《2025年零失误分层训练高中数学必修第二册人教版黑吉辽内蒙古专版》

13.【题型三】如图所示，在边长为 1 的正六边形 ABCDEF 中，O 是其中心，$\overrightarrow{BG}=\frac{1}{2}\overrightarrow{GC}$，设$\overrightarrow{AB}=a,\overrightarrow{AF}=b$.
(1)用$a,b$分别表示$\overrightarrow{AO}$及$\overrightarrow{AG}$;
(2)求$|\overrightarrow{AGI}|$及$\overrightarrow{AD}$与$\overrightarrow{AG}$的夹角$\theta$的余弦值.

答案： 13.解:
(1)连接FO，则$\overrightarrow{AO} = \overrightarrow{AF} + \overrightarrow{FO} = \overrightarrow{AF} + \overrightarrow{AB} = \mathbf{b} + \mathbf{a}$，$\overrightarrow{AG} = \overrightarrow{AB} + \overrightarrow{BG} = \overrightarrow{AB} + \frac{1}{3}\overrightarrow{BC} = \overrightarrow{AB} + \frac{1}{3}(\overrightarrow{AC} - \overrightarrow{AB}) = \frac{2}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$。
(2)$\because \mathbf{a} · \mathbf{b} = |\mathbf{a}| · |\mathbf{b}| · \cos 120^{\circ} = -\frac{1}{2}$，又$|\overrightarrow{AG}|^{2} = (\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b})^{2} = \frac{4}{9}|\mathbf{a}|^{2} + \frac{4}{9}\mathbf{a} · \mathbf{b} + \frac{1}{9}|\mathbf{b}|^{2} = \frac{16}{9} + \frac{8}{9} × (-\frac{1}{2}) + \frac{1}{9} = \frac{13}{9}$，$\therefore |\overrightarrow{AG}| = \frac{\sqrt{13}}{3}$。又$\overrightarrow{AD} = 2(\mathbf{a} + \mathbf{b})$，$|\overrightarrow{AD}| = 2$，$\therefore \overrightarrow{AD} · \overrightarrow{AG} = [2(\mathbf{a} + \mathbf{b})] · (\frac{4}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}) = \frac{2}{3} × (4|\mathbf{a}|^{2} + 5\mathbf{a} · \mathbf{b} + |\mathbf{b}|^{2}) = \frac{5}{3}$，$\therefore \cos \theta = \frac{\overrightarrow{AD} · \overrightarrow{AG}}{|\overrightarrow{AD}||\overrightarrow{AG}|} = \frac{\frac{5}{3}}{2 × \frac{\sqrt{13}}{3}} = \frac{5\sqrt{13}}{26}$，即$\overrightarrow{AD}$与$\overrightarrow{AG}$的夹角$\theta$的余弦值为$\frac{5\sqrt{13}}{26}$。

1. [多选] 已知$a,b,c$均为非零向量，下列命题错误的是 (ACD)

A.$\lambda\in R,\lambda(a + b)=a· b$
B.$(a· b)· c=a·(b· c)$可能成立
C.若$a· b=b· c$，则$a=c$
D.若$a· b=1$，则$|a|=1$或$|b|=1$

答案： 1.ACD[提示:对于实数$\lambda$，$\lambda(\mathbf{a} + \mathbf{b})$为向量，$\mathbf{a} · \mathbf{b}$为实数，故A错误;记$(\mathbf{a} · \mathbf{b}) · \mathbf{c} = \lambda\mathbf{c}$，$\mathbf{a} · (\mathbf{b} · \mathbf{c}) = \mu\mathbf{a}$，所以$\mathbf{a}$，$\mathbf{c}$不共线且$\lambda\mu \neq 0$时，$(\mathbf{a} · \mathbf{b}) · \mathbf{c} = \mathbf{a} · (\mathbf{b} · \mathbf{c})$不成立，故B正确;$\mathbf{a} · \mathbf{b} = \mathbf{b} · \mathbf{c} \Rightarrow \mathbf{b} · (\mathbf{a} - \mathbf{c}) = 0 \Rightarrow \mathbf{a} = \mathbf{c}$或$\mathbf{b} \perp (\mathbf{a} - \mathbf{c})$，故C错误;当$|\mathbf{a}| = 4$，$|\mathbf{b}| = \frac{1}{2}$，$\langle \mathbf{a}, \mathbf{b} \rangle = \frac{\pi}{3}$时，$\mathbf{a} · \mathbf{b} = 1$，不满足$|\mathbf{a}| = 1$或$|\mathbf{b}| = 1$，故D错误。]

2. 设$e_1,e_2$是夹角为$45^{\circ}$的两个单位向量，且$a=e_1+2e_2,b=2e_1+e_2$，则$|a + b|$的值为$3\sqrt{2} + \sqrt{2}$.

答案： 2.$3\sqrt{2} + \sqrt{2}$[提示:因为$\mathbf{a} + \mathbf{b} = 3(\mathbf{e}_{1} + \mathbf{e}_{2})$，所以$|\mathbf{a} + \mathbf{b}| = |3(\mathbf{e}_{1} + \mathbf{e}_{2})| = 3\sqrt{(\mathbf{e}_{1} + \mathbf{e}_{2})^{2}} = 3\sqrt{|\mathbf{e}_{1}|^{2} + |\mathbf{e}_{2}|^{2} + 2\mathbf{e}_{1} · \mathbf{e}_{2}} = 3\sqrt{|\mathbf{e}_{1}|^{2} + |\mathbf{e}_{2}|^{2} + 2|\mathbf{e}_{1}||\mathbf{e}_{2}|\cos 45^{\circ}} = 3\sqrt{2} + \sqrt{2}$。]

3. 在平行四边形 ABCD 中，若$\overrightarrow{BM}=\frac{1}{2}\overrightarrow{BC},\overrightarrow{DN}=\frac{1}{3}\overrightarrow{DC},\overrightarrow{AC}=x\overrightarrow{AM}+y\overrightarrow{AN}$，则$x + y$等于 (D)

A.$\frac{3}{5}$
B.$\frac{4}{5}$
C.1
D.$\frac{7}{5}$

答案： 3.D[提示:$\because$四边形ABCD是平行四边形，$\therefore \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$。$\because \overrightarrow{BM} = \frac{1}{2}\overrightarrow{BC}$，$\overrightarrow{DC} = 3\overrightarrow{DN}$，$\therefore \overrightarrow{AM} = \overrightarrow{AB} + \overrightarrow{BM} = \overrightarrow{AB} + \frac{1}{2}\overrightarrow{BC} = \overrightarrow{AB} + \frac{1}{2}\overrightarrow{AD}$，$\overrightarrow{AN} = \overrightarrow{AD} + \overrightarrow{DN} = \overrightarrow{AD} + \frac{1}{3}\overrightarrow{DC} = \overrightarrow{AD} + \frac{1}{3}\overrightarrow{AB}$，$\therefore \overrightarrow{AC} = x\overrightarrow{AM} + y\overrightarrow{AN}$，$\therefore \overrightarrow{AC} = x(\overrightarrow{AB} + \frac{1}{2}\overrightarrow{AD}) + y(\overrightarrow{AD} + \frac{1}{3}\overrightarrow{AB}) = (x + \frac{y}{3})\overrightarrow{AB} + (\frac{x}{2} + y)\overrightarrow{AD}$，$\because \overrightarrow{AB}$与$\overrightarrow{AD}$不共线，$\therefore \begin{cases} x + \frac{y}{3} = 1, \\ \frac{x}{2} + y = 1, \end{cases}$解得$\begin{cases} x = \frac{4}{5}, \\ y = \frac{3}{5}, \end{cases}$则$x + y = \frac{7}{5}$。]

4. 如图所示，在$\triangle ABC$中，D 是 BC 的中点，点 E 在 AB 上，且$BE = 2EA$，AD 与 CE 交于点 O. 若$\overrightarrow{AB}·\overrightarrow{AC}=6\overrightarrow{AO}·\overrightarrow{EC}$，求$\frac{AB}{AC}$的值.

答案： 4.解:设$\overrightarrow{AO} = \lambda\overrightarrow{AD}$，又$\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$，则$\overrightarrow{AO} = \frac{\lambda}{2}(\overrightarrow{AB} + \overrightarrow{AC}) = \frac{\lambda}{2}\overrightarrow{AB} + \frac{\lambda}{2}\overrightarrow{AC}$。设$\overrightarrow{EO} = \mu\overrightarrow{EC}$，则$\overrightarrow{AO} = \overrightarrow{AE} + \overrightarrow{EO} = \overrightarrow{AE} + \mu\overrightarrow{EC} = \overrightarrow{AE} + \mu(\overrightarrow{AC} - \overrightarrow{AE}) = (1 - \mu)\overrightarrow{AE} + \mu\overrightarrow{AC}$，又$\overrightarrow{BE} = 2\overrightarrow{EA}$，$\therefore \overrightarrow{AE} = \frac{1}{3}\overrightarrow{AB}$，$\therefore \overrightarrow{AO} = \frac{1 - \mu}{3}\overrightarrow{AB} + \mu\overrightarrow{AC}$，又$\because \overrightarrow{EC} = \overrightarrow{AC} - \overrightarrow{AE} = \overrightarrow{AC} - \frac{1}{3}\overrightarrow{AB}$，$\therefore \begin{cases} \frac{\lambda}{2} = \frac{1 - \mu}{3}, \\ \frac{\lambda}{2} = \mu, \end{cases}$解得$\begin{cases} \lambda = \frac{1}{2}, \\ \mu = \frac{1}{4}, \end{cases}$ $\therefore \overrightarrow{AO} = \frac{1}{4}\overrightarrow{AB} + \frac{1}{4}\overrightarrow{AC}$，又$\because \overrightarrow{EC} = \overrightarrow{AC} - \overrightarrow{AE} = \overrightarrow{AC} - \frac{1}{3}\overrightarrow{AB}$，$\therefore \overrightarrow{AB} · \overrightarrow{AC} = 6\overrightarrow{AO} · \overrightarrow{EC} = 6(\frac{1}{4}\overrightarrow{AB} + \frac{1}{4}\overrightarrow{AC}) · (\overrightarrow{AC} - \frac{1}{3}\overrightarrow{AB}) = -\frac{1}{2}\overrightarrow{AB}^{2} + \frac{3}{2}\overrightarrow{AC}^{2} = 3$，$|\overrightarrow{AB} · \overrightarrow{AC}|$，$\therefore |\overrightarrow{AB}| = \sqrt{3}|\overrightarrow{AC}|$，即$\frac{|\overrightarrow{AB}|}{|\overrightarrow{AC}|} = \sqrt{3}$。