《2025年一本知识大盘点高中数学全一册人教版》

例 5 如图，在三棱柱 $ ABC - A_1B_1C_1 $ 中，底面边长和侧棱长都相等，$ \angle BAA_1 = \angle CAA_1 = 60^{\circ} $，则异面直线 $ AB_1 $ 与 $ BC_1 $ 所成角的余弦值为(C)


A.$ \frac{\sqrt{3}}{3} $
B.$ \frac{\sqrt{3}}{6} $
C.$ \frac{\sqrt{6}}{6} $
D.$ \frac{\sqrt{6}}{3} $

答案： 例5
设 $\overrightarrow{AA_1} = \boldsymbol{c}$，$\overrightarrow{AB} = \boldsymbol{a}$，$\overrightarrow{AC} = \boldsymbol{b}$，棱长均为 1，
则$\boldsymbol{a} · \boldsymbol{b} = \frac{1}{2}, \quad \boldsymbol{b} · \boldsymbol{c} = \frac{1}{2}, \quad \boldsymbol{a} · \boldsymbol{c} = \frac{1}{2}$，
$\overrightarrow{AB_1} = \boldsymbol{a} + \boldsymbol{c}, \quad \overrightarrow{BC_1} = \overrightarrow{BC} + \overrightarrow{BB_1} = \boldsymbol{b} - \boldsymbol{a} + \boldsymbol{c}$，
$\overrightarrow{AB_1} · \overrightarrow{BC_1} = (\boldsymbol{a} + \boldsymbol{c}) · (\boldsymbol{b} - \boldsymbol{a} + \boldsymbol{c}) = \boldsymbol{a} · \boldsymbol{b} - \boldsymbol{a}^2 + \boldsymbol{a} · \boldsymbol{c} + \boldsymbol{b} · \boldsymbol{c} - \boldsymbol{a} · \boldsymbol{c} + \boldsymbol{c}^2$
$= \boldsymbol{a} · \boldsymbol{b} - \boldsymbol{a}^2 + \boldsymbol{b} · \boldsymbol{c} + \boldsymbol{c}^2 = \frac{1}{2} - 1 + \frac{1}{2} + 1 = 1$，
$|\overrightarrow{AB_1}| = \sqrt{(\boldsymbol{a} + \boldsymbol{c})^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$，
$|\overrightarrow{BC_1}| = \sqrt{(\boldsymbol{b} - \boldsymbol{a} + \boldsymbol{c})^2} = \sqrt{1 + 1 + 1 - 1 - 1 + 1} = \sqrt{2}$，
$\cos \langle \overrightarrow{AB_1}, \overrightarrow{BC_1} \rangle = \frac{\overrightarrow{AB_1} · \overrightarrow{BC_1}}{|\overrightarrow{AB_1}| |\overrightarrow{BC_1}|} = \frac{1}{\sqrt{2} × \sqrt{3}} = \frac{\sqrt{6}}{6}$，
异面直线 $AB_1$ 与 $BC_1$ 所成角的余弦值为 $\frac{\sqrt{6}}{6}$，
答案: C

例 6 如图，已知正四面体 $ ABCD $(所有棱长均相等)的棱长为 1，$ E $，$ F $，$ G $，$ H $ 分别是四面体 $ ABCD $ 各棱的中点，设 $ \overrightarrow{AB} = \boldsymbol{a} $，$ \overrightarrow{AC} = \boldsymbol{b} $，$ \overrightarrow{AD} = \boldsymbol{c} $，试用向量法解决下列问题：
(1) 求 $ |\overrightarrow{EF}| $；
(2) 求 $ \overrightarrow{EF} $，$ \overrightarrow{GH} $ 的夹角。

答案： 例6
(1) $\overrightarrow{BE} = \frac{1}{2} \overrightarrow{BC} = \frac{1}{2} (\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{2} (\boldsymbol{b} - \boldsymbol{a})$，
$\overrightarrow{AF} = \frac{1}{2} \overrightarrow{AD} = \frac{1}{2} \boldsymbol{c}$，
$\overrightarrow{EF} = \overrightarrow{EB} + \overrightarrow{BA} + \overrightarrow{AF} = -\frac{1}{2} (\boldsymbol{b} - \boldsymbol{a}) - \boldsymbol{a} + \frac{1}{2} \boldsymbol{c} = \frac{1}{2} (\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b})$，
$|\overrightarrow{EF}| = \frac{1}{2} \sqrt{(\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b})^2} = \frac{1}{2} \sqrt{\boldsymbol{c}^2 + \boldsymbol{a}^2 + \boldsymbol{b}^2 - 2\boldsymbol{a} · \boldsymbol{c} - 2\boldsymbol{b} · \boldsymbol{c} + 2\boldsymbol{a} · \boldsymbol{b}} = \frac{\sqrt{2}}{2}$。
(2) $\overrightarrow{EF} = \frac{1}{2} (\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b}), \quad |\overrightarrow{EF}| = \frac{\sqrt{2}}{2}$，
$\overrightarrow{GH} = \frac{1}{2} (\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a}), \quad |\overrightarrow{GH}| = \frac{\sqrt{2}}{2}$，
$\cos \langle \overrightarrow{EF}, \overrightarrow{GH} \rangle = \frac{\overrightarrow{EF} · \overrightarrow{GH}}{|\overrightarrow{EF}| |\overrightarrow{GH}|} = \frac{\frac{1}{2} (\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b}) · \frac{1}{2} (\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a})}{\frac{\sqrt{2}}{2} × \frac{\sqrt{2}}{2}}$
$= \frac{1}{2} [(\boldsymbol{c} - \boldsymbol{a})^2 - \boldsymbol{b}^2] = \frac{1}{2} (\boldsymbol{c}^2 + \boldsymbol{a}^2 - 2 \boldsymbol{c} · \boldsymbol{a} - \boldsymbol{b}^2)$
$= \frac{1}{2} (1 + 1 - 2 × 1 × 1 × \cos 60° - 1) = 0$，
$\overrightarrow{EF}$ 与 $\overrightarrow{GH}$ 的夹角为 $90°$。