题目内容
3.已知平面内三个向量:$\overrightarrow a=(3,2),\overrightarrow b=(-1,2),\overrightarrow c=(4,1)$.(Ⅰ)若$(\overrightarrow a+k\overrightarrow c)∥(2\overrightarrow b-\overrightarrow a)$,求实数k的值;
(Ⅱ)设$\overrightarrow d=(x,y)$,且满足$(\overrightarrow a+\overrightarrow b)⊥(\overrightarrow d-\overrightarrow c)$,$|\overrightarrow d-\overrightarrow c|=\sqrt{5}$,求$\overrightarrow d$.
分析 (1)利用向量共线定理即可得出.
(2)利用向量垂直与数量积的关系、数量积运算性质即可得出.
解答 解:(1)因为$\overrightarrow a+k\overrightarrow c=(3,2)+k(4,1)=(3+4k,2+k)$,$2\overrightarrow b-\overrightarrow a=(-5,2)$,
又$(\overrightarrow a+k\overrightarrow c)∥(2\overrightarrow b-\overrightarrow a)$,
∴-5(2+k)=2(3+4k),解得k=-$\frac{16}{13}$.
(2)∵$\overrightarrow{a}+\overrightarrow{b}$=(2,4),$\overrightarrow{d}-\overrightarrow{c}$=(x-4,y-1),
又$(\overrightarrow a+\overrightarrow b)⊥(\overrightarrow d-\overrightarrow c)$,$|\overrightarrow d-\overrightarrow c|=\sqrt{5}$,
∴$\left\{\begin{array}{l}{2(x-4)+4(y-1)=0}\\{(x-4)^{2}+(y-1)^{2}=5}\end{array}\right.$,解得$\left\{\begin{array}{l}{x=6}\\{y=0}\end{array}\right.$,或$\left\{\begin{array}{l}{x=2}\\{y=2}\end{array}\right.$.
故$\overrightarrow{d}$=(6,0)或(2,2).
点评 本题考查了向量共线定理、向量垂直与数量积的关系、数量积运算性质,考查了推理能力与计算能力,属于基础题.
| A. | 3+$\sqrt{5}$ | B. | 5+$\sqrt{5}$ | C. | 5 | D. | 6 |
| A. | 2 | B. | $\frac{9}{2}$ | C. | 2 | D. | $\frac{1}{2}$ |
| A. | $\frac{1}{2}$ | B. | $\frac{9}{16}$ | C. | $\frac{11}{16}$ | D. | $\frac{13}{16}$ |
| A. | -$\frac{4}{5}$ | B. | $\frac{4}{5}$ | C. | ±$\frac{4}{5}$ | D. | $\frac{3}{5}$ |