6.2.3 向量的数乘运算
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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 \(|\vec{b}|\)和\(|\vec{c}|\).
解 向量\(\vec{b}\),\(\vec{c}\)如下图,其中\(|\vec{b}|=|-2 \vec{a}|=2|\vec{a}|=4\), \(|\vec{c}|=\left|\dfrac{1}{2} \vec{a}\right|=\dfrac{1}{2}|\vec{a}|=1\).
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 \(4(\vec{a}+\vec{b})-3(\vec{a}-\vec{b})-\vec{a}\).
解 \(4(\vec{a}+\vec{b})-3(\vec{a}-\vec{b})-\vec{a}=4 \vec{a}+4 \vec{b}-3 \vec{a}+3 \vec{b}-\vec{a}=7 \vec{b}\).
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 \(\vec{b}=\lambda \vec{a}\).
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;
当\(λ=0\) 时,\(\lambda \vec{a}=\overrightarrow{0}\).
三点共线定理
若 \(\overrightarrow{O C}=x \overrightarrow{O A}+y \overrightarrow{O B}\)
(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\(C\)在\(AB\)同侧,则\(x+y<1\);
(3) 如图三,若点\(O\)和点\(C\)在\(AB\)异侧,则\(x+y>1\);
![](https://cdn.nlark.com/yuque/0/2023/png/21447551/1681470478144-a97b3617-5d52-46e5-9fc4-ca1becc666cf.png )
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】点\(C\)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 点\(C\)在线段\(AB\)上,且\(|\overrightarrow{A C}|=\dfrac{2}{3}|\overrightarrow{C B}|\),如图所示;
若\(\overrightarrow{A B}=\lambda \overrightarrow{B C}\),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.\(\lambda \vec{a}\)与\(\vec{a}\)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
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答案 \(C\)
解析 当\(\vec{a} \neq 0\),\(λ<0\)时,\(\lambda \vec{a}\)与\(\vec{a}\)反向,且\(\lambda|\vec{a}|<0,\),则\(A\),\(D\)错误.
又\(\because 0 \cdot \vec{a}\)的结果为\(\vec{0}\),则\(B\)错误.由运算律知\(C\)正确. -
答案 \(\vec{b}-\vec{c}\)
解析 原式\(=10 \vec{a}-8 \vec{b}+2 \vec{c}-3 \vec{a}+9 \vec{b}-3 \vec{c}-7 \vec{a}=\vec{b}-\vec{c}\). -
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\(\overrightarrow{A B}=\)( )
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】 在\(△ABC\)中,\(D\),\(E\)分别为边\(AB\),\(AC\)的中点,\(BE\)与\(CD\)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. \(\dfrac{1}{6} \vec{a}+\dfrac{1}{6} \vec{b}\)
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\(E\)作\(EF/ /AB\),
\(∵E\)、\(D\)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 \(\lambda=\dfrac{2}{3}\),
\(\therefore \overrightarrow{A P}=(1-\lambda) \vec{a}+\dfrac{\lambda}{2} \vec{b}=\dfrac{1}{3} \vec{a}+\dfrac{1}{3} \vec{b}\)
方法2 构造平行四边形法
过点\(P\)分别作\(PH//AB\),\(PG//AC\),则四边形\(AGPH\)是平行四边形,
则\(\overrightarrow{A P}=\overrightarrow{A G}+\overrightarrow{A H}=x \overrightarrow{A B}+y \overrightarrow{A C}=x \vec{a}+y \vec{b}\),
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】 点\(O\)在\(△ABC\)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,
作\(OD=4OC\),以\(OA\),\(OD\)为邻边作平行四边形\(OAED\),
连接\(AD\),\(OE\),交于点\(M\),\(OE\)交\(AC\)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 若\(\overrightarrow{A B}=\lambda \overrightarrow{C D}\),意味着\(AB||CD\)且\(AB=|λ|CD\);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 \(\overrightarrow{O P}-\overrightarrow{O Q}=\)( )
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 \(\overrightarrow{A P}=\)( )
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,\(E\)是\(AC\)的中点,\(F\)是\(BC\)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( )
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\(x+y=\)\(\underline{\quad \quad}\).
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\(P\)是\(△ABC\)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\(G\)是\(△ABC\)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
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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:\(A\). -
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 \(\overrightarrow{A P}=\overrightarrow{A B}+\overrightarrow{B P}=\overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}=\vec{a}+\dfrac{\vec{b}}{2}\),
选\(A\). -
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\(\overrightarrow{O A}=\dfrac{6}{5} \overrightarrow{O E}-\dfrac{3}{5} \overrightarrow{O F}\), \(\overrightarrow{O B}=\dfrac{6}{5} \overrightarrow{O F}-\dfrac{2}{5} \overrightarrow{O E}\),
又\(\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{O B}=\dfrac{6}{5} \overrightarrow{O E}-\dfrac{3}{5} \overrightarrow{O F}+\dfrac{6}{5} \overrightarrow{O F}-\dfrac{2}{5} \overrightarrow{O E}=\dfrac{4}{5} \overrightarrow{O E}+\dfrac{3}{5} \overrightarrow{O F}\),
所以\(m=\dfrac{4}{5}\),\(n=\dfrac{3}{5}\),故\(m+n=\dfrac{7}{5}\),
故选:\(C\).
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答案 \(-\dfrac{\sqrt{3}}{3}\)
解析 如图过\(C\)作\(CE⊥OB\)于\(E\),
因为\(AB\)是圆\(O\)的直径,\(C\)、\(D\)是圆\(O\)上的点,\(∠CBA=60°\),
所以\(E\)为\(OB\)的中点,连结\(OD\),则\(\overrightarrow{C E}=\dfrac{\sqrt{3}}{2} \overrightarrow{O D}\),
\(\therefore \overrightarrow{C D}=\overrightarrow{C O}+\overrightarrow{O D}=\overrightarrow{O A}-\overrightarrow{B C}+\dfrac{2}{\sqrt{3}} \overrightarrow{C E}\), \(\overrightarrow{C E}=\overrightarrow{C B}+\overrightarrow{B E}=-\overrightarrow{B C}+\dfrac{1}{2} \overrightarrow{O A}\),
\(\overrightarrow{C D}=\overrightarrow{O A}-\overrightarrow{B C}+\dfrac{2}{\sqrt{3}}\left(-\overrightarrow{B C}+\dfrac{1}{2} \overrightarrow{O A}\right)=\left(\dfrac{1}{\sqrt{3}}+1\right) \overrightarrow{O A}-\left(1+\dfrac{2}{\sqrt{3}}\right) \overrightarrow{B C}\)
又\(\overrightarrow{C D}=x \overrightarrow{O A}+y \overrightarrow{B C}\),
\(x+y=\left(\dfrac{1}{\sqrt{3}}+1\right)-\left(1+\dfrac{2}{\sqrt{3}}\right)=-\dfrac{\sqrt{3}}{3}\).
故答案为:\(-\dfrac{\sqrt{3}}{3}\). -
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),
设 \(\overrightarrow{A P}=t \overrightarrow{A D}+(1-t) \overrightarrow{A E}=t \overrightarrow{A D}+(1-t)(\overrightarrow{A B}+\overrightarrow{B E})\)
\(=t \overrightarrow{A D}+(1-t) \overrightarrow{A B}+\dfrac{1}{3}(1-t) \overrightarrow{B C}(0 \leq t \leq 1)\),
\(\because \overrightarrow{A D}=\overrightarrow{A C}+\overrightarrow{C D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}=\dfrac{1}{2} \overrightarrow{A B}+\overrightarrow{B C}\),
\(\therefore \overrightarrow{A P}=t\left(\dfrac{1}{2} \overrightarrow{A B}+\overrightarrow{B C}\right)+(1-t) \overrightarrow{A B}+\dfrac{1}{3}(1-t) \overrightarrow{B C}=\left(1-\dfrac{1}{2} t\right) \overrightarrow{A B}+\left(\dfrac{1}{3}+\dfrac{2}{3} t\right) \overrightarrow{B C}\).
又 \(\because \overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{B C}(\lambda, \mu \in \boldsymbol{R})\),
\(\therefore\left\{\begin{array}{l} 1-\dfrac{1}{2} t=\lambda \\ \dfrac{1}{3}+\dfrac{2}{3} t=\mu \end{array}\right.\),
\(\therefore \lambda^2+\mu=\left(1-\dfrac{1}{2} t\right)^2+\dfrac{1}{3}+\dfrac{2}{3} t=\dfrac{1}{4}\left(t-\dfrac{2}{3}\right)^2+\dfrac{11}{9}\),
\(∴\)当\(t=\dfrac{2}{3}\)时,\(λ^2+μ\)的最小值为 \(\dfrac{11}{9}\). -
答案 \(\dfrac{1}{4}\)
解析 延长\(BP\),交\(AC\)于\(D\),画出图形,如图所示;
\(\therefore \overrightarrow{B A}+\overrightarrow{B C}=2 \overrightarrow{B D}\),
又\(\because \overrightarrow{B A}+\overrightarrow{B C}=6 \overrightarrow{B P}\), \(\therefore \overrightarrow{B D}=3 \overrightarrow{B P}\), \(\therefore \dfrac{S_{\triangle A B P}}{S_{\triangle A D P}}=\dfrac{1}{2}\);
又\(D\)是\(AC\)的中点,
\(\therefore S_{\triangle A D P}=S_{\triangle C D P}\), \(\therefore \dfrac{S_{\triangle A B P}}{S_{\triangle A C P}}=\dfrac{S_{\triangle A B P}}{2 S_{\triangle A D P}}=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}\). -
答案 等边三角形
解析 \(∵G\)是\(△ABC\)的重心, \(\overrightarrow{G A}=-\dfrac{2}{3} \times \dfrac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{G B}=\dfrac{1}{3}(\overrightarrow{B A}+\overrightarrow{B C}), \overrightarrow{G C}=\dfrac{1}{3}(\overrightarrow{C A}+\overrightarrow{C B})\),
又 \(\overrightarrow{G B}=\dfrac{1}{3}(\overrightarrow{B A}+\overrightarrow{B C})\),
\(\therefore(a-b) \overrightarrow{A B}+(a-c) \overrightarrow{A C}+(b-c) \overrightarrow{B C}=\overrightarrow{0}\),
\(∴a-b=a-c=b-c\),
∴\(a=b=c\).
\(∴△ABC\)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\(k \vec{a}+\vec{b}\)与\(\vec{a}-\vec{b}\)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\(k \vec{a}+\vec{b}\)与\(\vec{a}-\vec{b}\)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\(E\)为\(AB\)边的中点,点\(F\)在\(AC\)边上,且\(CF=2FA\),\(BF\)交\(CE\)于点\(M\),设\(\overrightarrow{A M}=x \overrightarrow{A E}+y \overrightarrow{A F}\),则\(x+y=\)\(\underline{\quad \quad}\) .
解析 如图,\(E\),\(M\),\(C\)三点共线,\(∴\)存在实数\(λ\),使\(\overrightarrow{A M}=\lambda \overrightarrow{A E}+(1-\lambda) \overrightarrow{A C}\),
\(∵CF=2FA\),
\(∴AC=3AF\),\(\therefore \overrightarrow{A M}=\lambda \overrightarrow{A E}+3(1-\lambda) \overrightarrow{A F}\),
又\(\overrightarrow{A M}=x \overrightarrow{A E}+y \overrightarrow{A F}\);
\(\therefore\left\{\begin{array}{l}
\lambda=x \\
3(1-\lambda)=y
\end{array}\right.\),\(∴3(1-x)=y\)①;
同样,\(B\),\(M\),\(F\)三点共线,所以存在\(μ\),使\(\overrightarrow{A M}=\mu \overrightarrow{A B}+(1-u) \overrightarrow{A F}\),
\(∵E\)为\(AB\)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( )
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\(G\)是\(△ABC\)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
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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\(D\). -
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\(P\)作\(PE∥ON\),\(PF∥OM\),分别交\(OM\),\(ON\)于点\(E\),\(F\);
\(PE\)交\(AB\)于点\(P'\),过点\(P'\)作\(P'F'∥OM\)交\(ON\)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\(\overrightarrow{O P}=x^{\prime} \overrightarrow{O E}+y^{\prime} \overrightarrow{O F}=x^{\prime} \overrightarrow{O E}+y^{\prime \prime} \overrightarrow{O F^{\prime}}=u \overrightarrow{O A}+v \overrightarrow{O B}\),
而\(x'=x\),\(y″>y\),
\(∴u=u'\),\(v>v'\),\(∴u+v>u'+v'=1\).
即可判断出\(A\),\(∵1+2>1\),\(∴\)点\(P\)位于阴影区域内,故正确;
同理\(C\)正确;而\(BD\)不正确;
故选:\(AC\).
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答案 \(2\)
解析 由\(3 \overrightarrow{O A}=\overrightarrow{O B}+\lambda \overrightarrow{O C}\),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\(λ=2\). -
答案 \(3\)
解析 如图所示,
\(∵\)三点\(E\),\(G\),\(F\)共线,\(∴\)存在实数\(λ\)三点\(\overrightarrow{A G}=\lambda \overrightarrow{A E}+(1-\lambda) \overrightarrow{A F}\),
\(\because \overrightarrow{A E}=\alpha \overrightarrow{A B}, \overrightarrow{A F}=\beta \overrightarrow{A C}\), \(\therefore \overrightarrow{A G}=\lambda \alpha \overrightarrow{A B}+(1-\lambda) \beta \overrightarrow{A C}\).
\(∵G\)是\(△ABC\)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,\(AC\)与\(BD\)交于点\(O\), \(\overrightarrow{A B}=\vec{a}\), \(\overrightarrow{A D}=\vec{b}\),则下列运算正确的是( )
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\(A\)是\(BC\)的中点,\(\overrightarrow{O D}=2 \overrightarrow{D B}\),\(DC\)和\(OA\)交于点\(E\),则\(AO\)与\(OE\)的比值为( )
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( )
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\(\underline{\quad \quad}\).
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\(AB\)、\(AC\)于\(P\)、\(Q\)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\(\underline{\quad \quad}\).
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:\(AR=RT=TC\).
参考答案
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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:\(C\). -
答案 \(B\)
解析 对于①,若\(\overrightarrow{A D}=\overrightarrow{B C}\),则\(|\overrightarrow{A D}|=|\overrightarrow{B C}|\)且\(\overrightarrow{A D} / / \overrightarrow{B C}\),四边形\(ABCD\)是平行四边形,①正确;
对于②,若\(\overrightarrow{A D}=\dfrac{1}{3} \overrightarrow{B C}\), \(|\overrightarrow{A D}| \neq|\overrightarrow{B C}|\)且\(\overrightarrow{A D} / / \overrightarrow{B C}\),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,
由\(|\overrightarrow{A B}|=|\overrightarrow{A D}|\),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\((\overrightarrow{A B}+\overrightarrow{A D})^2=(\overrightarrow{A B}-\overrightarrow{A D})^2\),
即\(\overrightarrow{A B}^2+2 \overrightarrow{A B} \cdot \overrightarrow{A D}+\overrightarrow{A D}^2=\overrightarrow{A B}^2-2 \overrightarrow{A B} \cdot \overrightarrow{A D}+\overrightarrow{A D}^2\),
\(\therefore \overrightarrow{A B} \cdot \overrightarrow{A D}=0\)即\(AB⊥AD\),如图所示
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:\(B\). -
答案 \(B\)
解析 \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{b}\), \(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=-2 \vec{b}\),
\(\overrightarrow{A B}=-2 \overrightarrow{A D}\), \(\overrightarrow{A B}\)和\(\overrightarrow{A D}\)共线,且有公共点,所以\(A\),\(B\),\(D\)三点共线.
故选:\(B\). -
答案 \(C\)
解析 \(∵A\),\(B\),\(C\)三点共线, \(\therefore \overrightarrow{A B}\)与 \(\overrightarrow{B C}\)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\(m=-6\). -
答案 \(C\)
解析 \(∵O\),\(E\),\(A\)三点共线,且\(A\)是\(BC\)的中点;
\(\therefore \overrightarrow{O B}+\overrightarrow{O C}=2 \overrightarrow{O A}\);
设\(\overrightarrow{O E}=\dfrac{\lambda}{2}(\overrightarrow{O B}+\overrightarrow{O C})\),而\(\overrightarrow{O B}=\dfrac{3}{2} \overrightarrow{O D}\),
代入上式便可得出\(\overrightarrow{O E}=\dfrac{3 \lambda}{4} \overrightarrow{O D}+\dfrac{\lambda}{2} \overrightarrow{O C}\),
由\(C\),\(E\),\(D\)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\(AO\)与\(OE\)的比值为 \(\dfrac{5}{4}\).
故选:\(C\). -
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\(\overrightarrow{O P}\)与\(\overrightarrow{O C}\)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;
当\(P\)为\(OC\)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\(\mu=\dfrac{1}{2}\),
故\(A\)正确,\(B\)错误;
故选 \(AC\). -
答案 \(2\)
解析 \(\overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B}=3 \overrightarrow{A B}-\overrightarrow{A B}=2 \overrightarrow{A B}\). -
答案 \(1008(\vec{a}+\vec{b})\)
解析 设\(D\)为\(A_0\)和\(A_{2015}\)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 \(\overrightarrow{O A_1}+\overrightarrow{O A_{2014}}=2 \overrightarrow{O D}\),
……
\(\overrightarrow{O A_{1007}}+\overrightarrow{O A_{1008}}=2 \overrightarrow{O D}\),
\(\therefore \overrightarrow{O A_0}+\overrightarrow{O A_1}+\cdots+\overrightarrow{O A_{2015}}=\dfrac{2016}{2} \cdot 2 \overrightarrow{O D}=1008(\vec{a}+\vec{b})\) . -
答案 \(\dfrac{2}{3} \vec{b}+\dfrac{1}{3} \vec{a}\)
解析 过\(D\)作\(DE//AB\),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 \(\overrightarrow{A D}=\overrightarrow{A E}+\overrightarrow{A F}=\dfrac{2}{3} \overrightarrow{A C}+\dfrac{1}{3} \overrightarrow{A B}=\dfrac{2}{3} \vec{b}+\dfrac{1}{3} \vec{a}\).
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答案 \(\left[2, \dfrac{9}{4}\right]\)
解析 \(∵P\),\(G\),\(Q\)三点共线,\(∴\)存在\(m\),使 \(\overrightarrow{A G}=m \overrightarrow{A Q}+(1-m) \overrightarrow{A P}\),
又\(∵G\)是\(△ABC\)的重心,
\(\therefore \overrightarrow{A G}=\dfrac{1}{3}(\overrightarrow{A B}+\overrightarrow{A C})=\dfrac{1}{3}(y \overrightarrow{A Q}+x \overrightarrow{A P})\),
\(\therefore \dfrac{1}{3}(y \overrightarrow{A Q}+x \overrightarrow{A P})=m \overrightarrow{A Q}+(1-m) \overrightarrow{A P}\),
\(∴x+y=3\),
又\(\because \overrightarrow{A B}=x \overrightarrow{A P}\),\(∴1≤x≤2\),
故\(x y=x(3-x)=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\),
故\(2 \leq-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4} \leq \dfrac{9}{4}\),
故答案为:\(\left[2, \dfrac{9}{4}\right]\). -
答案 \(\dfrac{5}{2}\)
解析 如图所示.\(∵\)点\(P\)在\(△ABC\)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.
以\(PB\)、\(PC\)为邻边作平行四边形\(PBDC\),连接\(PD\)交\(BC\)于点\(O\).
则 \(\overrightarrow{P B}+\overrightarrow{P C}=\overrightarrow{P D}=2 \overrightarrow{P O}=-2 \overrightarrow{P A}\),
化为 \(\overrightarrow{P A}+\dfrac{1}{2} \overrightarrow{P B}+\dfrac{1}{2} \overrightarrow{P C}=\overrightarrow{0}\).
\(\because \overrightarrow{P A}+x \overrightarrow{P B}+y \overrightarrow{P C}=\overrightarrow{0}\), \(\therefore x=y=\dfrac{1}{2}\).
\(\therefore 2 x+3 y=\dfrac{5}{2}\).
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证明 设\(\overrightarrow{A B}=\vec{a}\), \(\overrightarrow{A D}=\vec{b}\), \(\overrightarrow{A R}=\vec{r}\), \(\overrightarrow{A T}=\vec{t}\),则 \(\overrightarrow{A C}=\vec{a}+\vec{b}\).
因为\(\overrightarrow{A R}\)与\(\overrightarrow{A C}\)共线,
所以存在实数\(n\),使得\(\vec{r}=n(\vec{a}+\vec{b})\),\(n∈R\).
因为\(\overrightarrow{E R}\)与\(\overrightarrow{E B}\)共线,所以存在实数\(m\),使得\(\overrightarrow{E R}=m\overrightarrow{EB}\),\(m∈R\).
而\(\overrightarrow{E B}=\overrightarrow{A B}-\overrightarrow{A E}=\vec{a}-\dfrac{1}{2} \vec{b}\),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\(\overrightarrow{A R}=\overrightarrow{R T}=\overrightarrow{T C}\).
故\(AR=RT=TC\).
【B组---提高题】
1.已知点\(O\)是\(△ABC\)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. \(\dfrac{3}{4}\)
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
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答案 \(A\)
解析 如图所示,延长\(OB\)到\(D\)使得\(BD=OB\),延长\(OC\)到\(E\)使得\(CE=2OC\),
\(∵\)满足\(\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\overrightarrow{0}\),
∴点\(O\)是\(△ADE\)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:\(A\). -
答案 \(\dfrac{\sqrt{17}}{2}\)
解析 如图,\(⊙O\)为单位圆,\(A\)、\(B\)、C在\(⊙O\)上,\(OA⊥OB\), \(\angle B O A=\dfrac{\pi}{2}\),
\(B'\)在\(OB\)的延长线上,\(OB'=2\),\(B″\)为\(OB\)中点,
\(A'\)为\(OA\)中点,\(A″\)在\(OB\)的延长线上,\(OA″=2\),
设\(\vec{a}=\overrightarrow{O A}\),\(\vec{b}=\overrightarrow{O B}\),\(C\)为\(⊙O\)上一点,\(\vec{c}=\overrightarrow{O C}\),
则\(\dfrac{O A^{\prime}}{O C^{\prime}}=\dfrac{O C}{O A^{\prime \prime}}=\dfrac{1}{2}\),\(∴△OCA'∽△OA″C\),\(∴CA'=2A″C\),
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\(P\)在\(△ABC\)所在平面内,则点\(G\)与\(P\)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,\(\overrightarrow{O C}=\dfrac{1}{3} \overrightarrow{O A}\),\(\overrightarrow{O D}=\dfrac{1}{2} \overrightarrow{O B}\),\(AD\)与\(BC\)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\(EF\)与\(AD\)重合时,\(λ=1\),\(\mu=\dfrac{1}{2}\),此时\(\dfrac{1}{\lambda}+\dfrac{2}{\mu}=5\);当\(EF\)与\(BC\)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.
参考答案
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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\(P\)为\(△ABC\)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, \(\overrightarrow{A P} \cdot \overrightarrow{B C}=\lambda\left(\dfrac{\overrightarrow{A B}}{|\overrightarrow{A B}| \cos B}+\dfrac{\overrightarrow{A C}}{|\overrightarrow{A C}| \cos C}\right) \cdot \overrightarrow{B C}=\lambda(-|\overrightarrow{B C}|+|\overrightarrow{B C}|)=0\),
\(\therefore \overrightarrow{A P} \perp \overrightarrow{B C}\),
\(∴\)点\(P\)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\(OH\)交\(AF\)于\(G\),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. -
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 \(\overrightarrow{O M}=m \vec{a}+n \vec{b}(m \in R, \quad n \in R)\),
由\(A\),\(D\),\(M\)三点共线,可知存在\(α(α∈R\),且\(α≠-1)\)使得 \(\overrightarrow{A M}=\alpha \overrightarrow{M D}\),
则\(\overrightarrow{O M}-\overrightarrow{O A}=\alpha(\overrightarrow{O D}-\overrightarrow{O M})\),
又\(\overrightarrow{O D}=\dfrac{1}{2} \overrightarrow{O B}\),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,
由\(B\),\(C\),\(M\)三点共线,可知存在\(β(β∈R\),且\(β≠-1)\)使得 \(\overrightarrow{C M}=\beta \overrightarrow{M B}\),
则 \(\overrightarrow{O M}-\overrightarrow{O C}=\beta(\overrightarrow{O B}-\overrightarrow{O M})\),又 \(\overrightarrow{O C}=\dfrac{1}{3} \overrightarrow{O A}\),
所以\(\overrightarrow{O M}=\dfrac{1}{3(\beta+1)} \vec{a}+\dfrac{\beta}{1+\beta} \vec{b}\), \(\therefore\left\{\begin{array}{l} m=\dfrac{1}{4(1+\beta)} \\ n=\dfrac{\beta}{1+\beta} \end{array}\right.\)
即\(3m+n=1\)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\(γ\)得\(\dfrac{1}{\lambda}+\dfrac{2}{\mu}=5\).