我們以點 $A$ 為原點 $(0,0)$，並設 $\overset{\large\rightharpoonup}{AB} = \overset{\large\rightharpoonup}{b}$、$\overset{\large\rightharpoonup}{AD} = \overset{\large\rightharpoonup}{d}$。 由向量加法可得： $$\overset{\large\rightharpoonup}{AC} = \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{BC} = \overset{\large\rightharpoonup}{b} + (\overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD}) = 2\overset{\large\rightharpoonup}{b} + \overset{\large\rightharpoonup}{d}$$ 且 $\overset{\large\rightharpoonup}{BD} = \overset{\large\rightharpoonup}{d} - \overset{\large\rightharpoonup}{b}$。 已知向量 $\overset{\large\rightharpoonup}{AC}$ 與 $\overset{\large\rightharpoonup}{BD}$ 垂直且等長： • 垂直條件（內積為 $0$）： $$\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{BD} = 0 \Rightarrow (2\overset{\large\rightharpoonup}{b} + \overset{\large\rightharpoonup}{d}) \cdot (\overset{\large\rightharpoonup}{d} - \overset{\large\rightharpoonup}{b}) = 0 \Rightarrow |\overset{\large\rightharpoonup}{d}|^2 - 2|\overset{\large\rightharpoonup}{b}|^2 + \overset{\large\rightharpoonup}{b}\cdot\overset{\large\rightharpoonup}{d} = 0 \ ext{ } \text{--- (1)}$$ • 等長條件（長度平方相等）： $$|\overset{\large\rightharpoonup}{AC}|^2 = |\overset{\large\rightharpoonup}{BD}|^2 \Rightarrow |2\overset{\large\rightharpoonup}{b} + \overset{\large\rightharpoonup}{d}|^2 = |\overset{\large\rightharpoonup}{d} - \overset{\large\rightharpoonup}{b}|^2 \Rightarrow 3|\overset{\large\rightharpoonup}{b}|^2 + 6\overset{\large\rightharpoonup}{b}\cdot\overset{\large\rightharpoonup}{d} = 0 \Rightarrow \overset{\large\rightharpoonup}{b}\cdot\overset{\large\rightharpoonup}{d} = -\frac{1}{2}|\overset{\large\rightharpoonup}{b}|^2 \ ext{ } \text{--- (2)}$$ 將 (2) 代入 (1) 可求得： $$|\overset{\large\rightharpoonup}{d}|^2 - 2|\overset{\large\rightharpoonup}{b}|^2 - \frac{1}{2}|\overset{\large\rightharpoonup}{b}|^2 = 0 \Rightarrow |\overset{\large\rightharpoonup}{d}|^2 = \frac{5}{2}|\overset{\large\rightharpoonup}{b}|^2$$ 設 $\angle BAD = \theta$。由向量夾角公式： $$\cos\theta = \frac{\overset{\large\rightharpoonup}{b}\cdot\overset{\large\rightharpoonup}{d}}{|\overset{\large\rightharpoonup}{b}||\overset{\large\rightharpoonup}{d}|} = \frac{- \frac{1}{2}|\overset{\large\rightharpoonup}{b}|^2}{|\overset{\large\rightharpoonup}{b}| \sqrt{\frac{5}{2}}|\overset{\large\rightharpoonup}{b}|} = -\frac{1}{\sqrt{10}}$$ 因為 $\cos\theta < 0$，可知 $\theta$ 為第二象限角，則 $\sin\theta = \sqrt{1 - \cos^2\theta} = \frac{3}{\sqrt{10}}$。 故可求得： $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/\sqrt{10}}{-1/\sqrt{10}} = -3$$ 答：-3。