《2025年小题狂做高中数学必修第二册人教版》

1. 已知向量$\boldsymbol{a}=(1,3)$,$\boldsymbol{b}=(-2,4)$,则下列结论正确的是 (C)

A.$\boldsymbol{a}·\boldsymbol{b}=14$
B.$|\boldsymbol{b}|=20$
C.$(\boldsymbol{a}-\boldsymbol{b})\perp\boldsymbol{a}$
D.$\cos\langle\boldsymbol{a},\boldsymbol{b}\rangle=\frac{\sqrt{2}}{5}$

答案： 1.C 对于A,$\boldsymbol{a} · \boldsymbol{b} = 1 × (-2) + 3 × 4 = 10$,故A错误;对于B,$|\boldsymbol{b}|^2 = (-2)^2 + 4^2 = 20$,则$|\boldsymbol{b}| = 2\sqrt{5}$,故B错误;对于C,由$\boldsymbol{a} - \boldsymbol{b} = (3, -1)$,得$(\boldsymbol{a} - \boldsymbol{b}) · \boldsymbol{a} = 3 × 1 + (-1) × 3 = 0$,所以$(\boldsymbol{a} - \boldsymbol{b}) \perp \boldsymbol{a}$,故C正确;对于D,$\cos \langle \boldsymbol{a}, \boldsymbol{b} \rangle = \frac{\boldsymbol{a} · \boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|} = \frac{10}{\sqrt{10} × 2\sqrt{5}} = \frac{\sqrt{2}}{2}$,故D错误.

2. 设$x,y\in\mathbf{R}$,向量$\boldsymbol{a}=(x,1)$,$\boldsymbol{b}=(1,y)$,$\boldsymbol{c}=(2,-4)$,且$\boldsymbol{a}\perp\boldsymbol{c}$,$\boldsymbol{b}//\boldsymbol{c}$,则$(\boldsymbol{a}+\boldsymbol{b})·(\boldsymbol{a}-\boldsymbol{c})=$ (C)

A.$-3$
B.$5$
C.$-5$
D.$15$

答案： 2.C 因为$\boldsymbol{a} \perp \boldsymbol{c}$,$\boldsymbol{b} // \boldsymbol{c}$,所以$\boldsymbol{a} · \boldsymbol{c} = 2x - 4 = 0$,$-4 - 2y = 0$,解得$x = 2$,$y = -2$.所以$\boldsymbol{a} + \boldsymbol{b} = (2, 1) + (1, -2) = (3, -1)$.$\boldsymbol{a} - \boldsymbol{c} = (2, 1) - (2, -4) = (0, 5)$.所以$(\boldsymbol{a} + \boldsymbol{b}) · (\boldsymbol{a} - \boldsymbol{c}) = 0 - 5 = -5$.

3. 已知向量$\boldsymbol{a}$,$\boldsymbol{b}$满足$|\boldsymbol{a}|=1$,$|\boldsymbol{b}|=2$,$\boldsymbol{a}-\boldsymbol{b}=(\sqrt{3},\sqrt{2})$,则$|2\boldsymbol{a}-\boldsymbol{b}|=$ (A)

A.$2\sqrt{2}$
B.$\sqrt{17}$
C.$\sqrt{15}$
D.$2\sqrt{5}$

答案： 3.A 由题意知$|\boldsymbol{a} - \boldsymbol{b}| = \sqrt{3 + 2} = \sqrt{5}$,且$(\boldsymbol{a} - \boldsymbol{b})^2 = \boldsymbol{a}^2 + \boldsymbol{b}^2 - 2\boldsymbol{a} · \boldsymbol{b} = 5 - 2\boldsymbol{a} · \boldsymbol{b} = 5$,所以$\boldsymbol{a} · \boldsymbol{b} = 0$,所以$(2\boldsymbol{a} - \boldsymbol{b})^2 = 4\boldsymbol{a}^2 + \boldsymbol{b}^2 - 4\boldsymbol{a} · \boldsymbol{b} = 4 + 4 = 8$,所以$|2\boldsymbol{a} - \boldsymbol{b}| = 2\sqrt{2}$.

4. 已知向量$\boldsymbol{a}=(x,3)$,$\boldsymbol{b}=(3,-\sqrt{3})$,且$\boldsymbol{a}$在$\boldsymbol{b}$上的投影向量为$\sqrt{3}\boldsymbol{b}$,则$x=$ (C)

A.$\sqrt{3}$
B.$2\sqrt{3}$
C.$5\sqrt{3}$
D.$4\sqrt{3}$

答案： 4.C 因为向量$\boldsymbol{a} = (x, 3)$,$\boldsymbol{b} = (3, -\sqrt{3})$,所以$\boldsymbol{a} · \boldsymbol{b} = 3x - 3\sqrt{3}$,$|\boldsymbol{b}| = \sqrt{3^2 + (-\sqrt{3})^2} = 2\sqrt{3}$,所以$\boldsymbol{a}$在$\boldsymbol{b}$上的投影向量为$\frac{\boldsymbol{a} · \boldsymbol{b}}{|\boldsymbol{b}|^2} · \boldsymbol{b} = \frac{3x - 3\sqrt{3}}{12} · \boldsymbol{b} = \sqrt{3}\boldsymbol{b}$,所以$\frac{3x - 3\sqrt{3}}{12} = \sqrt{3}$,解得$x = 5\sqrt{3}$.

5. (易错易混)已知向量$\boldsymbol{a}=(\sqrt{3},2\sqrt{3})$,$\boldsymbol{b}=(-\sqrt{3},\lambda)$,若$\boldsymbol{a}+\boldsymbol{b}$与$\boldsymbol{b}$的夹角为$\frac{2\pi}{3}$,则实数$\lambda=$ (A)

A.$-1$
B.$1$
C.$\pm1$
D.$\pm2$

答案： 5.A 因为$\boldsymbol{a} = (\sqrt{3}, 2\sqrt{3})$,$\boldsymbol{b} = (-\sqrt{3}, \lambda)$,所以$\boldsymbol{a} · \boldsymbol{b} = \sqrt{3} × (-\sqrt{3}) + 2\sqrt{3}\lambda = -3 + 2\sqrt{3}\lambda$,$\boldsymbol{a} + \boldsymbol{b} = (\sqrt{3}, 2\sqrt{3}) + (-\sqrt{3}, \lambda) = (0, 2\sqrt{3} + \lambda)$,$|\boldsymbol{a} + \boldsymbol{b}| = \sqrt{(2\sqrt{3} + \lambda)^2} = |2\sqrt{3} + \lambda|$.因为$(\boldsymbol{a} + \boldsymbol{b}) · \boldsymbol{b} = \boldsymbol{a} · \boldsymbol{b} + \boldsymbol{b}^2 = -3 + 2\sqrt{3}\lambda + 3 + \lambda^2 = \lambda^2 + 2\sqrt{3}\lambda$,又$(\boldsymbol{a} + \boldsymbol{b}) · \boldsymbol{b} = |\boldsymbol{a} + \boldsymbol{b}| |\boldsymbol{b}| \cos \frac{2\pi}{3} = |2\sqrt{3} + \lambda| × \sqrt{3 + \lambda^2} × (-\frac{1}{2})$,所以$\lambda^2 + 2\sqrt{3}\lambda = |2\sqrt{3} + \lambda| × \sqrt{3 + \lambda^2} × (-\frac{1}{2})$,解得$\lambda = 1$或$\lambda = -1$.因为$2\sqrt{3} + \lambda \neq 0$,所以$\lambda^2 + 2\sqrt{3}\lambda ＜ 0$,解得$-2\sqrt{3} ＜ \lambda ＜ 0$,所以$\lambda = -1$.
易错警示 含参的向量夹角问题,容易忽视对参数隐含的受限条件的处理.

6. 在$\triangle ABC$中,$AB\perp AC$,$AC = 6$,$\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AC}$,$E$是$BD$的中点,则$(\overrightarrow{AE}+\overrightarrow{BD})·\overrightarrow{CD}=$ (B)

A.$-8$
B.$-12$
C.$8$
D.$12$

答案：
6.B 解法1 如图,以A为原点,AB所在直线为x轴,AC所在直线为y轴建立平面直角坐标系,则$A(0, 0)$,$D(0, 2)$,$C(0, 6)$,设$B(b, 0)$,则$E(\frac{b}{2}, 1)$,$\overrightarrow{AE} = (\frac{b}{2}, 1)$,$\overrightarrow{BD} = (-b, 2)$,$\overrightarrow{CD} = (0, -4)$,所以$(\overrightarrow{AE} + \overrightarrow{BD}) · \overrightarrow{CD} = (-\frac{1}{2}b, 3) · (0, -4) = -12$.
　　　　　　　
解法2 因为$AB \perp AC$,$\overrightarrow{AD} = \frac{1}{3}\overrightarrow{AC}$,E是BD的中点,所以$\overrightarrow{AE}$和$\overrightarrow{BD}$在$\overrightarrow{CD}$上的投影向量分别为$\frac{1}{2}\overrightarrow{AD} = \frac{1}{6}\overrightarrow{AC}$和$\overrightarrow{AD} = \frac{1}{3}\overrightarrow{AC}$.又$\overrightarrow{CD} = -\frac{2}{3}\overrightarrow{AC}$,所以$(\overrightarrow{AE} + \overrightarrow{BD}) · \overrightarrow{CD} = (\frac{1}{6}\overrightarrow{AC} + \frac{1}{3}\overrightarrow{AC}) · (-\frac{2}{3}\overrightarrow{AC}) = \frac{1}{2} × (-\frac{2}{3})\overrightarrow{AC}^2 = -\frac{1}{3} × 36 = -12$.
方法总结 当题目中具有垂直条件时,可建立平面直角坐标系,用坐标将向量表示出来(“坐标法”),不建系的话也可用“基底法”.

7. (易错易混)已知平面向量$\boldsymbol{a}=(m,m + 2)$,$m\in\mathbf{R}$,$\boldsymbol{b}=(3,4)$,则下列结论正确的是 (AC)

A.$|\boldsymbol{a}|$的取值范围是$[\sqrt{2},+\infty)$
B.不存在一个实数$m$,使得$|\boldsymbol{a}+\boldsymbol{b}|=|\boldsymbol{a}-\boldsymbol{b}|$
C.若与$\boldsymbol{a}$共线的单位向量为$(\frac{3},{5},\frac{4},{5})$,则$m = 6$
D.若$\boldsymbol{a}$与$\boldsymbol{b}$的夹角为锐角,则$m＞-\frac{8},{7}$

答案： 7.AC 对于A,由题得$|\boldsymbol{a}| = \sqrt{m^2 + (m + 2)^2} = \sqrt{2m^2 + 4m + 4} = \sqrt{2(m + 1)^2 + 2}$,易知当$m = -1$时,$|\boldsymbol{a}|$取得最小值$\sqrt{2}$,故A正确;对于B,若$|\boldsymbol{a} + \boldsymbol{b}| = |\boldsymbol{a} - \boldsymbol{b}|$,则有$\boldsymbol{a}^2 + 2\boldsymbol{a} · \boldsymbol{b} + \boldsymbol{b}^2 = \boldsymbol{a}^2 - 2\boldsymbol{a} · \boldsymbol{b} + \boldsymbol{b}^2$,可得$\boldsymbol{a} · \boldsymbol{b} = 0$,则$3m + 4m + 8 = 0$,解得$m = -\frac{8}{7}$,故B错误;对于C,由$\boldsymbol{a}$与$(\frac{3}{5}, \frac{4}{5})$共线,得$\frac{3}{5}(m + 2) = \frac{4}{5}m$,解得$m = 6$,故C正确;对于D,由$\boldsymbol{a}$与$\boldsymbol{b}$的夹角为锐角,可得$\boldsymbol{a} · \boldsymbol{b} ＞ 0$且$\boldsymbol{a}$与$\boldsymbol{b}$不同向,所以$3m + 4m + 8 ＞ 0$且$3(m + 2) \neq 4m$,解得$m ＞ -\frac{8}{7}$且$m \neq 6$,故D错误.
易错警示 向量数量积大于0是夹角为锐角的必要不充分条件,因为向量同向共线时,数量积也大于0;向量的数量积为0是两个向量垂直的必要不充分条件,因为垂直的前提条件是非零向量.

8. 设非零向量$\boldsymbol{a}$,$\boldsymbol{b}$的夹角为$\theta$,定义运算$\boldsymbol{a}*\boldsymbol{b}=|\boldsymbol{a}||\boldsymbol{b}|\sin\theta$,下列说法正确的是 (AC)

A.若$\boldsymbol{a}=(1,1)$,$\boldsymbol{b}=(-1,1)$,则$\boldsymbol{a}*\boldsymbol{b}=2$
B.$\boldsymbol{a}*\boldsymbol{b}\geqslant|\boldsymbol{a}||\boldsymbol{b}|$
C.若$\boldsymbol{a}*\boldsymbol{b}=0$,则$\boldsymbol{a}//\boldsymbol{b}$
D.$\boldsymbol{a}*(\boldsymbol{b}+\boldsymbol{c})=\boldsymbol{a}*\boldsymbol{b}+\boldsymbol{a}*\boldsymbol{c}$

答案： 8.AC 对于A,$\boldsymbol{a} · \boldsymbol{b} = 0$,则$\boldsymbol{a} \perp \boldsymbol{b}$,所以$\theta = \frac{\pi}{2}$,所以$\boldsymbol{a} * \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \sin \theta = \sqrt{2} × \sqrt{2} × 1 = 2$,故A正确;对于B,因为$\sin \theta \leq 1$,所以$\boldsymbol{a} * \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \sin \theta \leq |\boldsymbol{a}| |\boldsymbol{b}|$,故B错误;对于C,若$\boldsymbol{a} * \boldsymbol{b} = 0$,故$\sin \theta = 0$,所以$\theta = 0$或$\pi$,所以$\boldsymbol{a} // \boldsymbol{b}$,故C正确;对于D,若$\boldsymbol{c} = -\boldsymbol{b}$,则$\boldsymbol{a} * (\boldsymbol{b} + \boldsymbol{c}) = 0$,$\boldsymbol{a} * \boldsymbol{b} + \boldsymbol{a} * (-\boldsymbol{b}) = |\boldsymbol{a}| |\boldsymbol{b}| \sin \theta + |\boldsymbol{a}| |\boldsymbol{b}| \sin (\pi - \theta) = 2|\boldsymbol{a}| |\boldsymbol{b}| \sin \theta$,不一定相等,故D错误.

9. 已知向量$\boldsymbol{a}=(\sin\theta,\cos\theta)$,$\boldsymbol{b}=(1,\sqrt{3})$,$\boldsymbol{c}=(3,\sqrt{3})$,则 (ACD)

A.若$\boldsymbol{a}\perp\boldsymbol{b}$,则$\tan\theta=-\sqrt{3}$
B.$\boldsymbol{c}$在$\boldsymbol{b}$上的投影向量为$\frac{1},{2}\boldsymbol{b}$
C.存在$\theta$,使得$\boldsymbol{a}$在$\boldsymbol{c}-\boldsymbol{b}$上投影向量的模为 1
D.$|\boldsymbol{a}-\boldsymbol{b}|$的取值范围为$[1,3]$

答案： 9.ACD 对于A,若$\boldsymbol{a} \perp \boldsymbol{b}$,则$\boldsymbol{a} · \boldsymbol{b} = \sin \theta + \sqrt{3} \cos \theta = 2\sin (\theta + \frac{\pi}{3}) = 0$,则$\theta + \frac{\pi}{3} = k\pi (k \in \mathbf{Z})$,即$\theta = k\pi - \frac{\pi}{3} (k \in \mathbf{Z})$,所以$\tan \theta = \tan (k\pi - \frac{\pi}{3}) = -\sqrt{3} (k \in \mathbf{Z})$,故A正确;对于B,$\boldsymbol{c}$在$\boldsymbol{b}$上的投影向量为$\frac{\boldsymbol{c} · \boldsymbol{b}}{|\boldsymbol{b}|^2} \boldsymbol{b} = \frac{3 + 3}{4} \boldsymbol{b} = \frac{3}{2} \boldsymbol{b}$,故B错误;对于C,$\boldsymbol{a}$在$\boldsymbol{c} - \boldsymbol{b}$上的投影向量的模为$\frac{|\boldsymbol{a} · (\boldsymbol{c} - \boldsymbol{b})|}{|\boldsymbol{c} - \boldsymbol{b}|} = \frac{|2\sin \theta|}{\sqrt{(3 - 1)^2 + (\sqrt{3} - 3)^2}} = \frac{|2\sin \theta|}{2\sqrt{2}} = |\sin \theta|$,若$|\sin \theta| = 1$,则$\theta = \frac{\pi}{2} + k\pi (k \in \mathbf{Z})$,所以存在$\theta$,使得$\boldsymbol{a}$在$\boldsymbol{c} - \boldsymbol{b}$上的投影向量的模为1,故C正确;对于D,$|\boldsymbol{a} - \boldsymbol{b}| = \sqrt{(\sin \theta - 1)^2 + (\cos \theta - \sqrt{3})^2} = \sqrt{5 - 2\sin \theta - 2\sqrt{3} \cos \theta} = \sqrt{5 - 4\sin (\theta + \frac{\pi}{3})}$,因为$-1 \leq \sin (\theta + \frac{\pi}{3}) \leq 1$,所以$1 \leq 5 - 4\sin (\theta + \frac{\pi}{3}) \leq 9$,所以$1 \leq |\boldsymbol{a} - \boldsymbol{b}| \leq 3$,故D正确.