《2025年热搜题高中数学必修第二册人教版》

1. [2024·榆林一中月考]设向量$a=(1,2)$，$b=(x,1)$，当向量$a + 2b$与$2a - b$平行时，$a· b=$(A)

A.$\frac{5}{2}$
B.2
C.1
D.$\frac{7}{2}$

答案： 1.A【解析】因为$a + 2b = (1 + 2x,4)$，$2a - b = (2 - x,3)$，$a + 2b$与$2a - b$平行，所以$3(1 + 2x) = 4(2 - x)$，所以$x = \frac{1}{2}$，所以$a · b = (1,2) · (\frac{1}{2},1) = 1 × \frac{1}{2} + 2 × 1 = \frac{5}{2}$. 故选A.

2. [2024·海口一中月考]已知向量$a = (\sqrt{3},1)$，$b$是不平行于$x$轴的单位向量，且$a· b = \sqrt{3}$，则$b=$(B)

A.$(\frac{\sqrt{3}}{2},\frac{1}{2})$
B.$(\frac{1}{2},\frac{\sqrt{3}}{2})$
C.$(\frac{1}{4},\frac{3\sqrt{3}}{4})$
D.$(1,0)$

答案： 2.B【解析】设$b = (x,y)$，其中$y \neq 0$，则$a · b = \sqrt{3}x + y = \sqrt{3}$.由$\begin{cases} x^{2} + y^{2} = 1, \\ \sqrt{3}x + y = \sqrt{3}, \end{cases}$解得$\begin{cases} x = \frac{1}{2}, \\ y = \frac{\sqrt{3}}{2}. \end{cases}$即$b = (\frac{1}{2},\frac{\sqrt{3}}{2})$. 故选B.

3. [2024·江苏启东中学月考]已知点$A(-1,1)$，$B(1,2)$，$C(-2,-1)$，$D(3,4)$，$\overrightarrow{CD}$方向上的单位向量为$e$，则向量$\overrightarrow{AB}$在$\overrightarrow{CD}$上的投影向量为$\frac{3\sqrt{2}}{2}e$。

答案： 3.$\frac{3\sqrt{2}}{2}e$【解析】由已知得$\overrightarrow{AB} = (2,1)$，$\overrightarrow{CD} = (5,5)$，因此$\overrightarrow{AB}$在$\overrightarrow{CD}$上的投影向量为$\frac{\overrightarrow{AB} · \overrightarrow{CD}}{|\overrightarrow{CD}|}e = \frac{15}{5\sqrt{2}}e = \frac{3\sqrt{2}}{2}e$.

4. [2024·信阳质检]已知向量$a = (4,3)$，$2a + b = (3,18)$，则$a$，$b$夹角的余弦值为(C)

A.$\frac{8}{65}$
B.$-\frac{8}{65}$
C.$\frac{16}{65}$
D.$-\frac{16}{65}$

答案： 4.C【解析】因为$a = (4,3)$，所以$2a = (8,6)$. 又因为$2a + b = (3,18)$，所以$b = (-5,12)$，所以$a · b = -20 + 36 = 16$.又因为$|a| = 5$，$|b| = 13$，所以$\cos \langle a,b \rangle = \frac{16}{5 × 13} = \frac{16}{65}$. 故选C.

5. [2024·荆州中学期中]若向量$a = (1,2)$，$b = (1,-1)$，则$2a + b$与$a - b$的夹角等于(C)

A.$-\frac{\pi}{4}$
B.$\frac{\pi}{6}$
C.$\frac{\pi}{4}$
D.$\frac{3\pi}{4}$

答案： 5.C【解析】设向量$2a + b$与$a - b$的夹角为$\theta$. 因为$a = (1,2)$，$b = (1,-1)$，所以$2a + b = (3,3)$，$a - b = (0,3)$，则$|2a + b| = 3\sqrt{2}$，$|a - b| = 3$，所以$\cos \theta = \frac{(2a + b) · (a - b)}{|2a + b| · |a - b|} = \frac{3 × 0 + 3 × 3}{3\sqrt{2} × 3} = \frac{\sqrt{2}}{2}$，因为$0 \leq \theta \leq \pi$，所以$\theta = \frac{\pi}{4}$. 故选C.

6. [2024·青岛二中月考]已知向量$a = (1,2)$，$b = (-2,-4)$，$|c| = \sqrt{5}$。若$(a + b)· c = \frac{5}{2}$，则$a$与$c$的夹角为$120^{\circ}$。

答案： 6.$120^{\circ}$【解析】设$a$与$c$的夹角为$\theta$. 由$a + b = (-1,-2) = -a$，$|a| = \sqrt{5}$，得$\cos \theta = \frac{a · c}{|a||c|} = - \frac{(a + b) · c}{|a||c|} = \frac{5}{5} × \frac{1}{2}$. 因为$0^{\circ} \leq \theta \leq 180^{\circ}$，所以$\theta = 120^{\circ}$.

7. [2024·安康一中月考]已知平面向量$a = (2,4)$，$b = (-1,2)$。若$c = a - (a· b)b$，则$|c|=$(D)

A.$4\sqrt{2}$
B.$2\sqrt{5}$
C.8
D.$8\sqrt{2}$

答案： 7.D【解析】因为$a · b = 2 × (-1) + 4 × 2 = 6$，所以$c = (2,4) - 6(-1,2) = (8,-8)$，所以$|c| = \sqrt{8^{2} + (-8)^{2}} = 8\sqrt{2}$.

8. [2024·鞍山一中期中]已知平面向量$a$与$b$的夹角为$60^{\circ}$，$a = (2,0)$，$|b| = 1$，则$|a + 2b|=$(B)

A.$\sqrt{3}$
B.$2\sqrt{3}$
C.4
D.12

答案： 8.B【解析】由$a = (2,0)$，得$|a| = 2$，因为$|b| = 1$，所以$a · b = 2 × 1 × \cos 60^{\circ} = 1$，故$|a + 2b| = \sqrt{a^{2} + 4a · b + 4b^{2}} = 2\sqrt{3}$.

9. [2024·长沙雅礼中学月考]已知向量$a = (1,0)$，$b = (\cos\theta,\sin\theta)$，$\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$，则$|a + b|$的取值范围是(D)

A.$[0,\sqrt{2}]$
B.$[1,\sqrt{2}]$
C.$[1,2]$
D.$[\sqrt{2},2]$

答案： 9.D【解析】$|a + b| = \sqrt{(1 + \cos \theta)^{2} + \sin^{2} \theta} = \sqrt{2 + 2\cos \theta}$. 因为$\theta \in [-\frac{\pi}{2},\frac{\pi}{2}]$，所以$\cos \theta \in [0,1]$，所以$|a + b| \in [\sqrt{2},2]$.

10. [2024·宁波一中月考]已知$a = (2,1)$与$b = (1,2)$，要使$|a + tb|$最小，则实数$t$的值为$-\frac{4}{5}$。

答案： 10.$-\frac{4}{5}$【解析】因为$a + tb = (2 + t,1 + 2t)$，所以$|a + tb| = \sqrt{(t + 2)^{2} + (1 + 2t)^{2}} = \sqrt{5(t + \frac{4}{5})^{2} + \frac{9}{5}}$，所以当$t = -\frac{4}{5}$时，$|a + tb|$有最小值$\frac{3\sqrt{5}}{5}$.

11. (多选)已知向量$a = (-3,2)$，$b = (-1,0)$，则下列选项正确的有(ABD)

A.$(a + b)· b = 4$
B.$(a - 3b)\perp b$
C.$|a - b| = \sqrt{2}|b|$
D.$a^2 = b^2 + 4a· b$

答案： 11.ABD【解析】对于A，$a + b = (-4,2)$，所以$(a + b) · b = -4 × (-1) + 0 = 4$，故A正确；对于B，$a - 3b = (0,2)$，所以$(a - 3b) · b = 0$，所以$(a - 3b) \perp b$，故B正确；对于C，$a - b = (-2,2)$，$|a - b| = \sqrt{(-2)^{2} + 2^{2}} = 2\sqrt{2}$，$|b| = 1$，所以$|a - b| = 2\sqrt{2}|b|$，故C错误；对于D，$a^{2} = (-3)^{2} + 2^{2} = 13$，$b^{2} + 4a · b = 1 + 4[(-3) × (-1) + 0] = 13$，即$a^{2} = b^{2} + 4a · b$，故D正确. 故选ABD.

12. [2024·淮北一中月考]设向量$a = (1,2)$，$b = (-1,1)$，若向量$a + \lambda b$与向量$a$垂直，则实数$\lambda$的值为(D)

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

答案： 12.D【解析】由已知得$a + \lambda b = (1 - \lambda,2 + \lambda)$. 因为向量$a + \lambda b$与向量$a$垂直，所以$(a + \lambda b) · a = 0$，即$(1 - \lambda) × 1 + (2 + \lambda) × 2 = 0$，解得$\lambda = -5$. 故选D.

13. [2024·衡阳八中月考]设$x$，$y\in R$，向量$a = (2,x)$，$b = (3,y)$，$c = (1,-1)$，且$a\perp c$，$b// c$，则$|a + b|=$(C)

A.5
B.4
C.$\sqrt{26}$
D.$2\sqrt{5}$

答案： 13.C【解析】由$a \perp c$，$b // c$，可得$\begin{cases} 2 - x = 0, \\ y = -3, \end{cases}$解得$\begin{cases} x = 2, \\ y = -3, \end{cases}$所以$a = (2,2)$，$b = (3,-3)$，所以$a + b = (5,-1)$，所以$|a + b| = \sqrt{5^{2} + (-1)^{2}} = \sqrt{26}$. 故选C.

14. [2024·合肥一中月考]已知点$A(2,3)$，若把向量$\overrightarrow{OA}$绕原点$O$按逆时针方向旋转$90^{\circ}$得到向量$\overrightarrow{OB}$，则点$B$的坐标为$(-3,2)$。

答案： 14.$(-3,2)$【解析】设点B的坐标为$(x,y)$，因为$\overrightarrow{OA} \perp \overrightarrow{OB}$，$|\overrightarrow{OA}| = |\overrightarrow{OB}|$，所以$\begin{cases} 2x + 3y = 0, \\ x^{2} + y^{2} = 13, \end{cases}$解得$\begin{cases} x = -3, \\ y = 2 \end{cases}$或$\begin{cases} x = 3, \\ y = -2 \end{cases}$（舍去）. 故点B的坐标为$(-3,2)$.

15. [2024·郑州一中月考]已知向量$a$，$b$满足$a\perp b$，$|a + b| = t|a|$，若$a + b$与$a - b$的夹角为$\frac{2\pi}{3}$，则$t=$2。

答案： 15.2【解析】因为$a \perp b$，$|a + b| = t|a|$，所以$|a + b| = |a - b| = t|a|$，则$\cos \frac{2\pi}{3} = -\frac{1}{2} = \frac{(a + b) · (a - b)}{|a + b||a - b|} = \frac{|a|^{2} - |b|^{2}}{t^{2}|a|^{2}}$，化简可得$2|b|^{2} = (2 + t^{2})|a|^{2}$，所以$|b| = \sqrt{\frac{t^{2} + 2}{2}}|a|$，再由$|a|^{2} + |b|^{2} = t^{2}|a|^{2}$，$t ＞ 0$，解得$t = 2$.