107指考數學甲多選第6題詳解

已知 $|\overset{\large\rightharpoonup}{a}| = |\overset{\large\rightharpoonup}{b}| = |\overset{\large\rightharpoonup}{c}| = 4$，$0 \le \theta \le \pi$。 由 $\overset{\large\rightharpoonup}{a} \times \overset{\large\rightharpoonup}{b} = \overset{\large\rightharpoonup}{c}$ 知 $\overset{\large\rightharpoonup}{c} \perp \overset{\large\rightharpoonup}{a}$ 且 $\overset{\large\rightharpoonup}{c} \perp \overset{\large\rightharpoonup}{b}$。 由長度公式：$|\overset{\large\rightharpoonup}{c}| = |\overset{\large\rightharpoonup}{a}| |\overset{\large\rightharpoonup}{b}| \sin \theta \implies 4 = 4 \times 4 \times \sin \theta \implies \sin \theta = \dfrac{1}{4}$。 因 $\sin \theta = \dfrac{1}{4}$，$\cos \theta = \pm \sqrt{1 - (1/4)^2} = \pm \dfrac{\sqrt{15}}{4} \neq \dfrac{1}{4}$。故 $(1)$ 錯誤。 $(2)$ 平行六面體體積 $V = |(\overset{\large\rightharpoonup}{a} \times \overset{\large\rightharpoonup}{b}) \cdot \overset{\large\rightharpoonup}{c}| = |\overset{\large\rightharpoonup}{c} \cdot \overset{\large\rightharpoonup}{c}| = |\overset{\large\rightharpoonup}{c}|^2 = 4^2 = 16$。故 $(2)$ 正確。 $(3)$ 由定義 $\overset{\large\rightharpoonup}{c} = \overset{\large\rightharpoonup}{a} \times \overset{\large\rightharpoonup}{b} \implies \overset{\large\rightharpoonup}{c} \perp \overset{\large\rightharpoonup}{a}$；由 $\overset{\large\rightharpoonup}{d} = \overset{\large\rightharpoonup}{a} \times \overset{\large\rightharpoonup}{c} \implies \overset{\large\rightharpoonup}{d} \perp \overset{\large\rightharpoonup}{a}$ 且 $\overset{\large\rightharpoonup}{d} \perp \overset{\large\rightharpoonup}{c}$。因此 $\overset{\large\rightharpoonup}{a}, \overset{\large\rightharpoonup}{c}, \overset{\large\rightharpoonup}{d}$ 兩兩互相垂直。故 $(3)$ 正確。 $(4)$ $|\overset{\large\rightharpoonup}{d}| = |\overset{\large\rightharpoonup}{a} \times \overset{\large\rightharpoonup}{c}| = |\overset{\large\rightharpoonup}{a}| |\overset{\large\rightharpoonup}{c}| \sin 90^\circ = 4 \times 4 \times 1 = 16 \neq 4$。故 $(4)$ 錯誤。 $(5)$ 設 $\overset{\large\rightharpoonup}{b}$ 與 $\overset{\large\rightharpoonup}{d}$ 夾角為 $\phi$，則 $\cos \phi = \dfrac{\overset{\large\rightharpoonup}{b} \cdot \overset{\large\rightharpoonup}{d}}{|\overset{\large\rightharpoonup}{b}| |\overset{\large\rightharpoonup}{d}|} = \dfrac{\overset{\large\rightharpoonup}{b} \cdot (\overset{\large\rightharpoonup}{a} \times \overset{\large\rightharpoonup}{c})}{4 \times 16} = \dfrac{(\overset{\large\rightharpoonup}{b} \times \overset{\large\rightharpoonup}{a}) \cdot \overset{\large\rightharpoonup}{c}}{64} = \dfrac{-\overset{\large\rightharpoonup}{c} \cdot \overset{\large\rightharpoonup}{c}}{64} = \dfrac{-16}{64} = -\dfrac{1}{4}$。而 $\cos \theta = \pm \dfrac{\sqrt{15}}{4} \neq -\dfrac{1}{4}$。故 $(5)$ 錯誤。 正確選項為 $(2)(3)$。