設球心為原點 $O(0,0,0)$，甲點 $A(1,0,0)$，乙點 $B(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{\sqrt{2}}{2})$，丙點為 $C(x,y,z)$。 甲、乙在球面上的最短路徑為大圓弧，其中點 $C$ 的方向向量 $\overset{\large\rightharpoonup}{OC}$ 必與 $\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB}$ 同向。 計算兩向量之和： $$\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB} = (1,0,0) + (\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{\sqrt{2}}{2}) = (\dfrac{3}{2}, \dfrac{1}{2}, \dfrac{\sqrt{2}}{2})$$ 計算該向量的長度： $$\left|\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB}\right| = \sqrt{(\dfrac{3}{2})^2 + (\dfrac{1}{2})^2 + (\dfrac{\sqrt{2}}{2})^2} = \sqrt{\dfrac{9}{4} + \dfrac{1}{4} + \dfrac{2}{4}} = \sqrt{3}$$ 因為丙地在地球表面上，故其長度 $\left|\overset{\large\rightharpoonup}{OC}\right| = 1$。我們將 $\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB}$ 單位化即得： $$\overset{\large\rightharpoonup}{OC} = \dfrac{1}{\sqrt{3}}(\dfrac{3}{2}, \dfrac{1}{2}, \dfrac{\sqrt{2}}{2}) = (\dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{3}}{6}, \dfrac{\sqrt{6}}{6})$$ 故丙地坐標為 $(\dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{3}}{6}, \dfrac{\sqrt{6}}{6})$。