6.3.2—6.3.4 平面向量的坐标运算
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必修第二册同步巩固,难度2颗星!
基础知识
正交分解及其坐标表示
① 正交分解
把一个向量分解为两个互相垂直的向量,叫做把向量作正交分解;
如上图,重力\(G\)分解成平行斜面的力\(F_1\)和垂直于斜面的压力\(F_2\).
② 向量的坐标表示
在平面内建立直角坐标系,以与\(x\)轴、\(y\)轴方向相同的两个单位向量 \(\vec{i}\) , \(\vec{j}\)为基底,
则平面内的任一向量\(\vec{a}\)可表示为\(\vec{a}=x \vec{i}+y \vec{j}=(x, y)\),
\((x ,y)\)称为向量\(\vec{a}\)的坐标,\(\vec{a}=(x, y)\)叫做向量\(\vec{a}\)的坐标表示.
向量\(\vec{a}=(x, y)\),可看成以原点为起点,点\((x ,y)\)为终点的向量.
【例】如下图,用基底\(\{\vec{i}, \vec{j}\}\)表示向量\(\overrightarrow{A B}\),并求出坐标.
解 \(\overrightarrow{A B}=\overrightarrow{A A_1}+\overrightarrow{A A_2}=2 \vec{i}+\vec{j}\),所以\(\overrightarrow{A B}=(2,1)\).
平面向量的坐标运算
设\(\vec{a}=\left(x_1, y_1\right)\) , \(\vec{b}=\left(x_2, y_2\right)\),则
(1) 向量的模 \(|\vec{a}|=\sqrt{x_1^2+y_1^2}\)
(2) 向量的加减法运算 \(\vec{a}+\vec{b}=\left(x_1+x_2, y_1+y_2\right)\), \(\vec{a}-\vec{b}=\left(x_1-x_2, y_1-y_2\right)\)
(3) 若\(A(x_1 ,y_1)\),\(B(x_2 ,y_2)\),则 \(\overrightarrow{A B}=\left(x_2-x_1, y_2-y_1\right)\)
(4) 实数与向量的积 \(\lambda \vec{a}=\lambda\left(x_1, y_1\right)=\left(\lambda x_1, \lambda y_1\right)\)
拓展 定比分点
线段\(P_1 P_2\)的端点\(P_1\) 、\(P_2\)的坐标分别是\((x_1 ,y_1)\),\((x_2 ,y_2)\),点\(P\)是直线\(P_1 P_2\)上的一点,
当\(\overrightarrow{P_1 P}=\lambda \overrightarrow{P P_2}\)时,点\(P\)的坐标是 \(\left(\dfrac{x_1+\lambda x_2}{1+\lambda} , \dfrac{y_1+\lambda y_2}{1+\lambda}\right)\).
【例】若\(\vec{a}=(-4,3)\), \(\vec{b}=(2,5)\),求 \(|\vec{a}|\), \(2 \vec{b}-3 \vec{a}\).
解 \(|\vec{a}|=\sqrt{(-4)^2+3^2}=5\), \(2 \vec{b}-3 \vec{a}=(4,10)-(-12,9)=(16,1)\).
平行向量
若\(\vec{a}(x_1 ,y_1)\),\(\vec{b}(x_2 ,y_2)\),其中\(\vec{b}≠\vec{0}\),则\(\vec{a}∥\vec{b} ⇔ x_1 y_2=x_2 y_1\).
证明 \(\vec{a}∥\vec{b}\)的充要条件是存在实数\(λ\),使得\(\vec{a}=λ\vec{b}\),所以\((x_1 ,y_1)=λ(x_2 ,y_2)\),
所以\(\left\{\begin{array}{l}
x_1=\lambda x_2 \\
y_1=\lambda y_2
\end{array}\right.\),消\(λ\)得\(x_1 y_2=x_2 y_1\).
【例】已知\(\vec{a}=(4,2)\),\(\vec{b}=(2,m)\),且\(\vec{a}||\vec{b}\),求\(m\).
解 \(∵\vec{a}||\vec{b}\),\(∴4m=4\),解得\(m=1\).
基本方法
【题型1】 平面向量的坐标运算
【典题1】 已知向量\(\vec{a}=(2,1)\),\(\vec{b}=(-1,k)\),若存在实数\(λ\),使得\(\vec{a}=λ\vec{b}\),则\(k\)和\(λ\)的值分别为( )
A.\(-\dfrac{1}{2},-2\) \(\qquad \qquad \qquad\) B. \(\dfrac{1}{2},-2\) \(\qquad \qquad \qquad\)C.\(-\dfrac{1}{2}, 2\) \(\qquad \qquad \qquad\) D.\(\dfrac{1}{2}, 2\)
解析 \(∵\vec{a}=λ\vec{b}\),\(∴(2,1)=(-λ,kλ)\),
\(\therefore\left\{\begin{array}{l}
-\lambda=2 \\
k \lambda=1
\end{array}\right.\),解得\(λ=-2\), \(k=-\dfrac{1}{2}\).
故选:\(A\).
点拨 实数与向量的积\(λ\vec{a}=λ(x_1 ,y_1)=(λx_1 ,λy_1)\).
【典题2】已知向量\(\vec{a}\),\(\vec{b}\)满足\(2\vec{a}-\vec{b}=(0,3)\),\(\vec{a}-2\vec{b}=(-3,0)\),\(λ\vec{a}+μ\vec{b}=(-1,1)\),则\(λ+μ=\)( )
A.\(-1\) \(\qquad \qquad \qquad\) B.\(0\) \(\qquad \qquad \qquad\) C.\(1\) \(\qquad \qquad \qquad\)D.\(2\)
解析 \(∵2\vec{a}-\vec{b}=(0,3)\),则\(4\vec{a}-2\vec{b}=(0,6)\),①,
又\(\vec{a}-2\vec{b}=(-3,0)\),②,
由① ②得:\(3\vec{a}=(3,6)\),即\(\vec{a}=(1,2)\),
同理,\(\vec{b}=(2,1)\),
又\(λ\vec{a}+μ\vec{b}=(λ+2μ,2λ+μ)=(-1,1)\),
即\(\left\{\begin{array}{l}
\lambda+2 \mu=-1 \\
2 \lambda+\mu=1
\end{array}\right.\),得\(\left\{\begin{array}{l}
\lambda=1 \\
\mu=-1
\end{array}\right.\),
故\(λ+μ=1+(-1)=0\),
故答案为:\(B\).
点拨 由\(2\vec{a}-\vec{b}=(0,3)\),\(\vec{a}-2\vec{b}=(-3,0)\)得\(\vec{a}\),\(\vec{b}\)的坐标类似解一个二元一次方程.
【巩固练习】
1.设向量\(\vec{a}=(1,1)\),\(\vec{b}=(3,-2)\),则\(3\vec{a}-2\vec{b}=\) ( )
A.\((-3,7)\) \(\qquad \qquad \qquad\) B.\((0,7)\) \(\qquad \qquad \qquad\)C.\((3,5)\) \(\qquad \qquad \qquad\) D.\((-3,5)\)
2.已知点\(A(0,3)\),\(B(-1,2)\),且\(\overrightarrow{B C}=(3,-4)\),则 \(\overrightarrow{A C}=\) ( )
A.\((2,5)\) \(\qquad \qquad \qquad\) B.\((2,-5)\) \(\qquad \qquad \qquad\) C.\((-2,-5)\) \(\qquad \qquad \qquad\) D.\((-2,5)\)
3.已知向量\(\vec{a}=(1,2)\),\(\vec{b}=(2,3)\),\(\vec{c}=(3,4)\),若\(\vec{c}=λ\vec{a}+μ\vec{b}\),则\(λ+μ=\) ( )
A.\(1\) \(\qquad \qquad \qquad\qquad\) B.\(-1\)\(\qquad \qquad \qquad \qquad\) C.\(-2\) \(\qquad \qquad \qquad \qquad\) D.\(3\)
参考答案
- 答案 \(A\)
解析 \(\vec{a}=(1,1)\),\(\vec{b}=(3,-2)\),
则\(3\vec{a}-2\vec{b}=(3,3)-(6,-4)=(-3,7)\).
故选:\(A\). - 答案 \(B\)
解析 \(\because\)点\(A(0,3)\),\(B(-1,2)\),且\(\overrightarrow{B C}=(3,-4)\),
\(\therefore \overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=(-1,-1)+(3,-4)=(2,-5)\).
故选:\(B\). - 答案 \(A\)
解析 由\(\vec{c}=λ\vec{a}+μ\vec{b}=(λ,2λ)+(2μ,3μ)=(λ+2μ,2λ+3μ)=(3,4)\),
所以\(λ+2μ=3\),\(2λ+3μ=4\),解得\(λ=-1\),\(μ=2\),所以\(λ+μ=1\),
故选:\(A\).
【题型2】 平行向量
【典题1】 已知向量\(\vec{a}=(-1,2)\),\(\vec{b}=(1,-2λ)\),若\((\vec{a}+3\vec{b})||(\vec{a}-\vec{b})\),则实数\(λ\)的值为( )
A.\(1\) \(\qquad \qquad \qquad\) B.\(0\) \(\qquad \qquad \qquad\) C. \(\dfrac{4}{3}\)\(\qquad \qquad \qquad\) D. \(-\dfrac{2}{3}\)
解析 根据题意,向量\(\vec{a}=(-1,2)\),\(\vec{b}=(1,-2λ)\),
则\(\vec{a}+3\vec{b}=(2,2-6λ)\),\(\vec{a}-\vec{b}=(-2,2+2λ)\),
若\((\vec{a}+3\vec{b})||(\vec{a}-\vec{b})\),则\(2(2+2λ)=-2×(2-6λ)\),解可得:\(λ=1\),
故选:\(A\).
点拨 若\(\vec{a}(x_1 ,y_1)\),\(\vec{b}(x_2 ,y_2)\),其中\(\vec{b}≠\vec{0}\),则\(\vec{a}∥\vec{b} ⇔ x_1 y_2=x_2 y_1\).
【典题2】已知点\(A(-1,-1)\),\(B(1,3)\),\(C(x,5)\),若对于平面上任意一点\(O\),都有\(\overrightarrow{O C}=\lambda \overrightarrow{O A}+(1-\lambda) \overrightarrow{O B}\),\(λ∈R\),则\(x=\)\(\underline{\quad \quad}\).
解析 \(\because A(-1,-1)\),\(B(1,3)\),\(C(x,5)\),且对于平面上任意一点\(O\),都有\(\overrightarrow{O C}=\lambda \overrightarrow{O A}+(1-\lambda) \overrightarrow{O B}\),\(λ∈R\),
\(\therefore A\),\(B\),\(C\)三点共线,即\(\overrightarrow{A B}\)与\(\overrightarrow{A C}\)共线,
\(\because \overrightarrow{A B}=(2,4)\), \(\overrightarrow{A C}=(x+1,6)\),
,解得:\(x=2\).
故答案为:\(2\).
【巩固练习】
1.已知平面向量\(\vec{a}\),\(\vec{b}\)满足\(\vec{a}=(-2,1)\),\(|\vec{b}|=3 \sqrt{5}\),\(\vec{a}\)与\(\vec{b}\)方向相同,则\(\vec{b}\)的坐标是( )
A.\((3,-6)\) \(\qquad \qquad \qquad\) B.\((6,-3)\) \(\qquad \qquad \qquad\) C.\((-6,3)\) \(\qquad \qquad \qquad\) D.\((-3,6)\)
2.已知向量\(\vec{m}=(\sqrt{3}, 2 \cos \theta+1)\), \(\vec{n}=(1,2 \sin \theta)\),且 \(\vec{m} / / \vec{n}\),则 \(\sin \left(\theta-\dfrac{\pi}{6}\right)=\) \(\underline{\quad \quad}\).
3.已知向量\(\vec{a}=(3,-2)\),\(\vec{b}=(x,y-1)\)且\(\vec{a}||\vec{b}\),若\(x\),\(y\)均为正数,则 \(\dfrac{3}{x}+\dfrac{2}{y}\)的最小值是\(\underline{\quad \quad}\).
参考答案
- 答案 \(C\)
解析 已知平面向量\(\vec{a}\),\(\vec{b}\)满足\(\vec{a}=(-2,1)\),\(|\vec{b}|=3 \sqrt{5}\),\(\vec{a}\)与\(\vec{b}\)方向相同,
则\(\vec{b}=(-2λ,λ)\),其中\(λ>0\),
则 \((-2 \lambda)^2+\lambda^2=45\),即\(λ=3\),即\(\vec{b}=(-6,3)\),
故选:\(C\). - 答案 \(\dfrac{1}{4}\)
解析 \(\because \vec{m} / / \vec{n}\),
\(\therefore 2 \sqrt{3} \sin \theta-2 \cos \theta-1=4 \sin \left(\theta-\dfrac{\pi}{6}\right)-1=0\),解得 \(\sin \left(\theta-\dfrac{\pi}{6}\right)=\dfrac{1}{4}\).
故答案为:\(\dfrac{1}{4}\). - 答案 \(8\)
解析 \(\because \vec{a}||\vec{b}\),\(\therefore -2x-3(y-1)=0\),化简得\(2x+3y=3\),
\(\therefore \dfrac{3}{x}+\dfrac{2}{y}=\left(\dfrac{3}{x}+\dfrac{2}{y}\right) \times \dfrac{1}{3}(2 x+3 y)=\dfrac{1}{3}\left(6+\dfrac{9 y}{x}+\dfrac{4 x}{y}+6\right)\)\(\geqslant \dfrac{1}{3}\left(12+2 \sqrt{\dfrac{9 y}{x} \cdot \dfrac{4 x}{y}}\right)=8\),
当且仅当 \(2 x=3 y=\dfrac{3}{2}\)时,等号成立;
\(\therefore \dfrac{3}{x}+\dfrac{2}{y}\)的最小值是\(8\).
【题型3】平面向量的坐标运用
【典题1】 平面上有\(A(2,1)\),\(B(-1,4)\),\(D(-2,3)\)三点,点\(C\)在直线\(AB\)上,且\(\overrightarrow{A C}=2 \overrightarrow{B C}\),连接\(DC\)并延长\(DC\)至\(E\),使 \(|\overrightarrow{C E}|=\dfrac{1}{2}|\overrightarrow{C D}|\),则点E的坐标为( )
A.\((-5,9)\) \(\qquad \qquad \qquad\) B.\((-3,9)\) \(\qquad \qquad \qquad\) C.\((-1,4)\) \(\qquad \qquad \qquad\) D.\((-3,7)\)
解析 因为\(A(2,1)\),\(B(-1,4)\),\(D(-2,3)\)三点,点\(C\)在直线\(AB\)上,且\(\overrightarrow{A C}=2 \overrightarrow{B C}\),
所以\(B\)为\(AC\)的中点,则\(C(-4,7)\),
连接\(DC\)并延长\(DC\)至\(E\),使\(|\overrightarrow{C E}|=\dfrac{1}{2}|\overrightarrow{C D}|\),即 \(\overrightarrow{C E}=-\dfrac{1}{2} \overrightarrow{C D}\),
设\(E(x,y)\),则 \((x+4, y-7)=-\dfrac{1}{2}(2,-4)\),
即\(\left\{\begin{array}{l}
x+4=-1 \\
y-7=2
\end{array}\right.\),解得\(\left\{\begin{array}{l}
x=-5 \\
y=9
\end{array}\right.\),
所以点\(E\)的坐标为\((-5,9)\).
故选:\(A\).
点拨 求点\(E\)的坐标使用待定系数法.
【典题2】已知直角梯形\(ABCD\)中,\(AD⊥AB\),\(AB=2AD=2CD\),过点\(C\)作\(CE⊥AB\),垂足为点\(E\),\(M\)为\(CE\)的中点,用向量的方法证明:
(1)\(DE||BC\);\(\qquad \qquad\) (2)\(D\),\(M\),\(B\)三点共线.
解析 (1)证明:如图,以\(E\)为原点,\(AB\)所在直线为\(x\)轴,\(EC\)所在直线为\(y\)轴建立平面直角坐标系,
令\(|\overrightarrow{A D}|=1\),则 \(|\overrightarrow{D C}|=1\), \(|\overrightarrow{A B}|=2\)
因为\(CE⊥AB\),\(AD⊥AB\),\(CD||AB\),\(AD=DC\),
易知四边形\(AECD\)为正方形.
所以可求得各点坐标分别为\(E(0,0)\),\(B(1,0)\),\(C(0,1)\),\(D(-1,1)\),\(A(-1,0)\),
因为\(\overrightarrow{E D}=(-1,1)-(0,0)=(-1,1)\), \(\overrightarrow{B C}=(0,1)-(1,0)=(-1,1)\),
所以\(\overrightarrow{E D}=\overrightarrow{B C}\),所以\(\overrightarrow{E D} / / \overrightarrow{B C}\),
又\(ED\)与\(BC\)无公共点,所以\(DE||BC\).
(2)证明:连接\(MB\),\(MD\).
因为\(M\)为\(CE\)的中点,所以\(M\left(0, \dfrac{1}{2}\right)\),
所以\(\overrightarrow{M D}=(-1,1)-\left(0, \dfrac{1}{2}\right)=\left(-1, \dfrac{1}{2}\right)\), \(\overrightarrow{M B}=(1,0)-\left(0, \dfrac{1}{2}\right)=\left(1,-\dfrac{1}{2}\right)\),
所以\(\overrightarrow{M D}=-\overrightarrow{M B}\),所以 \(\overrightarrow{M D} / / \overrightarrow{M B}\).
又\(MD\)与\(MB\)有公共点\(M\),所以\(D\),\(M\),\(B\)三点共线.
点拨 感受下向量法对于处理几何问题的威力,建系利用向量的坐标表示有时候往往使得问题更简便.
【巩固练习】
1.已知\(O\)是坐标原点,点\(A\)在第二象限,\(|\overrightarrow{O A}|=2\), \(\angle x O A=150^{\circ}\),求向量\(\overrightarrow{O A}\)的坐标为\(\underline{\quad \quad}\).
2.已知平行四边形\(ABCD\)的三个顶点\(A\),\(B\),\(C\)的坐标分别是\((-2,1)\),\((-1,3)\),\((3,4)\),则向量 \(\overrightarrow{B D}\)的坐标是\(\underline{\quad \quad}\).
3.已知对任意平面向量\(\overrightarrow{A B}=(x, y)\),把\(\overrightarrow{A B}\)绕其起点沿逆时针方向旋转\(θ\)角得到向量\(\overrightarrow{A P}=(x \cos \theta-y \sin \theta,\)\(x \sin \theta+y \cos \theta)\).如图所示,顶角\(∠Q=120^∘\)的等腰三角形\(PQR\)的顶点\(P\)、\(Q\)的坐标分别为\(P(1,0)\)、 \(Q(3, \sqrt{3})\),则顶点\(R\)的坐标为\(\underline{\quad \quad}\) .
4.如图,在直角梯形\(ABCD\)中,\(AB||DC\),\(AD⊥DC\),\(AD=DC=2AB\),\(E\)为\(AD\)的中点,若\(\overrightarrow{C A}=\lambda \overrightarrow{C E}+\mu \overrightarrow{D B}\),则\(λ+μ\)的值为\(\underline{\quad \quad}\).
参考答案
- 答案 \((-\sqrt{3}, 1)\)
解析 \(\because O\)是坐标原点,点\(A\)在第二象限,\(|\overrightarrow{O A}|=2\), \(\angle x O A=150^{\circ}\),
\(\therefore x_A=|\overrightarrow{O A}| \cdot \cos \angle x O A=2 \times \dfrac{-\sqrt{3}}{2}=-\sqrt{3}\),
\(y_A=|\overrightarrow{O A}| \cdot \sin \angle x O A=2 \times \dfrac{1}{2}=1\),即 \(A(-\sqrt{3}, 1)\), \(\therefore \overrightarrow{O A}=(-\sqrt{3}, 1)\).
故答案为: \((-\sqrt{3}, 1)\). - 答案 \((3,-1)\)
解析 \(\because\) 平行四边形\(ABCD\)的三个顶点\(A\),\(B\),\(C\)的坐标分别是\((-2,1)\),\((-1,3)\),\((3,4)\),
\(\therefore \overrightarrow{B A}=(-2,1)-(-1,3)=(-1,-2)\), \(\overrightarrow{A D}=\overrightarrow{B C}=(3,4)-(-1,3)=(4,1)\).
\(\therefore \overrightarrow{B D}=\overrightarrow{B A}+\overrightarrow{A D}=(-1,-2)+(4,1)=(3,-1)\). - 答案 \(\left(\dfrac{5}{2}, \dfrac{5 \sqrt{3}}{2}\right)\)
解析 设\(R(x,y)\),则\(\overrightarrow{Q R}=(x-3, y-\sqrt{3})\), \(\overrightarrow{Q P}=(-2,-\sqrt{3})\),
因为\(\angle Q=120^{\circ}\) ,所以\(\left\{\begin{array}{l} -2=(x-3) \cos 120^{\circ}-(y-\sqrt{3}) \sin 120^{\circ} \\ -\sqrt{3}=(x-3) \sin 120^{\circ}+(y-\sqrt{3}) \sin 120^{\circ} \end{array}\right.\),
解得\(x=\dfrac{5}{2}\),\(y=\dfrac{5 \sqrt{3}}{2}\),即顶点\(R\)的坐标为\(\left(\dfrac{5}{2}, \dfrac{5 \sqrt{3}}{2}\right)\).
故答案为:\(\left(\dfrac{5}{2}, \dfrac{5 \sqrt{3}}{2}\right)\). - 答案 \(\dfrac{8}{5}\)
解析 如图所示,建立直角坐标系.
不妨设\(AB=1\),则\(D(0,0)\),\(C(2,0)\),\(A(0,2)\),\(B(1,2)\),\(E(0,1)\).
\(\overrightarrow{C A}=(-2,2)\), \(\overrightarrow{C E}=(-2,1)\), \(\overrightarrow{D B}=(1,2)\),
\(\because\)若 \(\overrightarrow{C A}=\lambda \overrightarrow{C E}+\mu \overrightarrow{D B}\),\(\therefore (-2,2)=λ(-2,1)+μ(1,2)\),
\(\therefore\left\{\begin{array}{l} -2 \lambda+\mu=-2 \\ \lambda+2 \mu=2 \end{array}\right.\),解得 \(\lambda=\dfrac{6}{5}\), \(\mu=\dfrac{2}{5}\).则 \(\lambda+\mu=\dfrac{8}{5}\).
分层练习
【A组---基础题】
1.已知向量\(\vec{a}=(-1,2)\),\(\vec{b}=(3,-5)\),则\(3\vec{a}+2\vec{b}\)等于( )
A.\((3,-4)\) \(\qquad \qquad \qquad\) B.\((0,-4)\) \(\qquad \qquad \qquad\) C.\((3,6)\) \(\qquad \qquad \qquad\) D.\((0,6)\)
2.已知向量\(\vec{a}=\left(-2, \dfrac{3}{2}\right)\), \(2\vec{a}+3 \vec{b}=(5,-3)\),则\(\vec{b}=\) ( )
A.\((-3,2)\) \(\qquad \qquad \qquad\) B.\((3,-2)\) \(\qquad \qquad \qquad\) C.\((3,0)\) \(\qquad \qquad \qquad\) D.\((9,6)\)
3.已知平面向量\(\vec{a}=(m,-4)\),\(\vec{b}=(-1,m+3)\),若存在实数\(λ<0\),使得\(\vec{a}=λ\vec{b}\),则实数\(m\)的值为( )
A.\(1\) \(\qquad \qquad \qquad \qquad\) B. \(-\dfrac{12}{5}\) \(\qquad \qquad \qquad \qquad\) C.\(-1\) \(\qquad \qquad \qquad \qquad\) D.\(-4\)
4.已知向量\(\vec{a}=(-2,2)\),\(\vec{b}=(1,-2λ)\),若\((\vec{a}+3\vec{b})||(\vec{a}-\vec{b})\),则实数\(λ\)的值为( )
A.\(0\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{1}{2}\)\(\qquad \qquad \qquad \qquad\) C.\(1\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{4}{3}\)
5.已知平面直角坐标系内\(△ABC\)三个顶点的坐标分别为\(A(-1,1)\),\(B(2,3)\),\(C(-6,5)\),\(D\)为\(BC\)边的中点,则 \(\overrightarrow{A D}=\) ( )
A.\((-3,2)\) \(\qquad \qquad \qquad\) B.\((-1,3)\) \(\qquad \qquad \qquad\) C.\((-3,5)\) \(\qquad \qquad \qquad\) D.\((-2,4)\)
6.在平行四边形\(ABCD\)中,若\(\overrightarrow{A B}=(3,1)\),\(\overrightarrow{A C}=(2,4)\),则 \(\overrightarrow{B D}=\)\(\underline{\quad \quad}\).
7.已知\(O\)为坐标原点,\(\overrightarrow{P_1 P}=-2 \overrightarrow{P P_2}\),若\(P_1 (1,2)\)、\(P_2 (2,-1)\),则与 \(\overrightarrow{O P}\)共线的单位向量为\(\underline{\quad \quad}\).
8.已知向量 \(\vec{a}=(2 \sin \theta, 1)\), \(\vec{b}=(\cos \theta,-1)\), \(\theta \in\left(\dfrac{\pi}{2}, \pi\right)\),且\(\vec{a}||\vec{b}\),则\(\tan θ\)等于\(\underline{\quad \quad}\).
9.已知平行四边形\(ABCD\)中, \(\overrightarrow{E C}=2 \overrightarrow{D E}\), \(\overrightarrow{F C}=2 \overrightarrow{B F}\), \(\overrightarrow{F G}=2 \overrightarrow{G E}\).
(1)用 \(\overrightarrow{A B}\), \(\overrightarrow{A D}\)表示 \(\overrightarrow{A G}\);
(2)若 \(|\overrightarrow{A B}|=6\), \(|\overrightarrow{A D}|=3 \sqrt{2}\), \(\angle B A D=45^{\circ}\),如图建立直角坐标系,求\(\overrightarrow{G B}\)和\(\overrightarrow{DF}\)的坐标.
10.如图,在平面直角坐标系中,\(|\overrightarrow{O A}|=2|\overrightarrow{A B}|=2\), \(\overrightarrow{B C}=(-1, \sqrt{3})\), \(\angle O A B=\dfrac{2 \pi}{3}\).
(1)求点\(B\)的坐标; \(\qquad \qquad\) (2)求证:\(OC||AB\).
参考答案
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答案 \(A\)
解析 \(\because\)向量\(\vec{a}=(-1,2)\),\(\vec{b}=(3,-5)\),
\(\therefore 3\vec{a}+2\vec{b}=(-3,6)+(6, -10)=(3,-4)\),
故选:\(A\). -
答案 \(B\)
解析 设\(\vec{b}=(x,y)\),向量\(\vec{a}=\left(-2, \dfrac{3}{2}\right)\), \(2\vec{a}+3 \vec{b}=(5,-3)\),
即\((-4+3x,3+3y)=(5,-3)\),
\(\therefore\left\{\begin{array}{l} -4+3 x=5 \\ 3+3 y=-3 \end{array}\right.\),解得\(\left\{\begin{array}{l} x=3 \\ y=-2 \end{array}\right.\),则\(\vec{b}=(3,-2)\).
故选:\(B\). -
答案 \(A\)
解析 \(\because\)平面向量\(\vec{a}=(m,-4)\),\(\vec{b}=(-1,m+3)\),
存在实数\(λ<0\),使得\(\vec{a}=λ\vec{b}\),
\(\therefore (m,-4)=(-λ,λ(m+3))\),
\(\therefore\left\{\begin{array}{l} m=-\lambda \\ -4=\lambda(m+3) \end{array}\right.\),解得\(λ=4\)(舍) 或\(λ=-1\),
\(\therefore\)实数\(m=1\).
故选:\(A\). -
答案 \(B\)
解析 \(\because\) 向量\(\vec{a}=(-2,2)\),\(\vec{b}=(1,-2λ)\),
\(\therefore \vec{a}+3\vec{b}=(1,2-6λ)\),\(\vec{a}-\vec{b}=(-3,2+2λ)\).
\(\because (\vec{a}+3\vec{b})||(\vec{a}-\vec{b})\),\(\therefore 2+2λ-(2-6λ)×(-3)=0\), \(\therefore \lambda=\dfrac{1}{2}\),
故选:\(B\). -
答案 \(B\)
解析 \(\because B(2,3)\),\(C(-6,5)\),\(D\)为\(BC\)边的中点,\(\therefore D(-2,4)\),
\(\because A(-1,1)\), \(\therefore \overrightarrow{A D}=(-2+1,4-1)=(-1,3)\).
故选:\(B\). -
答案 \(A\)
解析 \(\overrightarrow{B D}=\overrightarrow{B A}+\overrightarrow{A D}=-\overrightarrow{A B}+\overrightarrow{B C}=-\overrightarrow{A B}+\overrightarrow{A C}-\overrightarrow{A B}=\overrightarrow{A C}-2 \overrightarrow{A B}\)
\(=(2-2×3,4-2×1)=(-4,2)\);
故选:\(A\). -
答案 \(\left(\dfrac{3}{5},-\dfrac{4}{5}\right)\)或 \(\left(-\dfrac{3}{5}, \dfrac{4}{5}\right)\)
解析 设\(P(x,y)\),则根据条件得\((x-1,y-2)=-2(2-x,-1-y)\),
\(\therefore\left\{\begin{array}{l} x-1=-2(2-x) \\ y-2=-2(-1-y) \end{array}\right.\),解得 \(\left\{\begin{array}{l} x=3 \\ y=-4 \end{array}\right.\), \(\therefore \overrightarrow{O P}=(3,-4)\),
\(\therefore\)与 \(\overrightarrow{O P}\)共线的单位向量为:\(\dfrac{\overrightarrow{O P}}{|\overrightarrow{O P}|}=\left(\dfrac{3}{5},-\dfrac{4}{5}\right)\)或\(-\dfrac{\overrightarrow{O P}}{|\overrightarrow{O P}|}=\left(-\dfrac{3}{5}, \dfrac{4}{5}\right)\). -
答案 \(-\dfrac{1}{2}\).
解析 根据题意,向量\(\vec{a}=(2 \sin \theta, 1)\), \(\vec{b}=(\cos \theta,-1)\),
若\(\vec{a}||\vec{b}\),则有\(2 \sin \theta \times(-1)=\cos \theta \times 1\),
解可得 \(-2 \sin \theta=\cos \theta\),则有\(\tan \theta=-\dfrac{1}{2}\);
故答案为:\(-\dfrac{1}{2}\). -
答案 (1) \(\overrightarrow{A G}=\dfrac{5}{9} \overrightarrow{A B}+\dfrac{7}{9} \overrightarrow{A D}\) (2) \(\overrightarrow{G B}=\left(\dfrac{1}{3},-\dfrac{7}{3}\right)\), \(\overrightarrow{D F}=(4,-2)\)
解析 (1)由题意可得\(\overrightarrow{A E}=\overrightarrow{A D}+\overrightarrow{D E}=\overrightarrow{A D}+\dfrac{1}{3} \overrightarrow{A B}\), \(\overrightarrow{A F}=\overrightarrow{A B}+\overrightarrow{B F}=\overrightarrow{A B}+\dfrac{1}{3} \overrightarrow{A D}\),
又\(\overrightarrow{F G}=2 \overrightarrow{G E}\),所以 \(\overrightarrow{A G}-\overrightarrow{A F}=2(\overrightarrow{A E}-\overrightarrow{A G})\),
所以 \(\overrightarrow{A G}=\dfrac{2}{3} \overrightarrow{A E}+\dfrac{1}{3} \overrightarrow{A F}=\dfrac{2}{3}\left(\overrightarrow{A D}+\dfrac{1}{3} \overrightarrow{A B}\right)+\dfrac{1}{3}\left(\overrightarrow{A B}+\dfrac{1}{3} \overrightarrow{A D}\right)=\dfrac{5}{9} \overrightarrow{A B}+\dfrac{7}{9} \overrightarrow{A D}\).
(2)过点\(D\)作\(AB\)的垂线交于点\(D'\),如图,
于是在\(Rt△ADD'\)中,由\(\angle B A D=45^{\circ}\),可知\(AD'=3\),
根据题意得各点坐标为\(A(0,0)\),\(B(6,0)\),\(C(9,3)\),\(D(3,3)\),\(E(5,3)\),\(F(7,1)\),
则\(\overrightarrow{A G}=\dfrac{5}{9} \overrightarrow{A B}+\dfrac{7}{9} \overrightarrow{A D}=\dfrac{5}{9}(6,0)+\dfrac{7}{9}(3,3)=\left(\dfrac{17}{3}, \dfrac{7}{3}\right)\),所以 \(G\left(\dfrac{17}{3}, \dfrac{7}{3}\right)\),
所以 \(\overrightarrow{A B}=(6,0)\), \(\overrightarrow{A G}=\left(\dfrac{17}{3}, \dfrac{7}{3}\right)\),
所以\(\overrightarrow{G B}=\overrightarrow{A B}-\overrightarrow{A G}=\left(\dfrac{1}{3},-\dfrac{7}{3}\right)\), \(\overrightarrow{D F}=(4,-2)\). -
答案 (1) \(\left(\dfrac{5}{2}, \dfrac{\sqrt{3}}{2}\right)\) (2)略.
解析 (1)由题意,因为\(\angle O A B=\dfrac{2 \pi}{3}\), \(|\overrightarrow{A B}|=1\),
故 \(\overrightarrow{A B}=\left(\cos \dfrac{\pi}{3}, \sin \dfrac{\pi}{3}\right)=\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\),
故 \(\overrightarrow{O B}=\overrightarrow{O A}+\overrightarrow{A B}=(2,0)+\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)=\left(\dfrac{5}{2}, \dfrac{\sqrt{3}}{2}\right)\),
即点\(B\)的坐标为 \(\left(\dfrac{5}{2}, \dfrac{\sqrt{3}}{2}\right)\).
证明:(2)由题意, \(\overrightarrow{O C}=\overrightarrow{O B}+\overrightarrow{B C}=\left(\dfrac{5}{2}, \dfrac{\sqrt{3}}{2}\right)+(-1, \sqrt{3})=\left(\dfrac{3}{2}, \dfrac{3 \sqrt{3}}{2}\right)\),
又\(\overrightarrow{A B}=\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\),故\(\overrightarrow{O C}=3 \overrightarrow{A B}\),且\(OC\),\(AB\)不共线,
故\(OC||AB\).
【B组---提高题】
1.在平面直角坐标系\(xOy\)中,已知点\(A(1,0)\)和点\(B(-3,4)\),若点\(C\)在\(∠AOB\)的平分线上,且\(|\overrightarrow{O C}|=\sqrt{5}\),则 \(\overrightarrow{O C}\)的坐标为\(\underline{\quad \quad}\).
2.如图,在等腰梯形\(ABCD\)中,\(AB∥CD\), \(A D=D C=C B=\dfrac{1}{2} A B=1\),\(F\)是\(BC\)的中点,点\(P\)在以\(A\)为圆心,\(AD\)为半径的圆弧\(DE\)上变动,\(E\)为圆弧\(DE\)与\(AB\)的交点,若\(\overrightarrow{A P}=\lambda \overrightarrow{E D}+\mu \overrightarrow{A F}\),其中\(λ,μ∈R\),则\(2λ+μ\)的取值范围是\(\underline{\quad \quad}\).
3.如图所示,已知矩形\(ABCD\)中,\(AB=2\),\(AD=1\), \(\overrightarrow{D M}=\dfrac{1}{3} \overrightarrow{D C}\), \(\overrightarrow{B N}=\dfrac{2}{3} \overrightarrow{B C}\),\(AC\)与\(MN\)相交于点\(E\).
(1)若 \(\overrightarrow{M N}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A D}\),求\(λ\)和\(μ\)的值;
(2)用向量 \(\overrightarrow{A M}\), \(\overrightarrow{A N}\)表示 \(\overrightarrow{A E}\).
参考答案
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答案 \((1,2)\)
解析 \(\overrightarrow{OB}\)方向上的单位向量为\(\overrightarrow{O B^{\prime}}=\left(-\dfrac{3}{5}, \dfrac{4}{5}\right)\),
由题意知, \(\overrightarrow{O C}\)方向上的向量为 \(\overrightarrow{O C^{\prime}}=\overrightarrow{O A}+\overrightarrow{O B^{\prime}}=(1,0)+\left(-\dfrac{3}{5}, \dfrac{4}{5}\right)=\left(\dfrac{2}{5}, \dfrac{4}{5}\right)\);
所以四边形\(OAC' B'\)是菱形,且\(OC'\)平分\(∠AOB'\),
即点\(C\)在直线\(OC'\)上;
又 \(\left|\overrightarrow{O C^{\prime}}\right|=\sqrt{\left(\dfrac{2}{5}\right)^2+\left(\dfrac{4}{5}\right)^2}=\dfrac{2 \sqrt{5}}{5}\), \(|\overrightarrow{O C}|=\sqrt{5}\);
所以 \(\overrightarrow{O C}=\dfrac{5}{2} \overrightarrow{O C^{\prime}}=(1,2)\).
故答案为\((1,2)\). -
答案 \([0,2]\)
解析 建立平面直角坐标系如图所示,
则\(A(0,0)\),\(E(1,0)\), \(D\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\),\(B(2,0)\), \(C\left(\dfrac{3}{2}, \dfrac{\sqrt{3}}{2}\right)\), \(F\left(\dfrac{7}{4}, \dfrac{\sqrt{3}}{4}\right)\);
设 \(P(\cos \alpha, \sin \alpha)\left(0^{\circ} \leq \alpha \leq 60^{\circ}\right)\),
由 \(\overrightarrow{A P}=\lambda \overrightarrow{E D}+\mu \overrightarrow{A F}\),
\(\therefore(\cos \alpha, \sin \alpha)=\lambda\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)+\mu\left(\dfrac{7}{4}, \dfrac{\sqrt{3}}{4}\right)\),
\(\therefore \cos \alpha=-\dfrac{1}{2} \lambda+\dfrac{7}{4} \mu\)⋯①, \(\sin \alpha=\dfrac{\sqrt{3}}{2} \lambda+\dfrac{\sqrt{3}}{4} \mu\)…②,
由①②解得 \(\lambda=-\dfrac{1}{4} \cos \alpha+\dfrac{7 \sqrt{3}}{12} \sin \alpha\), \(\mu=\dfrac{1}{2} \cos \alpha+\dfrac{\sqrt{3}}{6} \sin \alpha\),
\(\therefore 2 \lambda+\mu=2\left(-\dfrac{1}{4} \cos \alpha+\dfrac{7 \sqrt{3}}{12} \sin \alpha\right)+\left(\dfrac{1}{2} \cos \alpha+\dfrac{\sqrt{3}}{6} \sin \alpha\right)=\dfrac{4 \sqrt{3}}{3} \sin \alpha\),
\(α∈[0°,60°]\)时, \(\sin \alpha \in\left[0, \dfrac{\sqrt{3}}{2}\right]\),
\(\therefore \dfrac{4 \sqrt{3}}{3} \sin \alpha \in[0,2]\).
故答案为 \([0,2]\).
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答案 (1) \(\lambda=\dfrac{2}{3}\),\(\mu=-\dfrac{1}{3}\) (2) \(\overrightarrow{A E}=\dfrac{1}{3} \overrightarrow{A M}+\dfrac{2}{3} \overrightarrow{A N}\)
解析 以\(A\)点为原点,\(AB\)所在直线为\(x\)轴,\(AD\)所在直线为y轴,建立平面直角坐标系,
则\(A(0,0)\),\(D(0,1)\),\(B(2,0)\), \(M\left(\dfrac{2}{3}, 1\right)\), \(N\left(2, \dfrac{2}{3}\right)\),\(C(2,1)\),
(1) \(\overrightarrow{M N}=\left(\dfrac{4}{3},-\dfrac{1}{3}\right)=\lambda \overrightarrow{A B}+\mu \overrightarrow{A D}=(2 \lambda, \mu)\),解得:\(\lambda=\dfrac{2}{3}\),\(\mu=-\dfrac{1}{3}\) ;
(2)设 \(\overrightarrow{A E}=t \overrightarrow{A C}\), \(\overrightarrow{A C}=m \overrightarrow{A M}+n \overrightarrow{A N}\),
所以 \(\overrightarrow{A C}=(2,1)=\left(\dfrac{2}{3} m+2 n, m+\dfrac{2}{3} n\right)\),解得 \(m=\dfrac{3}{7}\), \(n=\dfrac{6}{7}\),
即 \(\overrightarrow{A C}=\dfrac{3}{7} \overrightarrow{A M}+\dfrac{6}{7} \overrightarrow{A N}\),所以 \(\overrightarrow{A E}=t \overrightarrow{A C}=\dfrac{3}{7} t \overrightarrow{A M}+\dfrac{6}{7} t \overrightarrow{A N}\),
又因为\(M\),\(E\),\(N\)三点共线,所以\(\dfrac{3}{7} t+\dfrac{6}{7} t=1\), \(t=\dfrac{7}{9},\),
所以 \(\overrightarrow{A E}=\dfrac{1}{3} \overrightarrow{A M}+\dfrac{2}{3} \overrightarrow{A N}\).