空間中有 $A, B, C, D$ 四點。已知 $\overline{AB} = 1$，$\overline{BC} = 2$，$\overline{CD} = 3$，$\angle ABC = \angle BCD = 120^{\circ}$，而 $\overline{AB}$ 與 $\overline{CD}$ 之夾角為 $60^{\circ}$，則 $\overline{AD}$ 之長為 $\text{____}$。

86 指考數學甲第 9 題

由向量加法性質，$\overline{AD}$ 的向量可表示為： $$\overset{\large\rightharpoonup}{AD} = \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{BC} + \overset{\large\rightharpoonup}{CD}$$ 兩邊平方求長度的平方： $$\left| \overset{\large\rightharpoonup}{AD} \right|^2 = \left| \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{BC} + \overset{\large\rightharpoonup}{CD} \right|^2$$ $$= \left| \overset{\large\rightharpoonup}{AB} \right|^2 + \left| \overset{\large\rightharpoonup}{BC} \right|^2 + \left| \overset{\large\rightharpoonup}{CD} \right|^2 + 2 \left( \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{BC} + \overset{\large\rightharpoonup}{BC} \cdot \overset{\large\rightharpoonup}{CD} + \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{CD} \right)$$ 利用向量夾角將內積項展開（注意起點要一致）： 1. $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{BC} = -\overset{\large\rightharpoonup}{BA} \cdot \overset{\large\rightharpoonup}{BC} = - \left| \overset{\large\rightharpoonup}{BA} \right| \left| \overset{\large\rightharpoonup}{BC} \right| \cos \angle ABC = -1 \times 2 \times \cos 120^{\circ}$ 2. $\overset{\large\rightharpoonup}{BC} \cdot \overset{\large\rightharpoonup}{CD} = -\overset{\large\rightharpoonup}{CB} \cdot \overset{\large\rightharpoonup}{CD} = - \left| \overset{\large\rightharpoonup}{CB} \right| \left| \overset{\large\rightharpoonup}{CD} \right| \cos \angle BCD = -2 \times 3 \times \cos 120^{\circ}$ 3. $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{CD} = \left| \overset{\large\rightharpoonup}{AB} \right| \left| \overset{\large\rightharpoonup}{CD} \right| \cos 60^{\circ} = 1 \times 3 \times \cos 60^{\circ}$ 代入長度與夾角值 $\cos 120^{\circ} = -\dfrac{1}{2}$，$\cos 60^{\circ} = \dfrac{1}{2}$： $$\left| \overset{\large\rightharpoonup}{AD} \right|^2 = 1^2 + 2^2 + 3^2 + 2 \left( -1 \times 2 \times \left(-\dfrac{1}{2}\right) - 2 \times 3 \times \left(-\dfrac{1}{2}\right) + 1 \times 3 \times \dfrac{1}{2} \right)$$ $$= 14 + 2 \left( 1 + 3 + \dfrac{3}{2} \right)$$ $$= 14 + 2 \left( \dfrac{11}{2} \right) = 14 + 11 = 25.$$ 解得 $\left| \overset{\large\rightharpoonup}{AD} \right| = 5$，即 $\overline{AD}$ 之長為 $5$。