已知 $A, B, C$ 為單位圓上三點，故 $|\overset{\rightharpoonup}{OA}| = |\overset{\rightharpoonup}{OB}| = |\overset{\rightharpoonup}{OC}| = 1$。 由 $2\overset{\rightharpoonup}{OA} + 3\overset{\rightharpoonup}{OB} + 4\overset{\rightharpoonup}{OC} = \overset{\rightharpoonup}{0}$ 可得： (1) $2\overset{\rightharpoonup}{OA} + 3\overset{\rightharpoonup}{OB} = -4\overset{\rightharpoonup}{OC}$，兩邊取長度得 $|2\overset{\rightharpoonup}{OA} + 3\overset{\rightharpoonup}{OB}| = |-4\overset{\rightharpoonup}{OC}| = 4|\overset{\rightharpoonup}{OC}| = 4$。正確。 (2) 將 (1) 式平方：$4|\overset{\rightharpoonup}{OA}|^2 + 9|\overset{\rightharpoonup}{OB}|^2 + 12\overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OB} = 16$ $4 + 9 + 12\overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OB} = 16 \implies 12\overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OB} = 3 \implies \overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OB} = \dfrac{1}{4} > 0$。錯誤。 (3) 同理可求其餘內積： $3\overset{\rightharpoonup}{OB} + 4\overset{\rightharpoonup}{OC} = -2\overset{\rightharpoonup}{OA} \implies 9 + 16 + 24\overset{\rightharpoonup}{OB} \cdot \overset{\rightharpoonup}{OC} = 4 \implies \overset{\rightharpoonup}{OB} \cdot \overset{\rightharpoonup}{OC} = -\dfrac{21}{24} = -\dfrac{7}{8}$ $2\overset{\rightharpoonup}{OA} + 4\overset{\rightharpoonup}{OC} = -3\overset{\rightharpoonup}{OB} \implies 4 + 16 + 16\overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OC} = 9 \implies \overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OC} = -\dfrac{11}{16}$ 內積愈小則夾角愈大。$-7/8 < -11/16 < 1/4$，故 $\angle BOC > \angle AOC > \angle AOB$。$\angle BOC$ 為最大角。錯誤。 (4) $AB^2 = |\overset{\rightharpoonup}{OB} - \overset{\rightharpoonup}{OA}|^2 = 1 + 1 - 2\overset{\rightharpoonup}{OA} \cdot \overset{\rightharpoonup}{OB} = 2 - 2(\dfrac{1}{4}) = \dfrac{3}{2}$。故 $AB = \sqrt{1.5} < 1.5 = \dfrac{3}{2}$。錯誤。 (5) 對 $2\overset{\rightharpoonup}{OA} + 3\overset{\rightharpoonup}{OB} + 4\overset{\rightharpoonup}{OC} = \overset{\rightharpoonup}{0}$ 取與 $\overset{\rightharpoonup}{OA}$ 的外積： $2(\overset{\rightharpoonup}{OA} \times \overset{\rightharpoonup}{OA}) + 3(\overset{\rightharpoonup}{OA} \times \overset{\rightharpoonup}{OB}) + 4(\overset{\rightharpoonup}{OA} \times \overset{\rightharpoonup}{OC}) = \overset{\rightharpoonup}{0}$ $|3(\overset{\rightharpoonup}{OA} \times \overset{\rightharpoonup}{OB})| = |-4(\overset{\rightharpoonup}{OA} \times \overset{\rightharpoonup}{OC})| \implies 3\sin \angle AOB = 4\sin \angle AOC$。正確。 正確選項為 $(1)(5)$。