6.2.3 向量的数乘运算

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必修第二册同步巩固,难度2颗星!

基础知识

向量数乘运算

一般地,我们规定实数\(λ\)与向量\(\vec{a}\)的积是一个向量,这种运算叫做向量的数乘,记作\(\lambda \vec{a}\)
它的长度与方向规定如下:
  (1) \(|\lambda \vec{a}|=|\lambda||\vec{a}|\)
  (2) 当\(λ>0\)时, \(\lambda \vec{a}\)的方向与\(\vec{a}\)的方向相同;当\(λ<0\) 时,\(\lambda \vec{a}\)的方向与\(\vec{a}\)方向相反.
 

【例】若已知向量\(\vec{a}\)如下图,且\(|\vec{a}|=2\),作出 \(\vec{b}=-2 \vec{a}\)\(\vec{c}=\dfrac{1}{2} \vec{a}\),并求出 \(|\vec{b}|\)\(|\vec{c}|\).
image.png
向量\(\vec{b}\)\(\vec{c}\)如下图,其中\(|\vec{b}|=|-2 \vec{a}|=2|\vec{a}|=4\)\(|\vec{c}|=\left|\dfrac{1}{2} \vec{a}\right|=\dfrac{1}{2}|\vec{a}|=1\).
image.png image.png
 

线性运算

向量的加、减、数乘运算统称为向量的线性运算.向量线性运算的结果仍是向量,
对于任意向量\(\vec{a}\)\(\vec{b}\)以及任意实数\(λ\)\(μ_1\)\(μ_2\),恒有 \(\lambda\left(\mu_1 \vec{a} \pm \mu_2 \vec{b}\right)=\lambda \mu_1 \vec{a} \pm \lambda \mu_2 \vec{b}\).
【例】 计算 \(4(\vec{a}+\vec{b})-3(\vec{a}-\vec{b})-\vec{a}\).
\(4(\vec{a}+\vec{b})-3(\vec{a}-\vec{b})-\vec{a}=4 \vec{a}+4 \vec{b}-3 \vec{a}+3 \vec{b}-\vec{a}=7 \vec{b}\).
 

两个向量共线

共线定理 非零向量\(\vec{a}\)与向量\(\vec{b}\)共线⇔有且只有一个实数\(λ\),使得 \(\vec{b}=\lambda \vec{a}\).
\(λ>0\)时 , \(\lambda \vec{a}\)的方向与\(\vec{a}\)的方向相同;
\(λ<0\) 时,\(\lambda \vec{a}\)的方向与\(\vec{a}\)方向相反;
\(λ=0\) 时,\(\lambda \vec{a}=\overrightarrow{0}\).
 

三点共线定理

\(\overrightarrow{O C}=x \overrightarrow{O A}+y \overrightarrow{O B}\)
  (1) 如图一,若\(A\)\(B\)\(C\)三点共线,则\(x+y=1\)
  (2) 如图二,若点\(O\)和点\(C\)\(AB\)同侧,则\(x+y<1\)
  (3) 如图三,若点\(O\)和点\(C\)\(AB\)异侧,则\(x+y>1\)

特殊的,在三角形$∆ABC$中,点$D$是$BC$的中点,则$\overrightarrow{A D}=\dfrac{1}{2} \overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A C}$.  

基本方法

【题型1】 向量的数乘

【典题1】 计算:\(3(6 \vec{a}+\vec{b})-9\left(\vec{a}+\dfrac{1}{3} \vec{b}\right)\).
解析 原式 \(=18 \vec{a}+3 \vec{b}-9 \vec{a}-3 \vec{b}=9 \vec{a}\).
 

【典题2】\(C\)在线段\(AB\)上,且\(|\overrightarrow{A C}|=\dfrac{2}{3}|\overrightarrow{C B}|\),若 \(\overrightarrow{A B}=\lambda \overrightarrow{B C}\),则\(λ=\)(  )
 A. \(\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) B. \(-\dfrac{2}{3}\)\(\qquad \qquad \qquad \qquad\) C.\(\dfrac{5}{3}\) \(\qquad \qquad \qquad \qquad\) D.\(-\dfrac{5}{3}\)
解析 \(C\)在线段\(AB\)上,且\(|\overrightarrow{A C}|=\dfrac{2}{3}|\overrightarrow{C B}|\),如图所示;
image.png
\(\overrightarrow{A B}=\lambda \overrightarrow{B C}\),即\(\overrightarrow{A B}=-\dfrac{5}{3} \overrightarrow{B C}\);所以\(\lambda=-\dfrac{5}{3}\)
故选:\(D\)
 

【巩固练习】

1.已知\(λ\)\(μ∈R\),则下面关系正确的 是(  )
 A.\(\lambda \vec{a}\)\(\vec{a}\)同向 \(\qquad \qquad \qquad \qquad \qquad \qquad\) B .\(0 \cdot \vec{a}=0\)
 C.\((\lambda+\mu) \vec{a}=\lambda \vec{a}+\mu \vec{a}\) \(\qquad \qquad \qquad \qquad\) D.若\(\vec{b}=\lambda \vec{a}\),则 \(|\vec{b}|=\lambda|\vec{a}|\)
 

2.计算 \(2(5 \vec{a}-4 \vec{b}+\vec{c})-3(\vec{a}-3 \vec{b}+\vec{c})-7 \vec{a}=\)\(\underline{\quad \quad}\)
 

3.点\(C\)在线段\(AB\)上,且\(\dfrac{A C}{C B}=\dfrac{3}{2}\),则 \(\overrightarrow{A C}=\) \(\underline{\quad \quad}\)\(\overrightarrow{A B}\)\(\overrightarrow{B C} =\)\(\underline{\quad \quad}\) \(\overrightarrow{A B}\)
 

参考答案

  1. 答案 \(C\)
    解析 \(\vec{a} \neq 0\)\(λ<0\)时,\(\lambda \vec{a}\)\(\vec{a}\)反向,且\(\lambda|\vec{a}|<0,\),则\(A\)\(D\)错误.
    \(\because 0 \cdot \vec{a}\)的结果为\(\vec{0}\),则\(B\)错误.由运算律知\(C\)正确.

  2. 答案 \(\vec{b}-\vec{c}\)
    解析 原式\(=10 \vec{a}-8 \vec{b}+2 \vec{c}-3 \vec{a}+9 \vec{b}-3 \vec{c}-7 \vec{a}=\vec{b}-\vec{c}\)

  3. 答案 \(\dfrac{3}{5}\)\(-\dfrac{2}{5}\)
    解析 \(\because \dfrac{A C}{C B}=\dfrac{3}{2}\)\(\therefore \overrightarrow{A C}=\dfrac{3}{5} \overrightarrow{A B}\)\(\overrightarrow{B C}=-\dfrac{2}{5} \overrightarrow{A B}\)
    故答案为: \(\dfrac{3}{5}\)\(-\dfrac{2}{5}\)
     

【题型2】 向量线性运算

【典题1】 如图,\(AB\)是圆\(O\)的一条直径,\(C\)\(D\)是半圆弧的两个三等分点,则\(\overrightarrow{A B}=\)(  )
image.png
 A.\(\overrightarrow{A C}-\overrightarrow{A D}\) \(\qquad \qquad\) B.\(2 \overrightarrow{A C}-2 \overrightarrow{A D}\) \(\qquad \qquad\) C.\(\overrightarrow{A D}-\overrightarrow{A C}\) \(\qquad \qquad\) D. \(2 \overrightarrow{A D}-2 \overrightarrow{A C}\)
解析 \(∵C\)\(D\)是半圆弧的两个三等分点,
\(∴CD//AB\),且\(AB=2CD\)
\(\therefore \overrightarrow{A B}=2 \overrightarrow{C D}=2(\overrightarrow{A D}-\overrightarrow{A C})=2 \overrightarrow{A D}-2 \overrightarrow{A C}\)
故选:\(D\)
 
【典题2】\(△ABC\)中,\(D\)\(E\)分别为边\(AB\)\(AC\)的中点,\(BE\)\(CD\)交于点\(P\),设\(\overrightarrow{A B}=\vec{a}\)\(\overrightarrow{A C}=\vec{b}\),则 \(\overrightarrow{A P}=\)(  )
 A. \(\dfrac{1}{3} \vec{a}+\dfrac{1}{3} \vec{b}\) \(\qquad \qquad\) B. \(\dfrac{2}{3} \vec{a}+\dfrac{2}{3} \vec{b}\) \(\qquad \qquad\) C.\(\dfrac{3}{4} \vec{a}+\dfrac{3}{4} \vec{b}\) \(\qquad \qquad\) D. \(\dfrac{1}{6} \vec{a}+\dfrac{1}{6} \vec{b}\)
image.png
解析 方法1 首尾相接法
\(\overrightarrow{A P}=\overrightarrow{A B}+\overrightarrow{B P}=\overrightarrow{A B}+\lambda \overrightarrow{B E}=\overrightarrow{A B}+\lambda(\overrightarrow{B A}+\overrightarrow{A E})=\overrightarrow{A B}+\lambda\left(\overrightarrow{B A}+\dfrac{1}{2} \overrightarrow{A C}\right)\)
\(=(1-\lambda) \overrightarrow{A B}+\dfrac{\lambda}{2} \overrightarrow{A C}=(1-\lambda) \vec{a}+\dfrac{\lambda}{2} \vec{b}\) ,其中\(\lambda=\dfrac{B P}{B E}\)
如图过点\(E\)\(EF/ /AB\)
\(∵E\)\(D\)是中点, \(\therefore E F=\dfrac{1}{2} A D=\dfrac{1}{2} B D\)
\(\therefore \dfrac{B P}{B E}=\dfrac{2}{3}\), 即 \(\lambda=\dfrac{2}{3}\)
\(\therefore \overrightarrow{A P}=(1-\lambda) \vec{a}+\dfrac{\lambda}{2} \vec{b}=\dfrac{1}{3} \vec{a}+\dfrac{1}{3} \vec{b}\)
image.png
方法2 构造平行四边形法
image.png
过点\(P\)分别作\(PH//AB\)\(PG//AC\),则四边形\(AGPH\)是平行四边形,
\(\overrightarrow{A P}=\overrightarrow{A G}+\overrightarrow{A H}=x \overrightarrow{A B}+y \overrightarrow{A C}=x \vec{a}+y \vec{b}\)
由方法\(1\)可得\(\dfrac{B P}{B E}=\dfrac{2}{3}\)
\(\because x=\dfrac{A G}{A B}=\dfrac{P E}{B E}=\dfrac{B E-B P}{B E}=1-\dfrac{B P}{B E}=1-\dfrac{2}{3}=\dfrac{1}{3}\)
同理可得\(y=\dfrac{1}{3}\)
\(\therefore \overrightarrow{A P}=\dfrac{1}{3} \vec{a}+\dfrac{1}{3} \vec{b}\).
点拨 用两个向量表示一个向量,方法很多,常见的有首尾相接法、构造平行四边形法或后面的坐标法,它们多多少少与平面几何内容扯上关系.
 

【典题3】\(O\)\(△ABC\)的内部,且满足 \(\overrightarrow{O A}+2 \overrightarrow{O B}+4 \overrightarrow{O C}=\overrightarrow{0}\),则\(△ABC\)的面积与\(△AOC\)的面积之比是\(\underline{\quad \quad}\) .
解析 如图所示,
image.png
\(OD=4OC\),以\(OA\)\(OD\)为邻边作平行四边形\(OAED\)
连接\(AD\)\(OE\),交于点\(M\)\(OE\)\(AC\)于点\(N\)
\(∵\)满足\(\overrightarrow{O A}+2 \overrightarrow{O B}+4 \overrightarrow{O C}=\overrightarrow{0}\)
\(\therefore \overrightarrow{O E}=-2 \overrightarrow{O B}\)\(\therefore \overrightarrow{O A}+4 \overrightarrow{O C}=-2 \overrightarrow{O B}\)
\(\therefore \overrightarrow{O N}=\dfrac{1}{5} \overrightarrow{O E}=-\dfrac{2}{5} \overrightarrow{O B}\)\(\therefore|\overrightarrow{O N}|=\dfrac{2}{5}|\overrightarrow{O B}|=\dfrac{2}{7}|\overrightarrow{B N}|\)
\(∴△ABC\)的面积与\(△AOC\)的面积之比是\(7:2\)
点拨 \(\overrightarrow{A B}=\lambda \overrightarrow{C D}\),意味着\(AB||CD\)\(AB=|λ|CD\);即在某些场景中,求两线段长度之比或两三角形面积之比,均可转化为两共线向量的关系.
 

【巩固练习】

1.如图,\(P\)\(Q\)是线段\(AB\)的三等分点,若\(\overrightarrow{O A}=\vec{a}\)\(\overrightarrow{O B}=\vec{b}\),则 \(\overrightarrow{O P}-\overrightarrow{O Q}=\)(  )
image.png
 A.\(\dfrac{1}{3}(\vec{a}-\vec{b})\) \(\qquad \qquad\) B.\(-\dfrac{1}{3}(\vec{a}-\vec{b})\) \(\qquad \qquad\) C.\(\dfrac{1}{3}(\vec{a}+\vec{b})\) \(\qquad \qquad\) D. \(-\dfrac{1}{3}(\vec{a}-\vec{b})\)
 

2.如图,在平行四边形\(ABCD\)中,设\(\overrightarrow{A B}=\vec{a}\)\(\overrightarrow{A D}=\vec{b}\)\(P\)为边\(BC\)的中点,则 \(\overrightarrow{A P}=\)(  )
image.png
 A. \(\vec{a}+\dfrac{\vec{b}}{2}\) \(\qquad \qquad \qquad \qquad\) B.\(\vec{a}-\dfrac{\vec{b}}{2}\) \(\qquad \qquad \qquad \qquad\) C. \(\vec{b}+\dfrac{\vec{a}}{2}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{\vec{a}}{2}-\vec{b}\)
 

3.如图,在\(▱OACB\)中,\(E\)\(AC\)的中点,\(F\)\(BC\)上的一点,且\(BC=3BF\),若\(\overrightarrow{O C}=m \overrightarrow{O E}+n \overrightarrow{O F}\),其中\(m\)\(n∈R\),则\(m+n\)的值为(  )
image.png
 A.\(1\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{7}{5}\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{7}{3}\)
 

4.如图,\(AB\)是圆\(O\)的直径,\(C\)\(D\)是圆\(O\)上的点,\(∠CBA=60°\)\(∠ABD=45°\)\(\overrightarrow{C D}=x \overrightarrow{O A}+y \overrightarrow{B C}\),则\(x+y=\)\(\underline{\quad \quad}\)
image.png
 

5.在梯形\(ABCD\)中,\(\overrightarrow{A B}=2 \overrightarrow{D C}\)\(\overrightarrow{B E}=\dfrac{1}{3} \overrightarrow{B C}\)\(P\)为线段\(DE\)上的动点(包括端点),且\(\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{B C}(\lambda, \mu \in \boldsymbol{R})\),则\(λ^2+μ\)的最小值为\(\underline{\quad \quad}\) .
 

6.已知点\(P\)\(△ABC\)内一点,且\(\overrightarrow{B A}+\overrightarrow{B C}=6 \overrightarrow{B P}\),则 \(\dfrac{S_{\triangle A B P}}{S_{\triangle A C P}}=\)\(\underline{\quad \quad}\) .
 

7.设\(G\)\(△ABC\)的重心,\(a\)\(b\)\(c\)分别是角\(A\)\(B\)\(C\)所对的边,若\(a \overrightarrow{G A}+b \overrightarrow{G B}+c \overrightarrow{G C}=\overrightarrow{0}\),则\(△ABC\)的形状是\(\underline{\quad \quad}\) .
 
 
 

参考答案

  1. 答案 \(A\)
    解析 \(∵P\)\(Q\)是线段\(AB\)的三等分点,\(\therefore \overrightarrow{P Q}=\dfrac{1}{3} \overrightarrow{A B}\)
    \(\therefore \overrightarrow{O P}-\overrightarrow{O Q}=\overrightarrow{Q P}=-\dfrac{1}{3} \overrightarrow{A B}=-\dfrac{1}{3}(\overrightarrow{O B}-\overrightarrow{O A})=\dfrac{1}{3}(\vec{a}-\vec{b})\)
    故选:\(A\)

  2. 答案 \(A\)
    解析 因为设\(\overrightarrow{A B}=\vec{a}\)\(\overrightarrow{A D}=\vec{b}\)\(P\)为边\(BC\)的中点,则 \(\overrightarrow{A P}=\overrightarrow{A B}+\overrightarrow{B P}=\overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}=\vec{a}+\dfrac{\vec{b}}{2}\)
    \(A\).

  3. 答案 \(C\)
    解析 因为\(\overrightarrow{O F}=\overrightarrow{O B}+\overrightarrow{B F}=\overrightarrow{O B}+\dfrac{1}{3} \overrightarrow{O A}\)\(\overrightarrow{O E}=\overrightarrow{O A}+\overrightarrow{A E}=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{O B}\)
    所以\(\overrightarrow{O A}=\dfrac{6}{5} \overrightarrow{O E}-\dfrac{3}{5} \overrightarrow{O F}\)\(\overrightarrow{O B}=\dfrac{6}{5} \overrightarrow{O F}-\dfrac{2}{5} \overrightarrow{O E}\)
    \(\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{O B}=\dfrac{6}{5} \overrightarrow{O E}-\dfrac{3}{5} \overrightarrow{O F}+\dfrac{6}{5} \overrightarrow{O F}-\dfrac{2}{5} \overrightarrow{O E}=\dfrac{4}{5} \overrightarrow{O E}+\dfrac{3}{5} \overrightarrow{O F}\)
    所以\(m=\dfrac{4}{5}\)\(n=\dfrac{3}{5}\),故\(m+n=\dfrac{7}{5}\)
    故选:\(C\)
    image.png

  4. 答案 \(-\dfrac{\sqrt{3}}{3}\)
    解析 如图过\(C\)\(CE⊥OB\)\(E\)
    image.png
    因为\(AB\)是圆\(O\)的直径,\(C\)\(D\)是圆\(O\)上的点,\(∠CBA=60°\)
    所以\(E\)\(OB\)的中点,连结\(OD\),则\(\overrightarrow{C E}=\dfrac{\sqrt{3}}{2} \overrightarrow{O D}\)
    \(\therefore \overrightarrow{C D}=\overrightarrow{C O}+\overrightarrow{O D}=\overrightarrow{O A}-\overrightarrow{B C}+\dfrac{2}{\sqrt{3}} \overrightarrow{C E}\)\(\overrightarrow{C E}=\overrightarrow{C B}+\overrightarrow{B E}=-\overrightarrow{B C}+\dfrac{1}{2} \overrightarrow{O A}\)
    \(\overrightarrow{C D}=\overrightarrow{O A}-\overrightarrow{B C}+\dfrac{2}{\sqrt{3}}\left(-\overrightarrow{B C}+\dfrac{1}{2} \overrightarrow{O A}\right)=\left(\dfrac{1}{\sqrt{3}}+1\right) \overrightarrow{O A}-\left(1+\dfrac{2}{\sqrt{3}}\right) \overrightarrow{B C}\)
    \(\overrightarrow{C D}=x \overrightarrow{O A}+y \overrightarrow{B C}\)
    \(x+y=\left(\dfrac{1}{\sqrt{3}}+1\right)-\left(1+\dfrac{2}{\sqrt{3}}\right)=-\dfrac{\sqrt{3}}{3}\)
    故答案为:\(-\dfrac{\sqrt{3}}{3}\)

  5. 答案 \(\dfrac{11}{9}\)
    解析 由题,梯形\(ABCD\)中,\(\overrightarrow{A B}=2 \overrightarrow{D C}\)\(\overrightarrow{B E}=\dfrac{1}{3} \overrightarrow{B C}\)\(P\)为线段\(DE\)上的动点(包括端点),
    \(\overrightarrow{A P}=t \overrightarrow{A D}+(1-t) \overrightarrow{A E}=t \overrightarrow{A D}+(1-t)(\overrightarrow{A B}+\overrightarrow{B E})\)
    \(=t \overrightarrow{A D}+(1-t) \overrightarrow{A B}+\dfrac{1}{3}(1-t) \overrightarrow{B C}(0 \leq t \leq 1)\)
    \(\because \overrightarrow{A D}=\overrightarrow{A C}+\overrightarrow{C D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}=\dfrac{1}{2} \overrightarrow{A B}+\overrightarrow{B C}\)
    \(\therefore \overrightarrow{A P}=t\left(\dfrac{1}{2} \overrightarrow{A B}+\overrightarrow{B C}\right)+(1-t) \overrightarrow{A B}+\dfrac{1}{3}(1-t) \overrightarrow{B C}=\left(1-\dfrac{1}{2} t\right) \overrightarrow{A B}+\left(\dfrac{1}{3}+\dfrac{2}{3} t\right) \overrightarrow{B C}\)
    \(\because \overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{B C}(\lambda, \mu \in \boldsymbol{R})\)
    \(\therefore\left\{\begin{array}{l} 1-\dfrac{1}{2} t=\lambda \\ \dfrac{1}{3}+\dfrac{2}{3} t=\mu \end{array}\right.\)
    \(\therefore \lambda^2+\mu=\left(1-\dfrac{1}{2} t\right)^2+\dfrac{1}{3}+\dfrac{2}{3} t=\dfrac{1}{4}\left(t-\dfrac{2}{3}\right)^2+\dfrac{11}{9}\)
    \(∴\)\(t=\dfrac{2}{3}\)时,\(λ^2+μ\)的最小值为 \(\dfrac{11}{9}\)

  6. 答案 \(\dfrac{1}{4}\)
    解析 延长\(BP\),交\(AC\)\(D\),画出图形,如图所示;
    image.png
    \(\therefore \overrightarrow{B A}+\overrightarrow{B C}=2 \overrightarrow{B D}\)
    \(\because \overrightarrow{B A}+\overrightarrow{B C}=6 \overrightarrow{B P}\)\(\therefore \overrightarrow{B D}=3 \overrightarrow{B P}\)\(\therefore \dfrac{S_{\triangle A B P}}{S_{\triangle A D P}}=\dfrac{1}{2}\)
    \(D\)\(AC\)的中点,
    \(\therefore S_{\triangle A D P}=S_{\triangle C D P}\)\(\therefore \dfrac{S_{\triangle A B P}}{S_{\triangle A C P}}=\dfrac{S_{\triangle A B P}}{2 S_{\triangle A D P}}=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}\)

  7. 答案 等边三角形
    解析 \(∵G\)\(△ABC\)的重心, \(\overrightarrow{G A}=-\dfrac{2}{3} \times \dfrac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{G B}=\dfrac{1}{3}(\overrightarrow{B A}+\overrightarrow{B C}), \overrightarrow{G C}=\dfrac{1}{3}(\overrightarrow{C A}+\overrightarrow{C B})\)
    \(\overrightarrow{G B}=\dfrac{1}{3}(\overrightarrow{B A}+\overrightarrow{B C})\)
    \(\therefore(a-b) \overrightarrow{A B}+(a-c) \overrightarrow{A C}+(b-c) \overrightarrow{B C}=\overrightarrow{0}\)
    \(∴a-b=a-c=b-c\)
    \(a=b=c\)
    \(∴△ABC\)的形状是等边三角形.
     

【题型3】向量共线问题

【典题1】 已知\(\vec{a}\)\(\vec{b}\)是两个不共线的向量,若向量\(k \vec{a}+\vec{b}\)\(\vec{a}-\vec{b}\)共线,则实数\(k=\)\(\underline{\quad \quad}\)
解析 \(∵ \vec{a}\)\(\vec{b}\)是两个不共线的向量,向量\(k \vec{a}+\vec{b}\)\(\vec{a}-\vec{b}\)共线,
\(\therefore k \vec{a}+\vec{b}=\lambda(\vec{a}-\vec{b})=\lambda \vec{a}-\lambda \vec{b}\)
\(∴k=λ\)\(1=-λ\)
则实数\(k=-1\)
点拨 共线定理 非零向量\(\vec{a}\)与向量\(\vec{b}\)共线⇔有且只有一个实数λ,使得 \(\vec{b}=\lambda \vec{a}\).
 

【典题2】在平面向量中有如下定理:设点\(O\)\(P\)\(Q\)\(R\)为同一平面内的点,则\(P\)\(Q\)\(R\)三点共线的充要条件是:存在实数\(t\),使 \(\overrightarrow{D P}=(1-t) \overrightarrow{O Q}+t \overrightarrow{O R}\).试利用该定理解答下列问题:
如图,在\(△ABC\)中,点\(E\)\(AB\)边的中点,点\(F\)\(AC\)边上,且\(CF=2FA\)\(BF\)\(CE\)于点\(M\),设\(\overrightarrow{A M}=x \overrightarrow{A E}+y \overrightarrow{A F}\),则\(x+y=\)\(\underline{\quad \quad}\)
image.png
解析 如图,\(E\)\(M\)\(C\)三点共线,\(∴\)存在实数\(λ\),使\(\overrightarrow{A M}=\lambda \overrightarrow{A E}+(1-\lambda) \overrightarrow{A C}\)
\(∵CF=2FA\)
\(∴AC=3AF\)\(\therefore \overrightarrow{A M}=\lambda \overrightarrow{A E}+3(1-\lambda) \overrightarrow{A F}\)
\(\overrightarrow{A M}=x \overrightarrow{A E}+y \overrightarrow{A F}\)
\(\therefore\left\{\begin{array}{l} \lambda=x \\ 3(1-\lambda)=y \end{array}\right.\)\(∴3(1-x)=y\)①;
同样,\(B\)\(M\)\(F\)三点共线,所以存在\(μ\),使\(\overrightarrow{A M}=\mu \overrightarrow{A B}+(1-u) \overrightarrow{A F}\)
\(∵E\)\(AB\)边的中点,\(∴AB=2AE\)
\(\therefore \overrightarrow{A M}=2 \mu \overrightarrow{A E}+(1-\mu) \overrightarrow{A F}\)
\(\therefore\left\{\begin{array}{l} x=2 \mu \\ y=1-\mu \end{array}\right.\)\(\therefore y=1-\dfrac{1}{2} x\)
∴联立①可得:\(x=\dfrac{4}{5}\)\(y=\dfrac{3}{5}\)
\(\therefore x+y=\dfrac{7}{5}\)
 

【巩固练习】

1.已知\(\vec{a}\)\(\vec{b}\)是不共线的向量,\(\overrightarrow{O A}=\lambda \vec{a}+\mu \vec{b}\)\(\overrightarrow{O B}=3 \vec{a}-2 \vec{b}\)\(\overrightarrow{O C}=2 \vec{a}+3 \vec{b}\),若\(A\)\(B\)\(C\)三点共线,则实数\(λ\)\(μ\)满足(  )
 A. \(λ=μ-1\) \(\qquad \qquad\) B. \(λ=μ+5\) \(\qquad \ \qquad\) C. \(λ=5-μ\) \(\qquad \qquad\) D.\(μ=13-5λ\)
 

2.(多选)如图,\(A\)\(B\)分别是射线\(OM\)\(ON\)上的点,下列以\(O\)为起点的向量中,终点落在阴影区域内的向量是(  )
image.png
 A.\(\overrightarrow{O A}+2 \overrightarrow{O B}\) \(\qquad \qquad\) B.\(\dfrac{1}{2} \overrightarrow{O A}+\dfrac{1}{3} \overrightarrow{O B}\) \(\qquad \qquad\) C.\(\dfrac{3}{4} \overrightarrow{O A}+\dfrac{1}{3} \overrightarrow{O B}\) \(\qquad \qquad\) D. \(\dfrac{3}{4} \overrightarrow{O A}+\dfrac{1}{5} \overrightarrow{O B}\)
 

3.已知\(3 \overrightarrow{O A}=\overrightarrow{O B}+\lambda \overrightarrow{O C}\),若\(A\)\(B\)\(C\)三点共线,则实数\(λ=\)\(\underline{\quad \quad}\) .
 

4.已知\(G\)\(△ABC\)的重心,直线\(EF\)过点\(G\)且与边\(AB\)\(AC\)分别交于点\(E\)\(F\)\(\overrightarrow{A E}=\alpha \overrightarrow{A B}\)\(\overrightarrow{A F}=\beta \overrightarrow{A C}\),则\(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\)的值为\(\underline{\quad \quad}\)
 

参考答案

  1. 答案 \(D\)
    解析 \(\because \overrightarrow{O A}=\lambda \vec{a}+\mu \vec{b}\)\(\overrightarrow{O B}=3 \vec{a}-2 \vec{b}\)\(\overrightarrow{O C}=2 \vec{a}+3 \vec{b}\)
    \(\therefore \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(3-\lambda) \vec{a}-(2+\mu) \vec{b}\)
    \(\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=2 \vec{a}+3 \vec{b}-(3 \vec{a}-2 \vec{b})=-\vec{a}+5 \vec{b}\)
    \(∵A\)\(B\)\(C\)三点共线,
    \(\therefore \overrightarrow{A B} / / \overrightarrow{B C}\),即\(\dfrac{3-\lambda}{-1}=\dfrac{-(2+\mu)}{5}\),化简可得\(μ=13-5λ\)
    故选\(D\).

  2. 答案 \(AC\)
    解析 由向量共线的充要条件可得:当点\(P\)在直线\(AB\)上时,存在唯一的一对有序实数\(u\)\(v\)
    使得\(\overrightarrow{O P}=u \overrightarrow{O A}+v \overrightarrow{O B}\)成立,且\(u+v=1\)
    可以证明当点\(P\)位于阴影区域内的充要条件是:
    满足\(\overrightarrow{O P}=u \overrightarrow{O A}+v \overrightarrow{O B}\),且\(u>0\)\(v>0\)\(u+v>1\)
    证明如下:如图所示,点\(P\)是阴影区域内的任意一点,
    过点\(P\)\(PE∥ON\)\(PF∥OM\),分别交\(OM\)\(ON\)于点\(E\)\(F\)
    \(PE\)\(AB\)于点\(P'\),过点\(P'\)\(P'F'∥OM\)\(ON\)于点\(F'\)
    则存在唯一一对实数\((x,y)\)\((u',v')\)
    使得\(\overrightarrow{O P^{\prime}}=x \overrightarrow{O E}+y \overrightarrow{O F^{\prime}}=u^{\prime} \overrightarrow{O A}+v^{\prime} \overrightarrow{O B}\),且\(u'+v'=1\)\(u'\)\(v'\)唯一;
    同理存在唯一一对实数\(x'\)\(y'\)使得\(\overrightarrow{O P}=x^{\prime} \overrightarrow{O E}+y^{\prime} \overrightarrow{O F}=x^{\prime} \overrightarrow{O E}+y^{\prime \prime} \overrightarrow{O F^{\prime}}=u \overrightarrow{O A}+v \overrightarrow{O B}\)
    \(x'=x\)\(y″>y\)
    \(∴u=u'\)\(v>v'\)\(∴u+v>u'+v'=1\)
    即可判断出\(A\)\(∵1+2>1\)\(∴\)\(P\)位于阴影区域内,故正确;
    同理\(C\)正确;而\(BD\)不正确;
    故选:\(AC\)
    image.png

  3. 答案 \(2\)
    解析 \(3 \overrightarrow{O A}=\overrightarrow{O B}+\lambda \overrightarrow{O C}\),整理得\(\overrightarrow{O A}=\dfrac{1}{3} \overrightarrow{O B}+\dfrac{\lambda}{3} \overrightarrow{O C}\)
    因为\(A\)\(B\)\(C\)三点共线,所以\(\dfrac{1}{3}+\dfrac{\lambda}{3}=1\),解得\(λ=2\)

  4. 答案 \(3\)
    解析 如图所示,
    image.png
    \(∵\)三点\(E\)\(G\)\(F\)共线,\(∴\)存在实数\(λ\)三点\(\overrightarrow{A G}=\lambda \overrightarrow{A E}+(1-\lambda) \overrightarrow{A F}\)
    \(\because \overrightarrow{A E}=\alpha \overrightarrow{A B}, \overrightarrow{A F}=\beta \overrightarrow{A C}\)\(\therefore \overrightarrow{A G}=\lambda \alpha \overrightarrow{A B}+(1-\lambda) \beta \overrightarrow{A C}\)
    \(∵G\)\(△ABC\)的重心, \(\therefore \overrightarrow{A G}=\dfrac{2}{3} \overrightarrow{A M}\)\(\overrightarrow{A M}=\dfrac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})\)
    \(\therefore \overrightarrow{A G}=\dfrac{1}{3} \overrightarrow{A B}+\dfrac{1}{3} \overrightarrow{A C}\)
    \(\therefore \lambda \alpha=\dfrac{1}{3}\)\((1-\lambda) \beta=\dfrac{1}{3}\)\(\therefore \dfrac{1}{\alpha}+\dfrac{1}{\beta}=3 \lambda+3(1-\lambda)=3\)
    故答案为:\(3\)
     

分层练习

【A组---基础题】

1.如图,在平行四边形\(ABCD\)中,\(AC\)\(BD\)交于点\(O\)\(\overrightarrow{A B}=\vec{a}\)\(\overrightarrow{A D}=\vec{b}\),则下列运算正确的是(  )
image.png
 A.\(\overrightarrow{B D}=\vec{a}+\vec{b}\) \(\qquad \qquad\) B.\(\overrightarrow{A C}=\vec{a}-\vec{b}\) \(\qquad \qquad\) C. \(\overrightarrow{O D}=\dfrac{1}{2}(\vec{b}-\vec{a})\) \(\qquad \qquad\) D. \(\overrightarrow{C O}=\dfrac{1}{2}(\vec{a}+\vec{b})\)
 

2.下列关于四边形\(ABCD\)判断正确的是(  )
 ①若\(\overrightarrow{A D}=\overrightarrow{B C}\),则四边形\(ABCD\)是平行四边形;
 ②若\(\overrightarrow{A D}=\dfrac{1}{3} \overrightarrow{B C}\),则四边形\(ABCD\)是梯形;
 ③若\(\overrightarrow{A C} \overrightarrow{A B}=\overrightarrow{D C}\),且\(|\overrightarrow{A B}|=|\overrightarrow{A D}|\),则四边形\(ABCD\)是菱形;
 ④若 \(|\overrightarrow{A B}+\overrightarrow{A D}|=|\overrightarrow{A B}-\overrightarrow{A D}|\),则四边形\(ABCD\)是矩形.
 A.②③④ \(\qquad \qquad \qquad \qquad\) B.①②③ \(\qquad \qquad \qquad \qquad\) C.①③④ \(\qquad \qquad \qquad \qquad\) D.①②③④
 

3.已知任意两个向量\(\vec{a}\)\(\vec{b}\)不共线,若\(\overrightarrow{O A}=\vec{a}+\vec{b}\)\(\overrightarrow{O B}=\vec{a}+2 \vec{b}\)\(\overrightarrow{O C}=2 \vec{a}-\vec{b}\)\(\overrightarrow{O D}=\vec{a}-\vec{b}\),则下列结论正确的是(  )
 A.\(A\)\(B\)\(C\)三点共线 \(\qquad \qquad \qquad \qquad\) B.\(A\)\(B\)\(D\)三点共线
 C.\(A\)\(C\)\(D\)三点共线 \(\qquad \qquad \qquad \qquad\) D.\(B\)\(C\)\(D\)三点共线
 

4.已知\(\vec{a}\)\(\vec{b}\)是平面内两个不共线向量, \(\overrightarrow{A B}=m \vec{a}+2 \vec{b}\)\(\overrightarrow{B C}=3 \vec{a}-\vec{b}\)\(A\)\(B\)\(C\)三点共线,则\(m=\) (  )
 A. \(-\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) C. \(-6\) \(\qquad \qquad \qquad \qquad\) D.\(6\)
 

5.如图,在\(△OBC\)中,点\(A\)\(BC\)的中点,\(\overrightarrow{O D}=2 \overrightarrow{D B}\)\(DC\)\(OA\)交于点\(E\),则\(AO\)\(OE\)的比值为(  )
image.png
 A.\(\dfrac{6}{5}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{5}{4}\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
 

6.(多选)如图所示,点\(A\)\(B\)\(C\)是圆\(O\)上的三点,线段\(OC\)与线段\(AB\)交于圆内一点\(P\),若\(\overrightarrow{A P}=\lambda \overrightarrow{A B}\)\(\overrightarrow{O C}=\mu \overrightarrow{O A}+3 \mu \overrightarrow{O B}\),则(  )
image.png
 A.\(P\)为线段\(OC\)的中点时,\(\mu=\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B.\(P\)为线段\(OC\)的中点时, \(\mu=\dfrac{1}{3}\)
 C.无论\(μ\)取何值,恒有\(\lambda=\dfrac{3}{4}\) \(\qquad \qquad \qquad \qquad\) D.存在\(μ∈R\)\(\lambda=\dfrac{1}{2}\)
 

7.点\(C\)在直线\(AB\)上,且\(\overrightarrow{A C}=3 \overrightarrow{A B}\),则\(\overrightarrow{BC}=\)\(\underline{\quad \quad}\) \(\overrightarrow{A B}\)
 

8.如图,\(O\)为直线\(A_0 A_{2015}\)外一点,若\(A_0\)\(A_1\)\(A_2\)\(A_3\)\(A_4\)\(A_5\),…,\(A_{2015}\)中任意相邻两点的距离相等,设\(\overrightarrow{O A_0}=\vec{a}\)\(\overrightarrow{O A_{2015}}=\vec{b}\),用\(\vec{a}\)\(\vec{b}\)表示\(\overrightarrow{O A_0}+\overrightarrow{O A_1}+\cdots+\overrightarrow{O A_{2015}}\),其结果为\(\underline{\quad \quad}\)
image.png
 

9.在\(△ABC\)中, \(\overrightarrow{A B}=\vec{a}\)\(\overrightarrow{A C}=\vec{b}\),若点\(D\)满足\(\overrightarrow{B D}=2 \overrightarrow{D C}\),则 \(\overrightarrow{A D}=\)\(\underline{\quad \quad}\).(用\(\vec{a}\)\(\vec{b}\)表示)
 

10.如图,过\(△ABC\)的重心\(G\)的直线分别交边\(AB\)\(AC\)\(P\)\(Q\)两点,且\(\overrightarrow{A B}=x \overrightarrow{A P}\)\(\overrightarrow{A C}=y \overrightarrow{A Q}\),则\(xy\)的取值范围是\(\underline{\quad \quad}\)
image.png
 

11.在\(△ABC\)中,\(E\)\(F\)分别为\(AB\)\(AC\)中点,\(P\)为线段\(EF\)上任意一点,实数\(x\)\(y\)满足\(\overrightarrow{P A}+x \overrightarrow{P B}+y \overrightarrow{P C}=\overrightarrow{0}\),设\(△ABC\)\(△PCA\)\(△PAB\)的面积分别为\(S\)\(S_1\)\(S_2\),记\(\dfrac{S_1}{S}=\lambda_1\)\(\dfrac{S_2}{S}=\lambda_2\),则\(λ_1 λ_2\)取得最大值时,\(2x+3y\)的值为\(\underline{\quad \quad}\)
 

12.如图所示,在\(◻ABCD\)中,\(AD\)\(DC\)边的中点分别为\(E\)\(F\),连接\(BE\)\(BF\),与\(AC\)分别交于点\(R\)\(T\).求证:\(AR=RT=TC\)
image.png
 
 
 

参考答案

  1. 答案 \(C\)
    解析 \(\overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B}=\vec{b}-\vec{a}\),故\(A\)错误,
    \(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A B}+\overrightarrow{A D}=\vec{a}+\vec{b}\),故\(B\)错误,
    \(\overrightarrow{O D}=\dfrac{1}{2} \overrightarrow{B D}=\dfrac{1}{2}(\vec{b}-\vec{a})\),故\(C\)正确,
    \(\overrightarrow{C O}=-\dfrac{1}{2} \overrightarrow{A C}=-\dfrac{1}{2}(\vec{a}+\vec{b})\),故\(D\)错误.
    故选:\(C\)

  2. 答案 \(B\)
    解析 对于①,若\(\overrightarrow{A D}=\overrightarrow{B C}\),则\(|\overrightarrow{A D}|=|\overrightarrow{B C}|\)\(\overrightarrow{A D} / / \overrightarrow{B C}\),四边形\(ABCD\)是平行四边形,①正确;
    对于②,若\(\overrightarrow{A D}=\dfrac{1}{3} \overrightarrow{B C}\)\(|\overrightarrow{A D}| \neq|\overrightarrow{B C}|\)\(\overrightarrow{A D} / / \overrightarrow{B C}\),四边形\(ABCD\)是梯形,②正确;
    对于③,由\(\overrightarrow{A B}=\overrightarrow{D C}\)得出四边形\(ABCD\)是平行四边形,
    \(|\overrightarrow{A B}|=|\overrightarrow{A D}|\),得出平行四边形\(ABCD\)是菱形,③正确;
    对于④,由\(|\overrightarrow{A B}+\overrightarrow{A D}|=|\overrightarrow{A B}-\overrightarrow{A D}|\),得\((\overrightarrow{A B}+\overrightarrow{A D})^2=(\overrightarrow{A B}-\overrightarrow{A D})^2\)
    \(\overrightarrow{A B}^2+2 \overrightarrow{A B} \cdot \overrightarrow{A D}+\overrightarrow{A D}^2=\overrightarrow{A B}^2-2 \overrightarrow{A B} \cdot \overrightarrow{A D}+\overrightarrow{A D}^2\)
    \(\therefore \overrightarrow{A B} \cdot \overrightarrow{A D}=0\)\(AB⊥AD\),如图所示
    image.png
    又四边形\(ABCD\)不一定是平行四边形,
    \(∴\)四边形\(ABCD\)不一定是矩形,④错误.
    综上,正确命题的序号为①②③.
    故选:\(B\)

  3. 答案 \(B\)
    解析 \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{b}\)\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=-2 \vec{b}\)
    \(\overrightarrow{A B}=-2 \overrightarrow{A D}\)\(\overrightarrow{A B}\)\(\overrightarrow{A D}\)共线,且有公共点,所以\(A\)\(B\)\(D\)三点共线.
    故选:\(B\)

  4. 答案 \(C\)
    解析 \(∵A\)\(B\)\(C\)三点共线, \(\therefore \overrightarrow{A B}\)\(\overrightarrow{B C}\)共线,
    \(∴\)存在\(λ\),使得\(\overrightarrow{A B}=\lambda \overrightarrow{B C}\)\(\therefore m \vec{a}+2 \vec{b}=3 \lambda \vec{a}-\lambda \vec{b}\), 且\(\vec{a}\)\(\vec{b}\)不共线,
    \(\therefore\left\{\begin{array}{l} m=3 \lambda \\ -\lambda=2 \end{array}\right.\),解得\(m=-6\).

  5. 答案 \(C\)
    解析 \(∵O\)\(E\)\(A\)三点共线,且\(A\)\(BC\)的中点;
    \(\therefore \overrightarrow{O B}+\overrightarrow{O C}=2 \overrightarrow{O A}\)
    \(\overrightarrow{O E}=\dfrac{\lambda}{2}(\overrightarrow{O B}+\overrightarrow{O C})\),而\(\overrightarrow{O B}=\dfrac{3}{2} \overrightarrow{O D}\)
    代入上式便可得出\(\overrightarrow{O E}=\dfrac{3 \lambda}{4} \overrightarrow{O D}+\dfrac{\lambda}{2} \overrightarrow{O C}\)
    \(C\)\(E\)\(D\)三点共线便可得到\(\dfrac{3 \lambda}{4}+\dfrac{\lambda}{2}=1\),解得\(\lambda=\dfrac{4}{5}\)
    \(\therefore \dfrac{5}{2} \overrightarrow{O E}=\overrightarrow{O B}+\overrightarrow{O C}\)
    \(\therefore 2 \overrightarrow{O A}=\dfrac{5}{2} \overrightarrow{O E}\),则\(AO\)\(OE\)的比值为 \(\dfrac{5}{4}\)
    故选:\(C\)

  6. 答案 \(AC\)
    解析 \(\overrightarrow{O P}=\overrightarrow{O A}+\overrightarrow{A P}=\overrightarrow{O A}+\lambda \overrightarrow{A B}=\overrightarrow{O A}+\lambda(\overrightarrow{O B}-\overrightarrow{O A})=(1-\lambda) \overrightarrow{O A}+\lambda \overrightarrow{O B}\)
    因为\(\overrightarrow{O P}\)\(\overrightarrow{O C}\)共线,所以\(\dfrac{1-\lambda}{\mu}=\dfrac{\lambda}{3 \mu}\),解得\(\lambda=\dfrac{3}{4}\),故\(C\)正确,\(D\)错误;
    \(P\)\(OC\)中点时,则\(\overrightarrow{O P}=\dfrac{1}{2} \overrightarrow{O C}\),则\(1-\lambda=\dfrac{1}{2} \mu\)\(\lambda=\dfrac{1}{2} \times 3 \mu\),解得\(\mu=\dfrac{1}{2}\)
    \(A\)正确,\(B\)错误;
    故选 \(AC\)

  7. 答案 \(2\)
    解析 \(\overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B}=3 \overrightarrow{A B}-\overrightarrow{A B}=2 \overrightarrow{A B}\)

  8. 答案 \(1008(\vec{a}+\vec{b})\)
    解析 \(D\)\(A_0\)\(A_{2015}\)的中点,
    由题意可得 \(\overrightarrow{O A_0}+\overrightarrow{O A_{2015}}=\vec{a}+\vec{b}=2 \overrightarrow{O D}\)
    同理可得 \(\overrightarrow{O A_1}+\overrightarrow{O A_{2014}}=2 \overrightarrow{O D}\)
    ……
    \(\overrightarrow{O A_{1007}}+\overrightarrow{O A_{1008}}=2 \overrightarrow{O D}\)
    \(\therefore \overrightarrow{O A_0}+\overrightarrow{O A_1}+\cdots+\overrightarrow{O A_{2015}}=\dfrac{2016}{2} \cdot 2 \overrightarrow{O D}=1008(\vec{a}+\vec{b})\) .

  9. 答案 \(\dfrac{2}{3} \vec{b}+\dfrac{1}{3} \vec{a}\)
    解析 \(D\)\(DE//AB\),作\(DF//AC\),易得\(AEDF\)是平行四边形,且 \(A E=\dfrac{2}{3} A C\)\(A F=\dfrac{1}{3} A B\)
    由向量的加法几何意义,有 \(\overrightarrow{A D}=\overrightarrow{A E}+\overrightarrow{A F}=\dfrac{2}{3} \overrightarrow{A C}+\dfrac{1}{3} \overrightarrow{A B}=\dfrac{2}{3} \vec{b}+\dfrac{1}{3} \vec{a}\).
    image.png

  10. 答案 \(\left[2, \dfrac{9}{4}\right]\)
    解析 \(∵P\)\(G\)\(Q\)三点共线,\(∴\)存在\(m\),使 \(\overrightarrow{A G}=m \overrightarrow{A Q}+(1-m) \overrightarrow{A P}\)
    \(∵G\)\(△ABC\)的重心,
    \(\therefore \overrightarrow{A G}=\dfrac{1}{3}(\overrightarrow{A B}+\overrightarrow{A C})=\dfrac{1}{3}(y \overrightarrow{A Q}+x \overrightarrow{A P})\)
    \(\therefore \dfrac{1}{3}(y \overrightarrow{A Q}+x \overrightarrow{A P})=m \overrightarrow{A Q}+(1-m) \overrightarrow{A P}\)
    \(∴x+y=3\)
    \(\because \overrightarrow{A B}=x \overrightarrow{A P}\)\(∴1≤x≤2\)
    \(x y=x(3-x)=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\)
    \(2 \leq-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4} \leq \dfrac{9}{4}\)
    故答案为:\(\left[2, \dfrac{9}{4}\right]\)

  11. 答案 \(\dfrac{5}{2}\)
    解析 如图所示.\(∵\)\(P\)\(△ABC\)的中位线\(EF\)上, \(\therefore \dfrac{S_{\triangle B P C}}{S}=\dfrac{1}{2}\)
    \(\therefore \dfrac{S_1+S_2}{S}=\dfrac{1}{2}\),即 \(S_1+S_2=\dfrac{1}{2} S\)
    \(\therefore \dfrac{1}{2} S \geq 2 \sqrt{S_1 S_2}\),当且仅当\(S_1=S_2=\dfrac{1}{4} S\)时取等号,此时\(S_1 S_2\)取得最大值\(\dfrac{1}{16} S^2\)
    此时点\(P\)为线段\(EF\)的中点.
    \(PB\)\(PC\)为邻边作平行四边形\(PBDC\),连接\(PD\)\(BC\)于点\(O\)
    \(\overrightarrow{P B}+\overrightarrow{P C}=\overrightarrow{P D}=2 \overrightarrow{P O}=-2 \overrightarrow{P A}\)
    化为 \(\overrightarrow{P A}+\dfrac{1}{2} \overrightarrow{P B}+\dfrac{1}{2} \overrightarrow{P C}=\overrightarrow{0}\)
    \(\because \overrightarrow{P A}+x \overrightarrow{P B}+y \overrightarrow{P C}=\overrightarrow{0}\)\(\therefore x=y=\dfrac{1}{2}\)
    \(\therefore 2 x+3 y=\dfrac{5}{2}\)
    image.png

  12. 证明\(\overrightarrow{A B}=\vec{a}\)\(\overrightarrow{A D}=\vec{b}\)\(\overrightarrow{A R}=\vec{r}\)\(\overrightarrow{A T}=\vec{t}\),则 \(\overrightarrow{A C}=\vec{a}+\vec{b}\)
    因为\(\overrightarrow{A R}\)\(\overrightarrow{A C}\)共线,
    所以存在实数\(n\),使得\(\vec{r}=n(\vec{a}+\vec{b})\)\(n∈R\)
    因为\(\overrightarrow{E R}\)\(\overrightarrow{E B}\)共线,所以存在实数\(m\),使得\(\overrightarrow{E R}=m\overrightarrow{EB}\)\(m∈R\)
    \(\overrightarrow{E B}=\overrightarrow{A B}-\overrightarrow{A E}=\vec{a}-\dfrac{1}{2} \vec{b}\),则\(\overrightarrow{E R}=m\left(\vec{a}-\dfrac{1}{2} \vec{b}\right)\)
    因为\(\overrightarrow{A R}=\overrightarrow{A E}+\overrightarrow{E R}\),所以\(n(\vec{a}+\vec{b})=\dfrac{1}{2} \vec{b}+m\left(\vec{a}-\dfrac{1}{2} \vec{b}\right)\),即 \((n-m) \vec{a}+\left(n+\dfrac{m-1}{2}\right) \vec{b}=\overrightarrow{0}\)
    因为向量\(\vec{a}\)\(\vec{b}\)不共线,于是有\(\left\{\begin{array}{l} n-m=0 \\ n+\dfrac{m-1}{2}=0 \end{array}\right.\),解得 \(m=n=\dfrac{1}{3}\),所以\(\overrightarrow{A R}=\dfrac{1}{3} \overrightarrow{A C}\)
    同理\(\overrightarrow{A T}=\dfrac{2}{3} \overrightarrow{A C}\).所以\(\overrightarrow{A R}=\overrightarrow{R T}=\overrightarrow{T C}\).
    \(AR=RT=TC\)
     
     

【B组---提高题】

1.已知点\(O\)\(△ABC\)内部一点,并且满足\(\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\overrightarrow{0}\)\(△BOC\)的面积为\(S_1\)\(△ABC\)的面积为\(S_2\),则 \(\dfrac{S_1}{S_2}=\)(  )
 A. \(\dfrac{1}{6}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{1}{3}\)\(\qquad \qquad \qquad \qquad\) C. \(\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{3}{4}\)
image.png
 

2.已知平面向量\(\vec{a}\)\(\vec{b}\)\(\vec{c}\)满足: \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\)\(\vec{a} \perp \vec{b}\),则\(|2 \vec{c}-\vec{a}|+\left|\dfrac{1}{2} \vec{c}-\vec{b}\right|\)的最小值为\(\underline{\quad \quad}\) .
 

参考答案

  1. 答案 \(A\)
    解析 如图所示,延长\(OB\)\(D\)使得\(BD=OB\),延长\(OC\)\(E\)使得\(CE=2OC\)
    \(∵\)满足\(\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\overrightarrow{0}\)
    ∴点\(O\)\(△ADE\)的重心.
    \(\therefore S_{\triangle O A D}=S_{\triangle O D E}=S_{\triangle O A E}\)
    \(S_{\triangle O A B}=\dfrac{1}{2} S_{\triangle O A D}\)\(S_{\triangle O A C}=\dfrac{1}{3} S_{\triangle O A E}, \quad S_{\triangle O B C}=\dfrac{1}{6} S_{\triangle O D E}\)
    \(\therefore S_1=\dfrac{1}{18} S_{\triangle A D E}\)\(S_2=S_{\triangle O A B}+S_{\triangle O A C}+S_{\triangle O B C}=\dfrac{1}{3} S_{\triangle A D E}\)
    \(\dfrac{S_1}{S_2}=\dfrac{1}{6}\)
    故选:\(A\)

  2. 答案 \(\dfrac{\sqrt{17}}{2}\)
    解析 如图,\(⊙O\)为单位圆,\(A\)\(B\)、C在\(⊙O\)上,\(OA⊥OB\)\(\angle B O A=\dfrac{\pi}{2}\)
    \(B'\)\(OB\)的延长线上,\(OB'=2\)\(B″\)\(OB\)中点,
    \(A'\)\(OA\)中点,\(A″\)\(OB\)的延长线上,\(OA″=2\)
    image.png
    \(\vec{a}=\overrightarrow{O A}\)\(\vec{b}=\overrightarrow{O B}\)\(C\)\(⊙O\)上一点,\(\vec{c}=\overrightarrow{O C}\)
    \(\dfrac{O A^{\prime}}{O C^{\prime}}=\dfrac{O C}{O A^{\prime \prime}}=\dfrac{1}{2}\)\(∴△OCA'∽△OA″C\)\(∴CA'=2A″C\)
    同理\(C B^{\prime \prime}=\dfrac{1}{2} C B^{\prime}\)
    \(2 \vec{c}-\vec{a}=2\left(\vec{c}-\dfrac{1}{2} \vec{a}\right)=2\left(\overrightarrow{O C}-\overrightarrow{O A^{\prime}}\right)=2 \overrightarrow{A^{\prime} C}\)
    \(\dfrac{1}{2} \vec{c}-\vec{b}=\dfrac{1}{2}(\vec{c}-2 \vec{b})=\dfrac{1}{2}(\vec{c}-2 \vec{b})=\dfrac{1}{2}\left(\overrightarrow{O C}-\overrightarrow{O B^{\prime}}\right)=\dfrac{1}{2} \overrightarrow{B^{\prime} C}\)
    \(\therefore|2 \vec{c}-\vec{a}|+\left|\dfrac{1}{2} \vec{c}-\vec{b}\right|=2\left|\overrightarrow{A^{\prime}} C\right|+\dfrac{1}{2}\left|\overrightarrow{B^{\prime}} C\right|=\left|B^{\prime \prime} C\right|+\left|C A^{\prime \prime}\right| \geq\left|B^{\prime \prime} A^{\prime \prime}\right|=\sqrt{\dfrac{1}{4}+4}=\dfrac{\sqrt{17}}{2}\).
     

【C组---拓展题】

1.在\(△ABC\)中,\(O\)是其外接圆的圆心,其两边中线的交点是\(G\),两条高线的交点是\(H\),给出下列结论或命题:
  (1)动点\(P\)满足\(\overrightarrow{A P}=\lambda\left(\dfrac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\dfrac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right)(\lambda \neq 0)\),则动点\(P\)的轨迹一定过点\(H\)
  (2)动点\(P\)\(△ABC\)所在平面内,则点\(G\)\(P\)重合时,使\(PA^2+PB^2+PC^2\)的值最小;
  (3)动点\(P\)满足\(\overrightarrow{A P}=\lambda\left(\dfrac{\overrightarrow{A B}}{|\overrightarrow{A B}| \cos B}+\dfrac{\overrightarrow{A C}}{|\overrightarrow{A C}| \cos C}\right)(\lambda \neq 0)\),则点\(P\)的轨迹一定过点\(O\)
  (4)\(GH=2OG\)
其中正确结论或命题的序号是\(\underline{\quad \quad}\).(填上所有正确结论或命题的序号)
 
 

2.如图所示,在\(△ABO\)中,\(\overrightarrow{O C}=\dfrac{1}{3} \overrightarrow{O A}\)\(\overrightarrow{O D}=\dfrac{1}{2} \overrightarrow{O B}\)\(AD\)\(BC\)相交于点\(M\).设\(\overrightarrow{O A}=\vec{a}\)\(\overrightarrow{O B}=\vec{b}\)
  (1)试用向量\(\vec{a}\)\(\vec{b}\)表示\(\overrightarrow{O M}\)
  (2)在线段\(AC\)上取点\(E\),在线段\(BD\)上取点\(F\),使\(EF\)过点\(M\).设\(\overrightarrow{O E}=\lambda \overrightarrow{O A}\)\(\overrightarrow{O F}=\mu \overrightarrow{O B}\),其中\(λ\)\(μ∈R\).当\(EF\)\(AD\)重合时,\(λ=1\)\(\mu=\dfrac{1}{2}\),此时\(\dfrac{1}{\lambda}+\dfrac{2}{\mu}=5\);当\(EF\)\(BC\)重合时,\(\lambda=\dfrac{1}{3}\)\(μ=1\),此时\(\dfrac{1}{\lambda}+\dfrac{2}{\mu}=5\);能否由此得出一般结论 不论\(E\)\(F\)在线段\(AC\)\(BD\)上如何变动,等式\(\dfrac{1}{\lambda}+\dfrac{2}{\mu}=5\)恒成立,请说明理由.
image.png
 
 

参考答案

  1. 答案 ②④
    解析 ①中,动点\(P\)满足\(\overrightarrow{A P}=\lambda\left(\dfrac{\overrightarrow{A B}}{|\overrightarrow{A B}|}+\dfrac{\overrightarrow{A C}}{|\overrightarrow{A C}|}\right)(\lambda \neq 0)\)
    \(∴AP\)平分\(∠BAC\)
    \(∴P\)点在\(∠BAC\)的角平分线上,不一定过点\(H\),故①错误.
    ②中,点\(P\)\(△ABC\)内的一点,且使得\(\overrightarrow{A P}^2+\overrightarrow{B P}^2+\overrightarrow{C P}^2\)取得最小值,
    根据重心的性质,可得②正确;
    ③中, \(\overrightarrow{A P} \cdot \overrightarrow{B C}=\lambda\left(\dfrac{\overrightarrow{A B}}{|\overrightarrow{A B}| \cos B}+\dfrac{\overrightarrow{A C}}{|\overrightarrow{A C}| \cos C}\right) \cdot \overrightarrow{B C}=\lambda(-|\overrightarrow{B C}|+|\overrightarrow{B C}|)=0\)
    \(\therefore \overrightarrow{A P} \perp \overrightarrow{B C}\)
    \(∴\)\(P\)一定在高线上,不一定过点\(O\).故③错.
    ④在三角形\(ABC\)的外接圆中,过点\(C\)作直径\(CM\),连\(MA\)\(MB\),则有\(MB\)平行且等于\(2OF\)
    因为\(MB⊥BC\)\(AD⊥BC\)\(MA⊥AC\)\(BE⊥AC\),所以四边形\(AMBH\)是平行四边形,
    因此\(AH=MB=2OF\),连接\(OH\)\(AF\)\(G\),三角形\(OFG\)与三角形\(HAG\)相似,
    可证\(G\)就是重心,所以\(GH=2OG\).故④正确.
    故答案为:②④.

  2. 答案 (1) \(\overrightarrow{O M}=\dfrac{1}{5} \vec{a}+\dfrac{2}{5} \vec{b}\);(2) 能得出结论.
    解析 (1)设 \(\overrightarrow{O M}=m \vec{a}+n \vec{b}(m \in R, \quad n \in R)\)
    \(A\)\(D\)\(M\)三点共线,可知存在\(α(α∈R\),且\(α≠-1)\)使得 \(\overrightarrow{A M}=\alpha \overrightarrow{M D}\)
    \(\overrightarrow{O M}-\overrightarrow{O A}=\alpha(\overrightarrow{O D}-\overrightarrow{O M})\)
    \(\overrightarrow{O D}=\dfrac{1}{2} \overrightarrow{O B}\),所以\(\overrightarrow{O M}=\dfrac{1}{\alpha+1} \vec{a}+\dfrac{\alpha}{2(1+\alpha)} \vec{b}\)
    \(\therefore\left\{\begin{array}{l} m=\dfrac{1}{1+\alpha} \\ n=\dfrac{\alpha}{2(1+\alpha)} \end{array}\right.\),即\(m+2n=1\)①,
    \(B\)\(C\)\(M\)三点共线,可知存在\(β(β∈R\),且\(β≠-1)\)使得 \(\overrightarrow{C M}=\beta \overrightarrow{M B}\)
    \(\overrightarrow{O M}-\overrightarrow{O C}=\beta(\overrightarrow{O B}-\overrightarrow{O M})\),又 \(\overrightarrow{O C}=\dfrac{1}{3} \overrightarrow{O A}\)
    所以\(\overrightarrow{O M}=\dfrac{1}{3(\beta+1)} \vec{a}+\dfrac{\beta}{1+\beta} \vec{b}\)\(\therefore\left\{\begin{array}{l} m=\dfrac{1}{4(1+\beta)} \\ n=\dfrac{\beta}{1+\beta} \end{array}\right.\)
    \(3m+n=1\)②,
    由①②得\(m=\dfrac{1}{5}\)\(n=\dfrac{2}{5}\),故 \(\overrightarrow{O M}=\dfrac{1}{5} \vec{a}+\dfrac{2}{5} \vec{b}\)
    (2)能得出结论.
    理由 由于\(E\)\(M\)\(F\)三点共线,则存在实数\(γ(γ∈R\),且\(γ≠-1)\)使得\(\overrightarrow{E M}=\gamma \overrightarrow{M F}\)
    于是\(\overrightarrow{O M}=\dfrac{\overrightarrow{O E}+\gamma \overrightarrow{O F}}{1+\gamma}\),又 \(\overrightarrow{O E}=\lambda \overrightarrow{O A}, \overrightarrow{O F}=\mu \overrightarrow{O B}\)
    所以\(\overrightarrow{O M}=\dfrac{\lambda \overrightarrow{O A}+\mu \gamma \overrightarrow{O B}}{1+\gamma}=\dfrac{\lambda}{1+\gamma} \vec{a}+\dfrac{\mu \gamma}{1+\gamma} \vec{b}\)
    所以\(\dfrac{1}{5} \vec{a}+\dfrac{2}{5} \vec{b}=\dfrac{\lambda}{1+\gamma} \vec{a}+\dfrac{\mu \gamma}{1+\gamma} \vec{b}\)
    从而\(\left\{\begin{array}{l} \dfrac{1}{5}=\dfrac{\lambda}{1+\gamma} \\ \dfrac{2}{5}=\dfrac{\mu \gamma}{1+\gamma} \end{array}\right.\),所以消去\(γ\)\(\dfrac{1}{\lambda}+\dfrac{2}{\mu}=5\)

posted @ 2023-05-04 10:10  贵哥讲数学  阅读(332)  评论(0编辑  收藏  举报
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