《2025年创新设计高考总复习数学浙江专版》

考点二 空间向量的数量积及其应用
例 2 如图，正四面体 $A - BCD$（所有棱长均相等）的棱长为 $1$，$E,F,G,H$ 分别是正四面体 $ABCD$ 中各棱的中点，设 $\overrightarrow{AB}=a$，$\overrightarrow{AC}=b$，$\overrightarrow{AD}=c$，试采用向量法解决下列问题：
(1) 求 $\overrightarrow{EF}$ 的模长；
(2) 求 $\overrightarrow{EF},\overrightarrow{GH}$ 的夹角。$\frac{\sqrt{2}}{2}$$90^{\circ}$

思维建模 由向量数量积的定义知，要求 $a$ 与 $b$ 的数量积，需已知 $|a|,|b|$ 和 $\langle a,b\rangle$，$a$ 与 $b$ 的夹角与方向有关，一定要根据方向正确判定夹角的大小，才能使 $a\cdot b$ 计算准确。

答案： 例2解　
(1)因为正四面体$A - BCD$的棱长为$1$，$E,F,G,H$分别是正四面体$A - BCD$中各棱的中点，$\overrightarrow{AB} = \boldsymbol{a},\overrightarrow{AC} = \boldsymbol{b},\overrightarrow{AD} = \boldsymbol{c}$，所以$\overrightarrow{BE} = \frac{1}{2}\overrightarrow{BC} = \frac{1}{2}(\overrightarrow{AC} - \overrightarrow{AB})=\frac{1}{2}(\boldsymbol{b} - \boldsymbol{a})$，$\overrightarrow{AF} = \frac{1}{2}\overrightarrow{AD} = \frac{1}{2}\boldsymbol{c}$，所以$\overrightarrow{EF} = \overrightarrow{EB} + \overrightarrow{BA} + \overrightarrow{AF}=-\frac{1}{2}(\boldsymbol{b} - \boldsymbol{a}) - \boldsymbol{a} + \frac{1}{2}\boldsymbol{c} = \frac{1}{2}(\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b})$，所以$|\overrightarrow{EF}|^{2} = \frac{1}{4}(\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b})^{2}=\frac{1}{4}(\boldsymbol{c}^{2} + \boldsymbol{a}^{2} + \boldsymbol{b}^{2} - 2\boldsymbol{a} \cdot \boldsymbol{c} + 2\boldsymbol{a} \cdot \boldsymbol{b} - 2\boldsymbol{b} \cdot \boldsymbol{c})=\frac{1}{4}(1 + 1 + 1 - 2 × 1 × 1 × \cos 60^{\circ} + 2 × 1 × 1 × \cos 60^{\circ} - 2 × 1 × 1 × \cos 60^{\circ}) = \frac{1}{2}$，故$|\overrightarrow{EF}| = \frac{\sqrt{2}}{2}$.
(2)在正四面体$A - BCD$中，$\overrightarrow{EF} = \frac{1}{2}(\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b}),|\overrightarrow{EF}| = \frac{\sqrt{2}}{2}$.同理，$\overrightarrow{GH} = \frac{1}{2}(\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a}),|\overrightarrow{GH}| = \frac{\sqrt{2}}{2}$.所以$\cos\langle\overrightarrow{EF},\overrightarrow{GH}\rangle = \frac{\overrightarrow{EF} \cdot \overrightarrow{GH}}{|\overrightarrow{EF}||\overrightarrow{GH}|}=\frac{\frac{1}{2}(\boldsymbol{c} - \boldsymbol{a} - \boldsymbol{b}) \cdot \frac{1}{2}(\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a})}{\frac{\sqrt{2}}{2} × \frac{\sqrt{2}}{2}}=\frac{1}{2}[(\boldsymbol{c} - \boldsymbol{a})^{2} - \boldsymbol{b}^{2}]=\frac{1}{2}(\boldsymbol{c}^{2} + \boldsymbol{a}^{2} - 2\boldsymbol{c} \cdot \boldsymbol{a} - \boldsymbol{b}^{2})=\frac{1}{2}(1 + 1 - 2 × 1 × 1 × \cos 60^{\circ} - 1) = 0$，所以$\overrightarrow{EF}$与$\overrightarrow{GH}$的夹角为$90^{\circ}$.

训练 2 如图所示，四棱柱 $ABCD - A_1B_1C_1D_1$ 中，底面为平行四边形，以顶点 $A$ 为端点的三条棱长都为 $1$，且两两夹角为 $60^{\circ}$。求：
(1) $AC_1$ 的长；
(2) $BD_1$ 与 $AC$ 夹角的余弦值。$\sqrt{6}$，$\frac{\sqrt{6}}{6}$

答案： 训练2解
(1)记$\overrightarrow{AB} = \boldsymbol{a},\overrightarrow{AD} = \boldsymbol{b},\overrightarrow{AA_{1}} = \boldsymbol{c}$，则$|\boldsymbol{a}| = |\boldsymbol{b}| = |\boldsymbol{c}| = 1$，$\langle\boldsymbol{a},\boldsymbol{b}\rangle = \langle\boldsymbol{b},\boldsymbol{c}\rangle = \langle\boldsymbol{c},\boldsymbol{a}\rangle = 60^{\circ}$，$\therefore \boldsymbol{a} \cdot \boldsymbol{b} = \boldsymbol{b} \cdot \boldsymbol{c} = \boldsymbol{c} \cdot \boldsymbol{a} = \frac{1}{2}$.$|\overrightarrow{AC_{1}}|^{2} = (\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c})^{2}=\boldsymbol{a}^{2} + \boldsymbol{b}^{2} + \boldsymbol{c}^{2} + 2(\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{b} \cdot \boldsymbol{c} + \boldsymbol{c} \cdot \boldsymbol{a})= 1 + 1 + 1 + 2 × (\frac{1}{2} + \frac{1}{2} + \frac{1}{2}) = 6$，$\therefore |\overrightarrow{AC_{1}}| = \sqrt{6}$，即$AC_{1}$的长为$\sqrt{6}$.
(2)$\because \overrightarrow{BD_{1}} = \boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a},\overrightarrow{AC} = \boldsymbol{a} + \boldsymbol{b}$，$\therefore |\overrightarrow{BD_{1}}| = \sqrt{2},|\overrightarrow{AC}| = \sqrt{3}$，$\overrightarrow{BD_{1}} \cdot \overrightarrow{AC} = (\boldsymbol{b} + \boldsymbol{c} - \boldsymbol{a}) \cdot (\boldsymbol{a} + \boldsymbol{b})=\boldsymbol{b}^{2} - \boldsymbol{a}^{2} + \boldsymbol{a} \cdot \boldsymbol{c} + \boldsymbol{b} \cdot \boldsymbol{c} = 1$，$\therefore \cos\langle\overrightarrow{BD_{1}},\overrightarrow{AC}\rangle = \frac{\overrightarrow{BD_{1}} \cdot \overrightarrow{AC}}{|\overrightarrow{BD_{1}}||\overrightarrow{AC}|} = \frac{\sqrt{6}}{6}$.$\therefore AC$与$BD_{1}$夹角的余弦值为$\frac{\sqrt{6}}{6}$.

考点三 利用空间向量证明平行与垂直
例 3 如图，已知 $PA\perp$ 平面 $ABCD$，四边形 $ABCD$ 为矩形，$PA = AD$，$M,N$ 分别为 $AB,PC$ 的中点，求证：
(1) $MN//$ 平面 $PAD$；
(2) 平面 $PMC\perp$ 平面 $PDC$。证明　(1)由题意，以$A$为坐标原点，$AB$，$AD$，$AP$所在的直线分别为$x$轴、$y$轴、$z$轴建立如图所示的空间直角坐标系$A - xyz$.设$PA = AD = a(a ＞ 0)$，$AB = b(b ＞ 0)$，则有$A(0,0,0),P(0,0,a),D(0,a,0),C(b,a,0)$，$B(b,0,0)$.因为$M,N$分别为$AB,PC$的中点，所以$M(\frac{b}{2},0,0)$，$N(\frac{b}{2},\frac{a}{2},\frac{a}{2})$，所以$\overrightarrow{MN} = (0,\frac{a}{2},\frac{a}{2})$，又$\overrightarrow{AP} = (0,0,a),\overrightarrow{AD} = (0,a,0)$，所以$\overrightarrow{MN} = \frac{1}{2}\overrightarrow{AP} + \frac{1}{2}\overrightarrow{AD}$.又$\overrightarrow{MN}\not\subset$平面$PAD$，所以$MN//$平面$PAD$.(2)结合(1)知，$M(\frac{b}{2},0,0),\overrightarrow{PC} = (b,a,-a)$，$\overrightarrow{PM} = (\frac{b}{2},0,-a),\overrightarrow{PD} = (0,a,-a)$.设平面$PMC$的法向量为$\boldsymbol{n_{1}} = (x_{1},y_{1},z_{1})$，则$\begin{cases}\boldsymbol{n_{1}} \cdot \overrightarrow{PC} = 0,\\\boldsymbol{n_{1}} \cdot \overrightarrow{PM} = 0,\end{cases}$即$\begin{cases}bx_{1} + ay_{1} - az_{1} = 0,\frac{b}{2}x_{1} - az_{1} = 0,\end{cases}$令$z_{1} = b$，则$x_{1} = 2a,y_{1} = - b$，得$\boldsymbol{n_{1}} = (2a,-b,b)$.设平面$PDC$的法向量为$\boldsymbol{n_{2}} = (x_{2},y_{2},z_{2})$，则$\begin{cases}\boldsymbol{n_{2}} \cdot \overrightarrow{PC} = 0,\\\boldsymbol{n_{2}} \cdot \overrightarrow{PD} = 0,\end{cases}$即$\begin{cases}bx_{2} + ay_{2} - az_{2} = 0,\\ay_{2} - az_{2} = 0,\end{cases}$得$x_{2} = 0$，令$z_{2} = 1$，则$y_{2} = 1$，得$\boldsymbol{n_{2}} = (0,1,1)$.因为$\boldsymbol{n_{1}} \cdot \boldsymbol{n_{2}} = 0 - b + b = 0$，所以$\boldsymbol{n_{1}}\perp\boldsymbol{n_{2}}$，故平面$PMC\perp$平面$PDC$.

答案：
例3证明　
(1)由题意，以$A$为坐标原点，$AB$，$AD$，$AP$所在的直线分别为$x$轴、$y$轴、$z$轴建立如图所示的空间直角坐标系$A - xyz$.设$PA = AD = a(a ＞ 0)$,$AB = b(b ＞ 0)$,则有$A(0,0,0),P(0,0,a),D(0,a,0),C(b,a,0)$，　　　　　　　　　 $B(b,0,0)$.因为$M,N$分别为$AB,PC$的中点，所以$M(\frac{b}{2},0,0)$，$N(\frac{b}{2},\frac{a}{2},\frac{a}{2})$，所以$\overrightarrow{MN} = (0,\frac{a}{2},\frac{a}{2})$，又$\overrightarrow{AP} = (0,0,a),\overrightarrow{AD} = (0,a,0)$，所以$\overrightarrow{MN} = \frac{1}{2}\overrightarrow{AP} + \frac{1}{2}\overrightarrow{AD}$.又$\overrightarrow{MN}\not\subset$平面$PAD$，所以$MN//$平面$PAD$.
(2)结合
(1)知，$M(\frac{b}{2},0,0),\overrightarrow{PC} = (b,a,-a)$，$\overrightarrow{PM} = (\frac{b}{2},0,-a),\overrightarrow{PD} = (0,a,-a)$.设平面$PMC$的法向量为$\boldsymbol{n_{1}} = (x_{1},y_{1},z_{1})$，则$\begin{cases}\boldsymbol{n_{1}} \cdot \overrightarrow{PC} = 0,\\\boldsymbol{n_{1}} \cdot \overrightarrow{PM} = 0,\end{cases}$即$\begin{cases}bx_{1} + ay_{1} - az_{1} = 0,\frac{b}{2}x_{1} - az_{1} = 0,\end{cases}$令$z_{1} = b$，则$x_{1} = 2a,y_{1} = - b$，得$\boldsymbol{n_{1}} = (2a,-b,b)$.设平面$PDC$的法向量为$\boldsymbol{n_{2}} = (x_{2},y_{2},z_{2})$，则$\begin{cases}\boldsymbol{n_{2}} \cdot \overrightarrow{PC} = 0,\\\boldsymbol{n_{2}} \cdot \overrightarrow{PD} = 0,\end{cases}$即$\begin{cases}bx_{2} + ay_{2} - az_{2} = 0,\\ay_{2} - az_{2} = 0,\end{cases}$得$x_{2} = 0$，令$z_{2} = 1$，则$y_{2} = 1$，得$\boldsymbol{n_{2}} = (0,1,1)$.因为$\boldsymbol{n_{1}} \cdot \boldsymbol{n_{2}} = 0 - b + b = 0$，所以$\boldsymbol{n_{1}}\perp\boldsymbol{n_{2}}$，故平面$PMC\perp$平面$PDC$.