2x}{\sqrt{3}}$。 已知 $\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AD} = |\overset{\large\rightharpoonup}{AC}|$。 $\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AD} = (\overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD}) \cdot \overset{\large\rightharpoonup}{AD} = \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AD} + |\overset{\large\rightharpoonup}{AD}|^2$ 因為 $\overset{\large\rightharpoonup}{AB} \perp \overset{\large\rightharpoonup}{AD}$，所以 $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AD} = 0$。 因此 $\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AD} = |\overset{\large\rightharpoonup}{AD}|^2 = \frac{x^2}{3} = |\overset{\large\rightharpoonup}{AC}| = \frac{2x}{\sqrt{3}}$。 得 $x^2 = 2\sqrt{3}x$，即 $x = 2\sqrt{3}$。 $\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AB} = (\overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD}) \cdot \overset{\large\rightharpoonup}{AB} = |\overset{\large\rightharpoonup}{AB}|^2 + 0 = x^2 = 12$。