106_07A_q05
106 指考數學甲 第 5 題
📅 106 年 📝 指考數學甲 第 5 題 題型:多選 課綱:99課綱
設 $\overset{\large\rightharpoonup}{u}$ 與 $\overset{\large\rightharpoonup}{v}$ 為兩非零向量,夾角為 $120^\circ$。若 $\overset{\large\rightharpoonup}{u}$ 與 $\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}$ 垂直,試選出正確的選項。
  1. $\overset{\large\rightharpoonup}{u}$ 的長度是 $\overset{\large\rightharpoonup}{v}$ 的長度的 $2$ 倍
  2. $\overset{\large\rightharpoonup}{v}$ 與 $\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}$ 的夾角為 $30^\circ$
  3. $\overset{\large\rightharpoonup}{u}$ 與 $\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}$ 的夾角為銳角
  4. $\overset{\large\rightharpoonup}{v}$ 與 $\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}$ 的夾角為銳角
  5. $\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}$ 的長度大於 $\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}$ 的長度
平面向量平面向量
解題手法設未知數〔AI 推測〕
答案

$(2)(3)$

詳解
設兩向量 $\overset{\large\rightharpoonup}{u}, \overset{\large\rightharpoonup}{v}$ 的夾角為 $120^\circ$。則內積 $\overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} = \left\|\overset{\large\rightharpoonup}{u}\right\| \left\|\overset{\large\rightharpoonup}{v}\right\| \cos 120^\circ = -\dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{u}\right\| \left\|\overset{\large\rightharpoonup}{v}\right\|$。 由題意 $\overset{\large\rightharpoonup}{u} \perp (\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v})$,得: $$\overset{\large\rightharpoonup}{u} \cdot (\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}) = \left\|\overset{\large\rightharpoonup}{u}\right\|^2 + \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} = 0$$ $$\left\|\overset{\large\rightharpoonup}{u}\right\|^2 - \dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{u}\right\| \left\|\overset{\large\rightharpoonup}{v}\right\| = 0$$ 因為 $\overset{\large\rightharpoonup}{u}$ 為非零向量,同除以 $\left\|\overset{\large\rightharpoonup}{u}\right\|$ 得: $$\left\|\overset{\large\rightharpoonup}{u}\right\| = \dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{v}\right\| \implies \left\|\overset{\large\rightharpoonup}{v}\right\| = 2\left\|\overset{\large\rightharpoonup}{u}\right\|$$ (1) 錯誤。$\overset{\large\rightharpoonup}{v}$ 的長度是 $\overset{\large\rightharpoonup}{u}$ 長度的 $2$ 倍。 (2) 正確。$\overset{\large\rightharpoonup}{v} \cdot (\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}) = \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = -\dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{u}\right\| \left\|\overset{\large\rightharpoonup}{v}\right\| + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = -\dfrac{1}{4} \left\|\overset{\large\rightharpoonup}{v}\right\|^2 + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = \dfrac{3}{4} \left\|\overset{\large\rightharpoonup}{v}\right\|^2$。 又 $\left\|\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}\right\|^2 = \left\|\overset{\large\rightharpoonup}{u}\right\|^2 + 2 \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = \dfrac{1}{4} \left\|\overset{\large\rightharpoonup}{v}\right\|^2 - \dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{v}\right\|^2 + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = \dfrac{3}{4} \left\|\overset{\large\rightharpoonup}{v}\right\|^2 \implies \left\|\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}\right\| = \dfrac{\sqrt{3}}{2} \left\|\overset{\large\rightharpoonup}{v}\right\|$。 設夾角為 $\theta$,則 $\cos \theta = \dfrac{\overset{\large\rightharpoonup}{v} \cdot (\overset{\large\rightharpoonup}{u}+\overset{\large\rightharpoonup}{v})}{\left\|\overset{\large\rightharpoonup}{v}\right\| \left\|\overset{\large\rightharpoonup}{u}+\overset{\large\rightharpoonup}{v}\right\|} = \dfrac{\frac{3}{4} \left\|\overset{\large\rightharpoonup}{v}\right\|^2}{\left\|\overset{\large\rightharpoonup}{v}\right\| \cdot \frac{\sqrt{3}}{2} \left\|\overset{\large\rightharpoonup}{v}\right\|} = \dfrac{\sqrt{3}}{2} \implies \theta = 30^\circ$。 (3) 正確。$\overset{\large\rightharpoonup}{u} \cdot (\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}) = \left\|\overset{\large\rightharpoonup}{u}\right\|^2 - \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} = \left\|\overset{\large\rightharpoonup}{u}\right\|^2 + \dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{u}\right\| \left\|\overset{\large\rightharpoonup}{v}\right\| > 0$,故夾角為銳角。 (4) 錯誤。$\overset{\large\rightharpoonup}{v} \cdot (\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}) = \overset{\large\rightharpoonup}{u} \cdot \overset{\large\rightharpoonup}{v} - \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = -\dfrac{1}{2} \left\|\overset{\large\rightharpoonup}{u}\right\| \left\|\overset{\large\rightharpoonup}{v}\right\| - \left\|\overset{\large\rightharpoonup}{v}\right\|^2 < 0$,故夾角為鈍角。 (5) 錯誤。$\left\|\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}\right\|^2 = \dfrac{3}{4} \left\|\overset{\large\rightharpoonup}{v}\right\|^2$,而 $\left\|\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}\right\|^2 = \left\|\overset{\large\rightharpoonup}{u}\right\|^2 - 2\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v} + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = \dfrac{1}{4}\left\|\overset{\large\rightharpoonup}{v}\right\|^2 + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 + \left\|\overset{\large\rightharpoonup}{v}\right\|^2 = \dfrac{9}{4}\left\|\overset{\large\rightharpoonup}{v}\right\|^2$。 故 $\left\|\overset{\large\rightharpoonup}{u} + \overset{\large\rightharpoonup}{v}\right\| < \left\|\overset{\large\rightharpoonup}{u} - \overset{\large\rightharpoonup}{v}\right\|$。 故選 $(2)(3)$。

題目來源:大學入學考試中心公開試題。

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