《2026年高中必刷题高中数学必修第二册人教版》

1. [河南郑州2025高一联考]若$\boldsymbol{a},\boldsymbol{b}$是平面内一组不共线的非零向量，则下列四组向量也可以作为一个基底的为(B)
①$\boldsymbol{a}-\boldsymbol{b}$和$2025\boldsymbol{b}-2025\boldsymbol{a}$；②$\boldsymbol{a}+\boldsymbol{b}$和$\boldsymbol{a}-\boldsymbol{b}$；③$3\boldsymbol{a}-2\boldsymbol{b}$和$2\boldsymbol{a}-3\boldsymbol{b}$；④$\boldsymbol{a}-3\boldsymbol{b}$和$6\boldsymbol{b}-2\boldsymbol{a}$.
A.①②
B.②③
C.③④
D.①④

答案： 1.B【解析】对于①，因为$2025\boldsymbol{b} - 2025\boldsymbol{a} = -2025(\boldsymbol{a} - \boldsymbol{b})$，所以$\boldsymbol{a} - \boldsymbol{b}$和$2025\boldsymbol{b} - 2025\boldsymbol{a}$是共线向量，不能作为一个基底；
对于②，设$\boldsymbol{a} + \boldsymbol{b} = \lambda(\boldsymbol{a} - \boldsymbol{b})$，可得$\begin{cases} \lambda = 1 \\ \lambda = -1 \end{cases}$，方程组无解，所以$\boldsymbol{a} + \boldsymbol{b}$和$\boldsymbol{a} - \boldsymbol{b}$不共线，可以作为一个基底；
对于③，设$3\boldsymbol{a} - 2\boldsymbol{b} = \mu(2\boldsymbol{a} - 3\boldsymbol{b})$，可得$\begin{cases} 2\mu = 3 \\ -3\mu = -2 \end{cases}$，方程组无解，所以$3\boldsymbol{a} - 2\boldsymbol{b}$和$2\boldsymbol{a} - 3\boldsymbol{b}$不共线，可以作为一个基底；
对于④，因为$\boldsymbol{a} - 3\boldsymbol{b} = -\frac{1}{2}(6\boldsymbol{b} - 2\boldsymbol{a})$，所以$\boldsymbol{a} - 3\boldsymbol{b}$和$6\boldsymbol{b} - 2\boldsymbol{a}$是共线向量，不能作为一个基底.故选B.

2. 如图，在$\triangle ABC$中，$\overrightarrow {BD}=2\overrightarrow {DC}$，点$E$是线段$AD$的中点，则$\overrightarrow {AC}=$(C)


A.$\frac {3}{4}\overrightarrow {AD}+\frac {1}{2}\overrightarrow {BE}$
B.$\frac {3}{4}\overrightarrow {AD}+\overrightarrow {BE}$
C.$\frac {5}{4}\overrightarrow {AD}+\frac {1}{2}\overrightarrow {BE}$
D.$\frac {5}{4}\overrightarrow {AD}+\overrightarrow {BE}$

答案：
2.C【解析】由题意知，$\overrightarrow{AC} = \overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AD} + \frac{1}{2}\overrightarrow{BD} = \overrightarrow{AD} + \frac{1}{2}(\overrightarrow{BE} + \overrightarrow{ED}) = \overrightarrow{AD} + \frac{1}{2}\overrightarrow{BE} + \frac{1}{2}\overrightarrow{ED} = \overrightarrow{AD} + \frac{1}{2}\overrightarrow{BE} + \frac{1}{2} × \frac{1}{2}\overrightarrow{AD} = \frac{5}{4}\overrightarrow{AD} + \frac{1}{2}\overrightarrow{BE}$.故选C.

3. (多选)[湖北黄冈2025高一期中]如图，已知$\triangle ABC$，$AB=1$，$AC=\sqrt {3}$，$BC=2$，其内有一点$G$，满足$\overrightarrow {GA}+\overrightarrow {GB}+\overrightarrow {GC}=\boldsymbol{0}$，过点$G$的直线分别交$AB$，$AC$于点$D$，$E$.设$\overrightarrow {AD}=\lambda \overrightarrow {AB}$，$\overrightarrow {AE}=\mu \overrightarrow {AC}(0＜\lambda ＜1$，$0＜\mu ＜1)$，则下列说法正确的是(BCD)


A.$\overrightarrow {AG}=\frac {1}{3}\overrightarrow {AB}+\frac {1}{4}\overrightarrow {AC}$
B.点$G$为$\triangle ABC$的重心
C.$\frac {1}{\lambda }+\frac {1}{\mu }=3$
D.$|\overrightarrow {CG}|=\frac {\sqrt {13}}{3}$

答案：
3.BCD【解析】对于A：由$\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = 0$得$\overrightarrow{GA} + (\overrightarrow{AG} + \overrightarrow{GB}) + (\overrightarrow{AG} + \overrightarrow{GC}) = 2\overrightarrow{AG} \Rightarrow \overrightarrow{AB} + \overrightarrow{AC} = 3\overrightarrow{AG} \Rightarrow \overrightarrow{AG} = \frac{1}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC}$，故A错误；
对于B：如图，取BC的中点为M，连接GM，由$\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = 0$得$\overrightarrow{GA} = -(\overrightarrow{GB} + \overrightarrow{GC}) = -2\overrightarrow{GM}$，所以点A，G，M共线，且G为靠近点M的三等分点，即G为$\triangle ABC$的重心，故B正确；
对于C：由$\overrightarrow{AG} = \frac{1}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC} = \frac{1}{3\lambda}\overrightarrow{AD} + \frac{1}{3\mu}\overrightarrow{AE},D,G,E$三点共线，得$\frac{1}{3\lambda} + \frac{1}{3\mu} = 1 \Rightarrow \frac{1}{\lambda} + \frac{1}{\mu} = 3$，故C正确；
对于D：因为$AC = \sqrt{3},BC = 2,AB = 1$，所以$AC^2 + AB^2 = BC^2$，所以$AB \perp AC$，
设AB的中点为N，连接GN，由G为$\triangle ABC$的重心，可知C，G，N三点共线，则$AN = \frac{1}{2}AB = \frac{1}{2}$，在$Rt \triangle ACN$中，有$CN = \sqrt{AC^2 + AN^2} = \frac{\sqrt{13}}{2}$，又G为重心，所以$CG = \frac{2}{3}CN = \frac{\sqrt{13}}{3}$，故D正确.故选BCD.

4. (多选)[吉林长春吉大附中实验学校2024高一期中]已知等边三角形$ABC$的边长为$2$，$\overrightarrow {BD}=2\overrightarrow {DC}$，$\overrightarrow {CE}=2\overrightarrow {EA}$，$AD$交$BE$于点$M$，则(AC)

A.$\overrightarrow {AD}=\frac {1}{3}\overrightarrow {AB}+\frac {2}{3}\overrightarrow {AC}$
B.$\overrightarrow {BM}=\frac {3}{4}\overrightarrow {BE}$
C.$S_{\triangle BMD}=8S_{\triangle AME}$
D.$S_{四边形CDME}=\frac {3\sqrt {3}}{7}$

答案：
4.AC【解析】对于A，因为$\overrightarrow{BD} = 2\overrightarrow{DC}$，所以$\overrightarrow{BD} = \frac{2}{3}\overrightarrow{BC}$，所以$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AB} + \frac{2}{3}\overrightarrow{BC} = \overrightarrow{AB} + \frac{2}{3}(\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{3}\overrightarrow{AB} + \frac{2}{3}\overrightarrow{AC}$，故A正确；
对于B，设$\overrightarrow{BM} = k\overrightarrow{BE}(k \in \mathbb{R})$，所以$\overrightarrow{AM} = \overrightarrow{AB} + \overrightarrow{BM} = \overrightarrow{AB} + k\overrightarrow{BE} = \overrightarrow{AB} + k(\frac{1}{3}\overrightarrow{AC} - \overrightarrow{AB}) = (1 - k)\overrightarrow{AB} + \frac{1}{3}k\overrightarrow{AC}$，
又A，M，D三点在一条直线上，故$\overrightarrow{AM} // \overrightarrow{AD}$，故$\frac{1 - k}{\frac{1}{3}} = \frac{\frac{1}{3}k}{\frac{2}{3}}$，解得$k = \frac{6}{7}$，即$\overrightarrow{BM} = \frac{6}{7}\overrightarrow{BE}$.
点悟：共线向量$\boldsymbol{b} = \lambda\boldsymbol{a}$的一种变形形式
故B错误；

对于C，设$S_{\triangle AME} = S$，由于$\overrightarrow{BM} = \frac{6}{7}\overrightarrow{BE}$，则
敲黑板：数乘向量不仅可以表示出方向关系，也可以得到线段比例关系，由此可得到面积之间的关系
$S_{\triangle ABE} = 7S_{\triangle AME} = 7S,S_{\triangle ABM} = 6S_{\triangle AME} = 6S$，又$\overrightarrow{AD} = \frac{1}{3}\overrightarrow{AB} + \frac{2}{3}\overrightarrow{AC} = \frac{3}{7}\overrightarrow{AD}$，所以$S_{\triangle ABD} = 14S \Rightarrow S_{\triangle BMD} = 8S$，故C正确；
对于D，因为$\overrightarrow{CE} = 2\overrightarrow{EA}$，所以$S_{\triangle BEC} = 2S_{\triangle ABE} = 14S$，所以$S_{四边形CDME} = S_{\triangle BEC} - S_{\triangle BMD} = 14S - 8S = 6S$，
由$S_{\triangle ABC} = S_{\triangle ABE} + S_{\triangle BEC} = 21S = \frac{\sqrt{3}}{4} × 2^2$，得$S = \frac{\sqrt{3}}{21}$，所以$S_{四边形CDME} = 6S = \frac{2\sqrt{3}}{7}$，故D错误.
故选AC.

5. 在等腰梯形$ABCD$中，已知$AB// DC$，$AB=2$，$BC=1$，$\angle DAB=60^{\circ }$，点$P$在线段$BC$上，且$\overrightarrow {AP}=m\overrightarrow {AB}+n\overrightarrow {AD}(m,n\in \mathbf{R})$，当点$P$为线段$BC$中点时，$m+n=$$\frac{5}{4}$；当点$P$在线段$BC$上运动时，$m+n$的最大值为$\frac{3}{2}$.

答案：
5.$\frac{5}{4} \frac{3}{2}$【解析】如图，连接AC.

因为$AB = 2,BC = 1,\angle DAB = 60°$，
所以$DC = AB - 2AD\cos 60° = 2 - 2 × 1 × \frac{1}{2} = 1$，故$\overrightarrow{DC} = \frac{1}{2}\overrightarrow{AB}$.
当点P为线段BC中点时，$\overrightarrow{AP} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC}) = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD} + \frac{1}{2}\overrightarrow{AB}) = \frac{3}{4}\overrightarrow{AB} + \frac{1}{2}\overrightarrow{AD}$，
又$\overrightarrow{AP} = m\overrightarrow{AB} + n\overrightarrow{AD}$，所以$m = \frac{3}{4},n = \frac{1}{2}$，所以$m + n = \frac{5}{4}$.
当点P在线段BC上运动时，$\overrightarrow{AP} = m\overrightarrow{AB} + n\overrightarrow{AD} = m\overrightarrow{AB} + n(\overrightarrow{AC} - \overrightarrow{DC}) = m\overrightarrow{AB} + n(\overrightarrow{AC} - \frac{1}{2}\overrightarrow{AB}) = (m - \frac{1}{2}n)\overrightarrow{AB} + n\overrightarrow{AC}$，
由B，P，C三点共线可知，$m - \frac{1}{2}n + n = 1$，即$m + \frac{n}{2} = 1$，所以$m + n = 1 + \frac{n}{2}$，
又$\overrightarrow{AP} = (1 - n)\overrightarrow{AB} + n\overrightarrow{AC}$，即$\overrightarrow{AP} - \overrightarrow{AB} = n(\overrightarrow{AC} - \overrightarrow{AB})$，所以$\overrightarrow{BP} = n\overrightarrow{BC}$，所以$0 \leq n \leq 1$，所以$m + n = 1 + \frac{n}{2} \leq \frac{3}{2}$，即$m + n$的最大值为$\frac{3}{2}$.

6. [重庆主城区七校2025高一期末]如图，在平行四边形$ABCD$中，$\overrightarrow {BE}=\frac {1}{2}\overrightarrow {BC}$，$\overrightarrow {CF}=2\overrightarrow {FD}$.
（1）若$\overrightarrow {EF}=x\overrightarrow {AB}+y\overrightarrow {AD}$，求$3x+2y$的值；
（2）若$|\overrightarrow {AB}|=6$，$|\overrightarrow {AD}|=4$，$\angle BAD=60^{\circ }$，求$\overrightarrow {AC}· \overrightarrow {EF}$的值.

答案： 6.【解】
(1)$\because \overrightarrow{BE} = \frac{1}{2}\overrightarrow{BC},\overrightarrow{CF} = 2\overrightarrow{FD},\therefore \overrightarrow{EF} = \overrightarrow{EC} + \overrightarrow{CF} = \frac{1}{2}\overrightarrow{BC} - \frac{2}{3}\overrightarrow{DC} = \frac{1}{2}\overrightarrow{AD} - \frac{2}{3}\overrightarrow{AB}$.
又$\overrightarrow{EF} = x\overrightarrow{AB} + y\overrightarrow{AD}$，$\therefore x = -\frac{2}{3},y = \frac{1}{2}$，故$3x + 2y = 3 × (-\frac{2}{3}) + 2 × \frac{1}{2} = -1$.
(2)$\because \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD},\therefore \overrightarrow{AC} · \overrightarrow{EF} = (\overrightarrow{AB} + \overrightarrow{AD}) · (\frac{1}{2}\overrightarrow{AD} - \frac{2}{3}\overrightarrow{AB}) = \frac{1}{2}\overrightarrow{AD}^2 - \frac{2}{3}\overrightarrow{AB}^2 - \frac{1}{6}\overrightarrow{AB} · \overrightarrow{AD}$.
又$\because |\overrightarrow{AB}| = 6,|\overrightarrow{AD}| = 4,\angle BAD = 60°$，
$\therefore \overrightarrow{AC} · \overrightarrow{EF} = \frac{1}{2} × 16 - \frac{2}{3} × 36 - \frac{1}{6} × 6 × 4 × \cos 60° = 8 - 24 - 2 = -18$，
故$\overrightarrow{AC} · \overrightarrow{EF}$的值为$-18$.

7. [浙江杭州四校2024高一联考]如图，在$\triangle ABC$中，$D$是$BC$的中点，$E$在边$AB$上且$BE=2EA$，$AD$与$CE$交于点$O$.
（1）用$\overrightarrow {AB},\overrightarrow {AC}$表示$\overrightarrow {DE}$；
（2）过点$O$作直线交线段$AB$于点$G$，交线段$AC$于点$H$，且$\overrightarrow {AG}=\frac {2}{3}\overrightarrow {AB}$，$\overrightarrow {AH}=t\overrightarrow {AC}$，求$t$的值.

答案：
7.【解】
(1)$\overrightarrow{DE} = \overrightarrow{AE} - \overrightarrow{AD} = \frac{1}{3}\overrightarrow{AB} - \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC}) = -\frac{1}{6}\overrightarrow{AB} - \frac{1}{2}\overrightarrow{AC}$.
(2)因为E，C，O三点共线，设$\overrightarrow{EO} = \lambda\overrightarrow{EC}(\lambda \in \mathbb{R})$，所以$\overrightarrow{AO} - \overrightarrow{AE} = \lambda(\overrightarrow{AC} - \overrightarrow{AE})$，$\overrightarrow{AO} = \lambda\overrightarrow{AC} + (1 - \lambda)\overrightarrow{AE}$.
由$\overrightarrow{BE} = 2\overrightarrow{EA}$得$\overrightarrow{AO} = \lambda\overrightarrow{AC} + (1 - \lambda)\frac{1}{3}\overrightarrow{AB}$.
设$\overrightarrow{AO} = \mu\overrightarrow{AD}(\mu \in \mathbb{R})$，所以$\overrightarrow{AO} = \frac{\mu}{2}\overrightarrow{AB} + \frac{\mu}{2}\overrightarrow{AC}$，
则有$\begin{cases} \lambda = \frac{\mu}{2} \\ 1 - \lambda = \frac{\mu}{3} \end{cases}$，解得$\lambda = \frac{1}{4}$，即$\overrightarrow{AO} = \frac{1}{4}\overrightarrow{AC} + \frac{1}{4}\overrightarrow{AB}$.
又G，O，H三点共线，设$\overrightarrow{AO} = m\overrightarrow{AG} + (1 - m)\overrightarrow{AH}(m \in \mathbb{R})$，又$\overrightarrow{AO} = m · \frac{2}{3}\overrightarrow{AB} + (1 - m) · \overrightarrow{AC}$，
所以$\begin{cases} \frac{2m}{3} = \frac{1}{4} \\ (1 - m)t = \frac{1}{4} \end{cases}$，解得$\begin{cases} m = \frac{3}{8} \\ t = \frac{2}{5} \end{cases}$.