建立空間直角坐標系，設正立方體同一面上的頂點為 $(0,0,1)$、$(1,0,1)$、$(1,1,1)$、$(0,1,1)$。 設 $A = (0,0,1)$，則其同面上的對頂點為 $B = (1,1,1)$。 正立方體的中心為 $O = \left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)$。 可得： $$\overset{\large\rightharpoonup}{OA} = A - O = \left(-\dfrac{1}{2}, -\dfrac{1}{2}, \dfrac{1}{2}\right)$$ $$\overset{\large\rightharpoonup}{OB} = B - O = \left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)$$ 計算兩向量的內積與長度： $$\overset{\large\rightharpoonup}{OA} \cdot \overset{\large\rightharpoonup}{OB} = \left(-\dfrac{1}{2}\right) \times \dfrac{1}{2} + \left(-\dfrac{1}{2}\right) \times \dfrac{1}{2} + \dfrac{1}{2} \times \dfrac{1}{2} = -\dfrac{1}{4}$$ $$\left|\overset{\large\rightharpoonup}{OA}\right| = \sqrt{\left(-\dfrac{1}{2}\right)^2 + \left(-\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2} = \dfrac{\sqrt{3}}{2}$$ $$\left|\overset{\large\rightharpoonup}{OB}\right| = \dfrac{\sqrt{3}}{2}$$ 因此： $$\cos \angle AOB = \dfrac{\overset{\large\rightharpoonup}{OA} \cdot \overset{\large\rightharpoonup}{OB}}{\left|\overset{\large\rightharpoonup}{OA}\right| \left|\overset{\large\rightharpoonup}{OB}\right|} = \dfrac{-\dfrac{1}{4}}{\dfrac{3}{4}} = -\dfrac{1}{3}$$ 答為 $-\dfrac{1}{3}$。