《2025年热搜题高中数学必修第二册人教版》

1. [2024·江西临川一中月考]在直角三角形ABC中，D是斜边AB的中点，P为线段CD的中点，则$\frac{|PA|^{2}+|PB|^{2}}{|PC|^{2}}=$(D)

A.2
B.4
C.5
D.10

答案： 1.D [解析]将$\triangle ABC$各边及$PA$，$PB$，$PC$均用向量表示，则$\frac{|PA|^2 + |PB|^2}{|PC|^2} = \frac{\overrightarrow{PA^2} + \overrightarrow{PB^2}}{\overrightarrow{PC^2}} = \frac{(\overrightarrow{PC} + \overrightarrow{CA})^2 + (\overrightarrow{PC} + \overrightarrow{CB})^2}{\overrightarrow{PC^2}} = \frac{2|\overrightarrow{PC|^2} + 2\overrightarrow{PC} · (\overrightarrow{CA} + \overrightarrow{CB}) + |\overrightarrow{AB}|^2}{\overrightarrow{PC|^2}} = \frac{2|\overrightarrow{PC|^2} + 2\overrightarrow{PC} · 2\overrightarrow{CD} + 4|\overrightarrow{CD|^2}}{\overrightarrow{PC|^2}} = \frac{2|\overrightarrow{PC|^2} + 2\overrightarrow{PC} · 4\overrightarrow{CP} + 16|\overrightarrow{PC|^2}}{|\overrightarrow{PC|^2}} = -6 + 16 = 10$. 故选D.

2. [2024·衡水一中月考]已知H为△ABC的垂心，若$\overrightarrow{AH}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}$，则$\sin\angle BAC=$(C)

A.$\frac{\sqrt{15}}{5}$
B.$\frac{\sqrt{10}}{5}$
C.$\frac{\sqrt{6}}{3}$
D.$\frac{\sqrt{3}}{3}$

答案： 2.C [解析]依题意，$\overrightarrow{BH} = \overrightarrow{BA} + \overrightarrow{AH} = -\frac{2}{3}\overrightarrow{AB} + \frac{2}{5}\overrightarrow{AC}$，同理$\overrightarrow{CH} = \overrightarrow{CA} + \overrightarrow{AH} = \frac{1}{3}\overrightarrow{AB} - \frac{3}{5}\overrightarrow{AC}$. 由$H$为$\triangle ABC$的垂心，得$\overrightarrow{BH} · \overrightarrow{AC} = 0$，即$(-\frac{2}{3}\overrightarrow{AB} + \frac{2}{5}\overrightarrow{AC}) · \overrightarrow{AC} = 0$，则$\frac{2}{5}|\overrightarrow{AC}|^2 = \frac{2}{3}|\overrightarrow{AC}||\overrightarrow{AB}|\cos\angle BAC$，即$\cos\angle BAC = \frac{3|\overrightarrow{AC|}}{5|\overrightarrow{AB|}}$，同理有$\overrightarrow{CH} · \overrightarrow{AB} = 0$，即$(\frac{1}{3}\overrightarrow{AB} - \frac{3}{5}\overrightarrow{AC}) · \overrightarrow{AB} = 0$，则$\frac{1}{3}|\overrightarrow{AB}|^2 = \frac{3}{5}|\overrightarrow{AC}||\overrightarrow{AB}|\cos\angle BAC$，即$\cos\angle BAC = \frac{5|\overrightarrow{AB|}}{9|\overrightarrow{AC|}}$，解得$\cos^2\angle BAC = \frac{1}{3}$，则$\sin^2\angle BAC = 1 - \cos^2\angle BAC = 1 - \frac{1}{3} = \frac{2}{3}$. 又$\angle BAC \in (0, \pi)$，所以$\sin\angle BAC = \frac{\sqrt{6}}{3}$. 故选C.

3. [2024·扬州中学月考]在△ABC中，$\angle BAC = 60^{\circ}$，$\angle BAC$的平分线AD交边BC于点D，已知$AB = 3$，且$\overrightarrow{AD}=\frac{1}{3}\overrightarrow{AC}+\frac{2}{3}\overrightarrow{AB}$，则AD的长为(C)

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

答案：
3.C [解析]如图，过$D$作$DE // AC$交$AB$于点$E$，作$DF // AB$交$AC$于点$F$，则$\overrightarrow{AD} = \overrightarrow{AE} + \overrightarrow{AF}$. 因为$\overrightarrow{AD} = \frac{1}{3}\overrightarrow{AC} + \frac{2}{3}\overrightarrow{AB}$，所以$\overrightarrow{AE} = \frac{2}{3}\overrightarrow{AB}$，$\overrightarrow{AF} = \frac{1}{3}\overrightarrow{AC}$，所以$\frac{BD}{BC} = \frac{AF}{AC}$，即$\frac{BD}{DC} = \frac{1}{2}$. 又$AD$是$\angle BAC$的平分线，所以$\frac{AB}{AC} = \frac{BD}{CD} = \frac{1}{2}$，而$AB = 3$，所以$AC = 6$，所以$\overrightarrow{AB} · \overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}|\cos\angle BAC = 3 × 6 × \cos 60° = 9$，所以$\overrightarrow{AD^2} = (\frac{1}{3}\overrightarrow{AC} + \frac{2}{3}\overrightarrow{AB})^2 = \frac{1}{9}\overrightarrow{AC^2} + \frac{4}{9}\overrightarrow{AC} · \overrightarrow{AB} + \frac{4}{9}\overrightarrow{AB^2} = \frac{1}{9} × 6^2 + \frac{4}{9} × 9 + \frac{4}{9} × 3^2 = 12$，所以$|\overrightarrow{AD}| = 2\sqrt{3}$. 故选C.

4. [2024·重庆巴蜀中学月考]如图，在平行四边形ABCD中，E，F分别是AD，AB的中点，G为BE与DF的交点。若$\overrightarrow{AB}=\boldsymbol{a}$，$\overrightarrow{AD}=\boldsymbol{b}$，求证：A，G，C三点共线。

答案： 4.证明：$\overrightarrow{BE} = \overrightarrow{AE} - \overrightarrow{AB} = \frac{1}{2}\mathbf{b} - \mathbf{a}$，$\overrightarrow{DF} = \overrightarrow{AF} - \overrightarrow{AD} = \frac{1}{2}\mathbf{a} - \mathbf{b}$. 因为$D$，$G$，$F$三点共线，所以$\overrightarrow{DG} = \lambda\overrightarrow{DF}$，$\lambda \in \mathbf{R}$，即$\overrightarrow{AG} = \overrightarrow{AD} + \lambda\overrightarrow{DF} = \frac{1}{2}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}$. 因为$B$，$G$，$E$三点共线，所以$\overrightarrow{BG} = \mu\overrightarrow{BE}$，$\mu \in \mathbf{R}$，即$\overrightarrow{AG} = \overrightarrow{AB} + \mu\overrightarrow{BE} = (1 - \mu)\mathbf{a} + \frac{1}{2}\mu\mathbf{b}$，则$\begin{cases} \frac{1}{2}\lambda = 1 - \mu, \\ 1 - \lambda = \frac{1}{2}\mu, \end{cases}$解得$\lambda = \frac{2}{3}$，所以$\overrightarrow{AG} = \frac{1}{3}(\mathbf{a} + \mathbf{b}) = \frac{1}{3}\overrightarrow{AC}$，所以$A$，$G$，$C$三点共线.

5. [2024·黄冈中学单元检测]在等腰三角形ABC中，$AC\perp BC$，$AC = BC$，D是BC的中点，E是AB上一点，且$AE = 2EB$。求证：$AD\perp CE$。

答案：
5.证明：建立如图所示的平面直角坐标系，设$CA = CB = 2$，则$A(2, 0)$，$B(0, 2)$，$C(0, 0)$，设$E(x, y)$. 因为$D$为$BC$的中点，所以$D(0, 1)$. 因为$AE = 2EB$，所以$\overrightarrow{AE} = \frac{2}{3}\overrightarrow{AB}$，所以$(x - 2, y) = \frac{2}{3}(-2, 2)$，所以$\begin{cases} x - 2 = -\frac{4}{3}, \\ y = \frac{4}{3}, \end{cases}$解得$\begin{cases} x = \frac{2}{3}, \\ y = \frac{4}{3}, \end{cases}$所以$E(\frac{2}{3}, \frac{4}{3})$. 所以$\overrightarrow{AD} · \overrightarrow{CE} = (-2, 1) · (\frac{2}{3}, \frac{4}{3}) = -\frac{4}{3} + \frac{4}{3} = 0$，所以$\overrightarrow{AD} \perp \overrightarrow{CE}$，所以$AD \perp CE$.

6. [2024·江苏启东中学月考]已知△ABC是等腰直角三角形，$\angle ABC = 90^{\circ}$，D是BC边的中点，$BE\perp AD$，垂足为E，延长BE交AC于点F，连接DF，求证：$\angle ADB=\angle FDC$。

答案：
6.证明：如图，以$B$为原点，$BC$，$BA$所在直线分别为$x$轴、$y$轴建立平面直角坐标系. 设$A(0, 2)$，$C(2, 0)$，则$D(1, 0)$，$\overrightarrow{AC} = (2, -2)$. 设$\overrightarrow{AF} = \lambda\overrightarrow{AC}$，则$\overrightarrow{BF} = \overrightarrow{BA} + \overrightarrow{AF} = (0, 2) + (2\lambda, -2\lambda) = (2\lambda, 2 - 2\lambda)$. 又$\overrightarrow{DA} = (-1, 2)$，$\overrightarrow{BF} \perp \overrightarrow{DA}$，所以$\overrightarrow{BF} · \overrightarrow{DA} = 0$，所以$-2\lambda + 2(2 - 2\lambda) = 0$，解得$\lambda = \frac{2}{3}$，所以$\overrightarrow{BF} = (\frac{4}{3}, \frac{2}{3})$，所以$\overrightarrow{DF} = \overrightarrow{BF} - \overrightarrow{BD} = (\frac{1}{3}, \frac{2}{3})$，又$\overrightarrow{DC} = (1, 0)$，所以$\cos\angle FDC = \frac{\overrightarrow{DF} · \overrightarrow{DC}}{|\overrightarrow{DF}||\overrightarrow{DC}|} = \frac{\sqrt{5}}{5}$. 又$\cos\angle ADB = \frac{\overrightarrow{DA} · \overrightarrow{DB}}{|\overrightarrow{DA}||\overrightarrow{DB}|} = \frac{\sqrt{5}}{5}$，$\angle ADB$，$\angle FDC \in (0, \pi)$，所以$\angle ADB = \angle FDC$.

7. [2024·晋城一中月考]一个物体受到同一平面内三个力$\boldsymbol{F}_{1}$，$\boldsymbol{F}_{2}$，$\boldsymbol{F}_{3}$的作用，沿北偏东$45^{\circ}$方向移动了8m，已知$|\boldsymbol{F}_{1}| = 2N$，方向为北偏东$30^{\circ}$，$|\boldsymbol{F}_{2}| = 4N$，方向为北偏东$60^{\circ}$，$|\boldsymbol{F}_{3}| = 6N$，方向为北偏西$30^{\circ}$，则这三个力的合力所做的功为(D)

A.24J
B.$24\sqrt{2}$J
C.$24\sqrt{3}$J
D.$24\sqrt{6}$J

答案：
7.D [解析]如图所示，建立平面直角坐标系，则$F_1 = (1, \sqrt{3})$，$F_2 = (2\sqrt{3}, 2)$，$F_3 = (-3, 3\sqrt{3})$，则$\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = (2\sqrt{3} - 2, 2 + 4\sqrt{3})$. 由已知得位移$\mathbf{s} = (4\sqrt{2}, 4\sqrt{2})$，故合力$\mathbf{F}$所做的功为$W = \mathbf{F} · \mathbf{s} = (2\sqrt{3} - 2) × 4\sqrt{2} + (2 + 4\sqrt{3}) × 4\sqrt{2} = 6\sqrt{3} × 4\sqrt{2} = 24\sqrt{6} (J)$. 故选D.