設正六邊形邊長為 $1$。各選項向量內積計算如下： - $(1)$ $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AB} = |\overset{\large\rightharpoonup}{AB}|^2 = 1$ - $(2)$ $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} = |\overset{\large\rightharpoonup}{AB}| \cdot |\overset{\large\rightharpoonup}{AC}| \cdot \cos 30^\circ = 1 \cdot \sqrt{3} \cdot \dfrac{\sqrt{3}}{2} = 1.5$ - $(3)$ $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AD} = |\overset{\large\rightharpoonup}{AB}| \cdot |\overset{\large\rightharpoonup}{AD}| \cdot \cos 60^\circ = 1 \cdot 2 \cdot \dfrac{1}{2} = 1$ - $(4)$ $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AE} = |\overset{\large\rightharpoonup}{AB}| \cdot |\overset{\large\rightharpoonup}{AE}| \cdot \cos 90^\circ = 0$ - $(5)$ $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AF} = |\overset{\large\rightharpoonup}{AB}| \cdot |\overset{\large\rightharpoonup}{AF}| \cdot \cos 120^\circ = 1 \cdot 1 \cdot \left(-\dfrac{1}{2}\right) = -0.5$ 比較可知內積最大者為 $(2)$。