《2025年小题狂做高中数学必修第二册人教版》

10. 已知$A(1,\sqrt{3}+1)$,$B(0,1)$,$C(\sqrt{3},2)$,则$\angle ABC=$$\frac{\pi}{6}$.

答案： 10.$\frac{\pi}{6}$ 依题意得,$\overrightarrow{BA} = (1, \sqrt{3})$,$\overrightarrow{BC} = (\sqrt{3}, 1)$,则$\overrightarrow{BA} · \overrightarrow{BC} = 1 × \sqrt{3} + \sqrt{3} × 1 = 2\sqrt{3}$,又$|\overrightarrow{BA}| = \sqrt{1^2 + (\sqrt{3})^2} = 2$,$|\overrightarrow{BC}| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$,所以$\cos \langle \overrightarrow{BA}, \overrightarrow{BC} \rangle = \frac{\overrightarrow{BA} · \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} = \frac{2\sqrt{3}}{2 × 2} = \frac{\sqrt{3}}{2}$,因为$0 \leq \langle \overrightarrow{BA}, \overrightarrow{BC} \rangle \leq \pi$,所以$\langle \overrightarrow{BA}, \overrightarrow{BC} \rangle = \frac{\pi}{6}$.

11. 已知$P$是菱形$ABCD$所在平面内的一点,若菱形的边长为定值,且$(\overrightarrow{PA}+\overrightarrow{PD})·(\overrightarrow{PB}+\overrightarrow{PC})$的最小值为$-8$,则该菱形的边长为$2\sqrt{2}$.

答案：
11.$2\sqrt{2}$ 设菱形ABCD的对角线交于点O,由$AC \perp BD$,可建立如图所示的平面直角坐标系.设$B(-a, 0)$,$D(a, 0)$,$A(0, b)$,$C(0, -b)$,$P(x, y)$,则$\overrightarrow{PA} = (-x, b - y)$,$\overrightarrow{PD} = (a - x, -y)$,$\overrightarrow{PB} = (-a - x, -y)$,$\overrightarrow{PC} = (-x, -b - y)$,所以$\overrightarrow{PA} + \overrightarrow{PD} = (a - 2x, b - 2y)$,$\overrightarrow{PB} + \overrightarrow{PC} = (-a - 2x, -b - 2y)$,则$(\overrightarrow{PA} + \overrightarrow{PD}) · (\overrightarrow{PB} + \overrightarrow{PC}) = (a - 2x)(-a - 2x) + (b - 2y)(-b - 2y) = 4x^2 + 4y^2 - a^2 - b^2 \geq -a^2 - b^2 = -|\overrightarrow{AB}|^2$,所以$-|\overrightarrow{AB}|^2 = -8$,所以$|\overrightarrow{AB}| = 2\sqrt{2}$.
　　　　　　　

12. 已知$A(2,0)$,$B(0,2)$,$C(\cos\alpha,\sin\alpha)$,$\alpha\in(0,\pi)$.若$|\overrightarrow{AC}|=|\overrightarrow{BC}|$,则$\alpha$的值为$\frac{\pi}{4}$;若$|\overrightarrow{OA}+\overrightarrow{OC}|=\sqrt{7}$,则$\overrightarrow{OB}$与$\overrightarrow{OC}$的夹角为$\frac{\pi}{6}$($O$为坐标原点).

答案： 12.$\frac{\pi}{4} \frac{\pi}{6}$ 由题$\overrightarrow{AC} = (\cos \alpha - 2, \sin \alpha)$,$\overrightarrow{BC} = (\cos \alpha, \sin \alpha - 2)$,所以$|\overrightarrow{AC}|^2 = (\cos \alpha - 2)^2 + \sin^2 \alpha = 5 - 4\cos \alpha$,$|\overrightarrow{BC}|^2 = \cos^2 \alpha + (\sin \alpha - 2)^2 = 5 - 4\sin \alpha$.又$|\overrightarrow{AC}| = |\overrightarrow{BC}|$,所以$5 - 4\cos \alpha = 5 - 4\sin \alpha$,得$\sin \alpha = \cos \alpha$,又因为$0 ＜ \alpha ＜ \pi$,所以$\alpha = \frac{\pi}{4}$.$|\overrightarrow{OA} + \overrightarrow{OC}| = \sqrt{7}$,两边平方得到$|\overrightarrow{OA}|^2 + |\overrightarrow{OC}|^2 + 2\overrightarrow{OA} · \overrightarrow{OC} = 7$.又$|\overrightarrow{OA}| = 4$,$|\overrightarrow{OC}| = 1$,所以$\overrightarrow{OA} · \overrightarrow{OC} = 1 = 2\cos \alpha$,即$\cos \alpha = \frac{1}{2}$,因为$0 ＜ \alpha ＜ \pi$,所以$\alpha = \frac{\pi}{3}$.设$\overrightarrow{OB}$与$\overrightarrow{OC}$的夹角为$\theta$,则$\cos \theta = \frac{\overrightarrow{OB} · \overrightarrow{OC}}{|\overrightarrow{OB}| · |\overrightarrow{OC}|} = \frac{2\sin \alpha}{2 × \frac{1}{2}} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,又$\theta \in (0, \pi)$,所以$\theta = \frac{\pi}{6}$,即$\overrightarrow{OB}$与$\overrightarrow{OC}$的夹角为$\frac{\pi}{6}$.

13. 如图,在四边形$OBCD$中,$\overrightarrow{OB}=2\overrightarrow{DC}$,$\overrightarrow{OA}=2\overrightarrow{AD}$,$OB\perp OD$,且$|\overrightarrow{AD}|=|\overrightarrow{CD}|=1$.
(1)用$\overrightarrow{OA}$,$\overrightarrow{OB}$表示$\overrightarrow{BC}$;
(2)若点$P$在线段$AC$上,且$\overrightarrow{AC}=2\overrightarrow{AP}$,求$\overrightarrow{BC}$与$\overrightarrow{BP}$夹角$\theta$的余弦值.

答案：
13.解:
(1)由$\overrightarrow{OB} = 2\overrightarrow{DC}$,$\overrightarrow{OA} = 2\overrightarrow{AD}$,得$\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OD} + \overrightarrow{DC} = -\overrightarrow{OB} + \frac{3}{2}\overrightarrow{OA} - \frac{1}{2}\overrightarrow{OB} = \frac{3}{2}\overrightarrow{OA} - \frac{1}{2}\overrightarrow{OB}$.
(2)因为$OB \perp OD$,所以以O为坐标原点,OB,OD所在直线分别为x轴,y轴,建立平面直角坐标系,如图所示.因为$|\overrightarrow{AD}| = |\overrightarrow{CD}| = 1$,所以$|\overrightarrow{OA}| = |\overrightarrow{OB}| = 2$,所以$A(2, 0)$,$B(0, 2)$,$C(3, 1)$,所以$\overrightarrow{AC} = (1, 1)$,$\overrightarrow{BC} = (3, -1)$.由$\overrightarrow{AC} = 2\overrightarrow{AP}$,可得$P(\frac{5}{2}, \frac{1}{2})$,故$\overrightarrow{BP} = (\frac{5}{2}, -\frac{3}{2})$,所以$\cos \theta = \frac{\overrightarrow{BC} · \overrightarrow{BP}}{|\overrightarrow{BC}| |\overrightarrow{BP}|} = \frac{\frac{15}{2} + \frac{3}{2}}{\sqrt{10} × \frac{\sqrt{34}}{2}} = \frac{9\sqrt{85}}{85}$.
　　　　　　　

14. 已知$x + 2y = 1$,用向量方法证明:$x^{2}+y^{2}\geqslant\frac{1},{5}$.

答案： 14.证明:构造向量$\boldsymbol{m} = (1, 2)$,$\boldsymbol{n} = (x, y)$,$\theta$为向量$\boldsymbol{m}$,$\boldsymbol{n}$的夹角.因为$\boldsymbol{m} · \boldsymbol{n} = |\boldsymbol{m}| |\boldsymbol{n}| \cos \theta$,所以$\boldsymbol{m} · \boldsymbol{n} \leq |\boldsymbol{m}| |\boldsymbol{n}|$,当且仅当$\cos \theta = 1$,即$\boldsymbol{m}$,$\boldsymbol{n}$同向时取等号,所以$x + 2y \leq \sqrt{1^2 + 2^2} · \sqrt{x^2 + y^2}$,当且仅当$\frac{x}{1} = \frac{y}{2}$,即$x = \frac{1}{5}$,$y = \frac{2}{5}$时取等号,所以$x^2 + y^2 \geq \frac{1}{5}$.