《2025年小题狂做高中数学必修第二册人教版》

9. 在平面直角坐标系$xOy$中,已知点$A(1,1)$,$B(2,3)$,$C(3,2)$,$\overrightarrow{OP}=m\overrightarrow{AB}+n\overrightarrow{AC}(m,n\in\mathbf{R})$,则 (AC)

A.若$\overrightarrow{OP}//\overrightarrow{BC}$,则$m + n = 0$
B.若点$P$在$BC$上,则$m + n = 1$
C.若$\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=\boldsymbol{0}$,则$m - n = 0$
D.若$\overrightarrow{AP}$与$\overrightarrow{BC}$共线,则$m + n = -1$

答案： 9. AC 由题知，$\overrightarrow{AB} = (1,2)$，$\overrightarrow{BC} = (1,-1)$，$\overrightarrow{AC} = (2,1)$，所以$\overrightarrow{OP} = m\boldsymbol{a} + n\boldsymbol{b} = (m + 2n,2m + n)$. 对于$A$，因为$\overrightarrow{OP} // \overrightarrow{BC}$，所以$2m + n + m + 2n = 0$，即$m + n = 0$，$A$正确；对于$B$，$\overrightarrow{BP} = \overrightarrow{OP} - \overrightarrow{OB} = (m + 2n - 2,2m + n - 3)$，因为点$P$在$BC$上，所以$\overrightarrow{BP} // \overrightarrow{BC}$，所以$2m + n - 3 + m + 2n - 2 = 0$，即$m + n = \frac{5}{3}$，$B$错误；对于$C$，$\overrightarrow{PA} = (1 - m - 2n,1 - 2m - n)$，$\overrightarrow{PB} = (2 - m - 2n,3 - 2m - n)$，$\overrightarrow{PC} = (3 - m - 2n,2 - 2m - n)$，因为$\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = (0,0)$，即$(6 - 3m - 6n,6 - 6m - 3n) = (0,0)$，所以$\begin{cases}6 - 3m - 6n = 0,\\6 - 6m - 3n = 0,\end{cases}$解得$m = n = \frac{2}{3}$，所以$m - n = 0$，$C$正确；对于$D$，$\overrightarrow{AP} = (m + 2n - 1,2m + n - 1)$，$\overrightarrow{BC} = (1,-1)$，由$\overrightarrow{AP}$与$\overrightarrow{BC}$共线，得$(m + 2n - 1) × (-1) - (2m + n - 1) = 0$，整理得$m + n = \frac{2}{3}$，$D$错误.

10. 已知点$A(3,3)$,$B(-1,1)$,$C(0,-1)$,则$\overrightarrow{BA}-2\overrightarrow{BC}=$$(2,6)$.

答案： 10. $(2,6)$ 由题意可知$\overrightarrow{BA} = (4,2)$，$\overrightarrow{BC} = (1,-2)$，则$\overrightarrow{BA} - 2\overrightarrow{BC} = (2,6)$.

11. 已知$O$为坐标原点,点$A(x_1,y_1)$,$B(x_2,y_2)$,$C$在线段$AB$上,且$\frac{|\overrightarrow{AC}|}{|\overrightarrow{AB}|}=\lambda(0\lt\lambda\lt1)$,则点$C$的坐标为$((1 - \lambda)x_1 + \lambda x_2,(1 - \lambda)y_1 + \lambda y_2)$.

答案： 11. $((1 - \lambda)x_1 + \lambda x_2,(1 - \lambda)y_1 + \lambda y_2)$ 依题意$\overrightarrow{AC} = \lambda\overrightarrow{AB}$，则有$(x_C - x_1,y_C - y_1) = \lambda(x_2 - x_1,y_2 - y_1)$，即$x_C - x_1 = \lambda(x_2 - x_1)$，$y_C - y_1 = \lambda(y_2 - y_1)$，整理得$x_C = (1 - \lambda)x_1 + \lambda x_2$，$y_C = (1 - \lambda)y_1 + \lambda y_2$.

12. 在$\triangle ABC$中,$A(1,1)$,$B(-2,3)$,$C(3,-1)$,线段$AB$上的点$M$满足$AM = 2MB$,线段$AC$上的点$N$满足$CN = 3NA$,直线$MN$与$BC$相交于点$P$,则$\overrightarrow{AP}$的坐标为$(-4,\frac{14}{5})$.

答案： 12. $(-4,\frac{14}{5})$ 由题知，$\overrightarrow{AB} = (-3,2)$，$\overrightarrow{AC} = (2,-2)$，$\overrightarrow{AM} = \frac{2}{3}\overrightarrow{AB} = (-2,\frac{4}{3})$，$\overrightarrow{AN} = \frac{1}{4}\overrightarrow{AC} = (\frac{1}{2},-\frac{1}{2})$，则$M(-1,\frac{7}{3})$，$N(\frac{3}{2},\frac{1}{2})$. 由$P$，$B$，$C$三点共线，设$\overrightarrow{PB} = \lambda\overrightarrow{BC} = (5\lambda,-4\lambda)$，则$P(-2 - 5\lambda,3 + 4\lambda)$，所以$\overrightarrow{PM} = (5\lambda + 1,-4\lambda - \frac{2}{3})$，$\overrightarrow{PN} = (5\lambda + \frac{7}{2},-4\lambda - \frac{5}{2})$. 因为$P$，$M$，$N$三点共线，所以$\overrightarrow{PN} // \overrightarrow{PM}$，则$(5\lambda + 1)(-4\lambda - \frac{5}{2}) = (-4\lambda - \frac{2}{3})(5\lambda + \frac{7}{2})$，解得$\lambda = \frac{1}{5}$，所以$P(-3,\frac{19}{5})$，则$\overrightarrow{AP} = (-4,\frac{14}{5})$.

13. 已知向量$\boldsymbol{a}=(3,2)$,$\boldsymbol{b}=(-1,2)$,$\boldsymbol{c}=(4,1)$.
(1)若$\boldsymbol{a}=\lambda\boldsymbol{b}-\mu\boldsymbol{c}(\lambda,\mu\in\mathbf{R})$,求$\lambda+\mu$的值;
(2)若向量$\boldsymbol{a}+k\boldsymbol{b}$与向量$2\boldsymbol{b}-\boldsymbol{c}$共线,求实数$k$的值.

答案： 13. 解：
(1)因为$\boldsymbol{b} = (-\lambda,2\lambda)$，$\boldsymbol{\mu} \boldsymbol{c} = (4\mu,\mu)$，又$\boldsymbol{a} = \lambda\boldsymbol{b} - \mu\boldsymbol{c}$，所以$\begin{cases}-\lambda - 4\mu = 3,\\2\lambda - \mu = 2,\end{cases}$解得$\begin{cases}\lambda = \frac{5}{9},\\\mu = -\frac{8}{9}.\end{cases}$所以$\lambda + \mu = -\frac{1}{3}$.
(2)因为$\boldsymbol{a} + k\boldsymbol{b} = (3 - k,2 + 2k)$，$2\boldsymbol{b} - \boldsymbol{c} = (-6,3)$，且$\boldsymbol{a} + k\boldsymbol{b}$与$2\boldsymbol{b} - \boldsymbol{c}$共线，所以$3(3 - k) = -6(2 + 2k)$，解得$k = -\frac{7}{3}$.

14. (创新·新定义)已知集合$S_n=\{X|X=(x_1,x_2,·s,x_n),x_i\in\mathbf{N}^*,i = 1,2,·s,n\}(n\geqslant2)$.对于$A=(a_1,a_2,·s,a_n)$,$B=(b_1,b_2,·s,b_n)\in S_n$,给出如下定义:①$\overrightarrow{AB}=(b_1 - a_1,b_2 - a_2,·s,b_n - a_n)$;②$\lambda(a_1,a_2,·s,a_n)=(\lambda a_1,\lambda a_2,·s,\lambda a_n)(\lambda\in\mathbf{R})$;③$A$与$B$之间的距离为$d(A,B)=\sum_{i = 1}^{n}|a_i - b_i|$ (说明:“$(a_1,a_2,·s,a_n)=(b_1,b_2,·s,b_n)$”的充要条件是“$a_i = b_i(i = 1,2,·s,n)$”). 若$A$,$B$,$C\in S_n$,且存在$\lambda\gt0$,使得$\overrightarrow{AB}=\lambda\overrightarrow{BC}$,则下列关系式正确的是 (A)

A.$d(A,B)+d(B,C)=d(A,C)$
B.$d(A,B)+d(A,C)=d(B,C)$
C.$d(A,B)+d(B,C)\lt d(A,C)$
D.$d(A,B)+d(B,C)\gt d(A,C)$

答案： 14. A 设$C = (c_1,c_2,·s,c_n)$，由$\overrightarrow{AB} = \lambda\overrightarrow{BC}$，得$b_i - a_i = \lambda(c_i - b_i)$，又$\lambda ＞ 0$，所以$b_i - a_i$与$c_i - b_i$同号，所以$|c_i - a_i| = |(b_i - a_i) + (c_i - b_i)| = |b_i - a_i| + |c_i - b_i|$，所以$\sum_{i = 1}^{n}|c_i - a_i| + \sum_{i = 1}^{n}|c_i - b_i| = \sum_{i = 1}^{n}|c_i - a_i|$，即$d(A,B) + d(B,C) = d(A,C)$.