在長方形 $ABCD$ 中，邊 $AB$ 與邊 $AD$ 互相垂直，故其向量內積為 $0$： $$\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AD} = 0$$ 我們可以將對角線向量 $\overset{\large\rightharpoonup}{AC}$ 分解為兩個相鄰邊向量的和： $$\overset{\large\rightharpoonup}{AC} = \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD}$$ 已知條件為 $\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AD} = \left\|\overset{\large\rightharpoonup}{AC}\right\|$。我們將其展開： $$\left( \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD} \right) \cdot \overset{\large\rightharpoonup}{AD} = \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AD} + \left\|\overset{\large\rightharpoonup}{AD}\right\|^2 = 0 + \left\|\overset{\large\rightharpoonup}{AD}\right\|^2 = \left\|\overset{\large\rightharpoonup}{AD}\right\|^2$$ 因此可得： $$\left\|\overset{\large\rightharpoonup}{AD}\right\|^2 = \left\|\overset{\large\rightharpoonup}{AC}\right\| \ \ \text{(式一)}$$ 由於 $ABCD$ 為長方形，且 $\angle CAB = 30^\circ$，在直角三角形 $ABC$（$\angle B = 90^\circ$）中，可得對邊 $BC = AC \sin 30^\circ = \dfrac{1}{2} AC$。 因為 $AD = BC$，故： $$\left\|\overset{\large\rightharpoonup}{AD}\right\| = \dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{AC}\right\|$$ 將此關係代入（式一）： $$\left( \dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{AC}\right\| \right)^2 = \left\|\overset{\large\rightharpoonup}{AC}\right\| \implies \dfrac{1}{4} \left\|\overset{\large\rightharpoonup}{AC}\right\|^2 = \left\|\overset{\large\rightharpoonup}{AC}\right\|$$ 因為長方形對角線長度 $\left\|\overset{\large\rightharpoonup}{AC}\right\| > 0$，等號兩邊同除以 $\left\|\overset{\large\rightharpoonup}{AC}\right\|$ 得： $$\left\|\overset{\large\rightharpoonup}{AC}\right\| = 4$$ 進而可求得各邊邊長： - $\left\|\overset{\large\rightharpoonup}{AD}\right\| = \dfrac{1}{2} \times 4 = 2$ - $\left\|\overset{\large\rightharpoonup}{AB}\right\| = \left\|\overset{\large\rightharpoonup}{AC}\right\| \cos 30^\circ = 4 \times \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ 現在，我們計算所求內積 $\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AB}$： $$\overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AB} = \left( \overset{\large\rightharpoonup}{AB} + \overset{\large\rightharpoonup}{AD} \right) \cdot \overset{\large\rightharpoonup}{AB} = \left\|\overset{\large\rightharpoonup}{AB}\right\|^2 + \overset{\large\rightharpoonup}{AD} \cdot \overset{\large\rightharpoonup}{AB} = \left\|\overset{\large\rightharpoonup}{AB}\right\|^2 + 0 = (2\sqrt{3})^2 = 12$$ 答案為 $12$。