《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

对点训练1
(1)下列各式计算正确的有 (C)
①$( - 7)6\boldsymbol{a} = - 42\boldsymbol{a}$;②$7(\boldsymbol{a} + \boldsymbol{b}) - 8\boldsymbol{b} = 7\boldsymbol{a} + 15\boldsymbol{b}$;
③$\boldsymbol{a} - 2\boldsymbol{b} + \boldsymbol{a} + 2\boldsymbol{b} = 2\boldsymbol{a}$;④$4(2\boldsymbol{a} + \boldsymbol{b}) = 8\boldsymbol{a} + 4\boldsymbol{b}$.
A.1个 B.2个 C.3个 D.4个
(2)若$\boldsymbol{a} = \boldsymbol{b} + \boldsymbol{c}$,化简$3(\boldsymbol{a} + 2\boldsymbol{b}) - 2(3\boldsymbol{b} + \boldsymbol{c}) - 2(\boldsymbol{a} + \boldsymbol{b})$的结果为 (A)
A. $- \boldsymbol{a}$ B. $- 4\boldsymbol{b}$ C. $\boldsymbol{c}$ D. $\boldsymbol{a} - \boldsymbol{b}$

答案： 对点训练1:
(1)C
(2)A
(1)①③④正确,②错$,7(\boldsymbol{a} + \boldsymbol{b})$
$ -8\boldsymbol{b} = 7\boldsymbol{a} + 7\boldsymbol{b} - 8\boldsymbol{b} = 7\boldsymbol{a} - \boldsymbol{b}.$
$(2)3(\boldsymbol{a} + 2\boldsymbol{b}) - 2(3\boldsymbol{b} + \boldsymbol{c}) - 2(\boldsymbol{a} + \boldsymbol{b}) = (3 - 2)\boldsymbol{a} + (6 - 6 -$
$2)\boldsymbol{b} - 2\boldsymbol{c} = \boldsymbol{a} - 2(\boldsymbol{b} + \boldsymbol{c}) = \boldsymbol{a} - 2\boldsymbol{a} = -\boldsymbol{a}.$

例2. 如图所示，四边形$OADB$是以向量$\overrightarrow{OA} = \boldsymbol{a}$,$\overrightarrow{OB} = \boldsymbol{b}$为邻边的平行四边形，又$\overrightarrow{BM} = \frac{1}{3}\overrightarrow{BC}$,$\overrightarrow{CN} = \frac{1}{3}\overrightarrow{CD}$,试用$\boldsymbol{a}$,$\boldsymbol{b}$表示$\overrightarrow{OM}$,$\overrightarrow{ON}$,$\overrightarrow{MN}$.
【分析】 用$\boldsymbol{a}$,$\boldsymbol{b}$表示$\overrightarrow{BM}\to$表示$\overrightarrow{OM}$,$\overrightarrow{ON}\to\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM}$
▶[归纳提升]

答案： 例$2:\overrightarrow{BM} = \frac{1}{3}\overrightarrow{BC} = \frac{1}{6}\overrightarrow{BA} = \frac{1}{6}(\overrightarrow{OA} - \overrightarrow{OB})$
$ = \frac{1}{6}(\boldsymbol{a} - \boldsymbol{b}),$
$\therefore\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \boldsymbol{b} + \frac{1}{6}\boldsymbol{a} - \frac{1}{6}\boldsymbol{b} = \frac{1}{6}\boldsymbol{a} + \frac{5}{6}\boldsymbol{b}.$
$\because\overrightarrow{CN} = \frac{1}{3}\overrightarrow{CD} = \frac{1}{6}\overrightarrow{OD},$
$\therefore\overrightarrow{ON} = \overrightarrow{OC} + \overrightarrow{CN} = \frac{1}{2}\overrightarrow{OD} + \frac{1}{6}\overrightarrow{OD} = \frac{2}{3}\overrightarrow{OD} = \frac{2}{3}(\overrightarrow{OA} + \overrightarrow{OB}) =$
$\frac{2}{3}\boldsymbol{a} + \frac{2}{3}\boldsymbol{b},$

对点训练2
(1)如图，在正方形$ABCD$中，$Q$为$BC$上一点，$AQ$交$BD$于$E$，$A$且$E$,$F$为$BD$的两个三等分点，则 (BCD)
A. $\overrightarrow{AE} + \overrightarrow{AF} + \overrightarrow{AC} = 0$
B. $\overrightarrow{AE} = \frac{2}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AD}$
C. $\overrightarrow{AF} = \frac{1}{3}\overrightarrow{AB} + \frac{2}{3}\overrightarrow{AD}$
D. $\overrightarrow{FQ} = \frac{2}{3}\overrightarrow{AB} - \frac{1}{6}\overrightarrow{AD}$
(2)如图所示，已知在$\triangle ABC$中，$\overrightarrow{AD} = \frac{2}{3}\overrightarrow{AB}$,$DE// BC$,$DE$交$AC$于点$E$,$BC$边上的中线$AM$交$DE$于点$N$,设$\overrightarrow{AB} = \boldsymbol{a}$,$\overrightarrow{AC} = \boldsymbol{b}$,用$\boldsymbol{a}$,$\boldsymbol{b}$表示向量$\overrightarrow{AE}$,$\overrightarrow{DE}$,$\overrightarrow{AM}$,$\overrightarrow{AN}$.

答案： 对点训练2:
(1)BCD
(2)见解析
【解析】
(1)因为$\overrightarrow{AE} + \overrightarrow{AF} = \overrightarrow{AC},$所以$\overrightarrow{AE} + \overrightarrow{AF} - \overrightarrow{AC} = 0,$故A
错误$.\overrightarrow{AE} = \overrightarrow{AB} + \overrightarrow{BE} = \overrightarrow{AB} + \frac{1}{3}\overrightarrow{BD} = \overrightarrow{AB} + \frac{1}{3}(\overrightarrow{AD} - \overrightarrow{AB}) = \frac{2}{3}\overrightarrow{AB} +$
$\frac{1}{3}\overrightarrow{AD},$故B正确$.\overrightarrow{AF} = \overrightarrow{AB} + \overrightarrow{BF} = \overrightarrow{AB} + \frac{2}{3}\overrightarrow{BD} = \overrightarrow{AB} + \frac{2}{3}(\overrightarrow{AD} -$
$\overrightarrow{AB}) = \frac{1}{3}\overrightarrow{AB} + \frac{2}{3}\overrightarrow{AD},$故C正确. 因为E为BD上靠近B的三等
分点,所以$\overrightarrow{BE} = \frac{1}{2}\overrightarrow{ED},$利用相似性质可得$\overrightarrow{BQ} = \frac{1}{2}\overrightarrow{AD},$则$\overrightarrow{FQ} =$
$\overrightarrow{BQ} - \overrightarrow{BF} = \frac{1}{2}\overrightarrow{AD} - \frac{2}{3}\overrightarrow{BD} = \frac{1}{2}\overrightarrow{AD} - \frac{2}{3}(\overrightarrow{AD} - \overrightarrow{AB}) = \frac{2}{3}\overrightarrow{AB} - \frac{1}{6}\overrightarrow{AD}.$
故D正确. 故选BCD.
$(2)\because DE // BC,\overrightarrow{AD} = \frac{2}{3}\overrightarrow{AB} = \frac{2}{3}\boldsymbol{a},\therefore\overrightarrow{AE} = \frac{2}{3}\overrightarrow{AC} = \frac{2}{3}\boldsymbol{b},$
$\because \triangle ADE \sim \triangle ABC,\therefore\overrightarrow{DE} = \frac{2}{3}\overrightarrow{BC} = \frac{2}{3}(\boldsymbol{b} - \boldsymbol{a}).$
$\because \triangle ADN \sim \triangle ABM,$且$\overrightarrow{AD} = \frac{2}{3}\overrightarrow{AB},\therefore\overrightarrow{AN} = \frac{2}{3}\overrightarrow{AM}.$
又$\because\overrightarrow{AM} = \overrightarrow{AB} + \overrightarrow{BM} = \boldsymbol{a} + \frac{1}{2}\overrightarrow{BC} = \boldsymbol{a} + \frac{1}{2}(\boldsymbol{b} - \boldsymbol{a}) = \frac{\boldsymbol{a} + \boldsymbol{b}}{2}$
$\therefore\overrightarrow{AN} = \frac{1}{3}(\boldsymbol{a} + \boldsymbol{b}).$