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5.1向量

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向量

平面向量

向量是高中阶段引入的一个重要的数学研究对象.

基本概念

定义 1.1.1（向量）

平面上的一条有向线段被称为向量，通常用粗体字母\(\displaystyle \vv{a},\vv{b},\vv{c}\)或\(\displaystyle \vv{a},\vv{b},\vv{c}\)等表示。向量的模长被定义为该线段的长度，模长为零的向量称为零向量，记作\(\displaystyle \vv{0}\)。模长为1的向量称为单位向量。

向量\(\displaystyle \vv{a},\vv{b}\)平行当且仅当它们的方向相同或相反，记作\(\displaystyle \vv{a}\parallel \vv{b}\)，此时也称向量\(\displaystyle \vv{a},\vv{b}\)共线。零向量与任意向量平行。 向量\(\displaystyle \vv{a},\vv{b}\)垂直当且仅当它们所在的直线互相垂直，记作\(\displaystyle \vv{a}\perp \vv{b}\)。 向量\(\displaystyle \vv{a}\)和\(\displaystyle \vv{b}\)相等当且仅当它们的模长相等且方向相同，记作\(\displaystyle \vv{a}=\vv{b}\)。

向量的线性运算

规定向量的加法与数乘运算如下：

定义 1.1.2（向量的加法运算）

设\(\displaystyle \vv{a},\vv{b}\)是两个向量，在平面中作出\(\displaystyle \vv{OA}=\vv{a},\vv{AB}=\vv{b}\)，定义向量\(\displaystyle \vv{a},\vv{b}\)的和\(\displaystyle \vv{a}+\vv{b}\)为\(\displaystyle \vv{OB}\).

定义 1.1.3（向量的数乘运算）

设\(\displaystyle \lambda\)是一个实数，\(\displaystyle \vv{a}\)是一个向量，则\(\displaystyle \lambda\vv{a}\)称为向量\(\displaystyle \vv{a}\)与实数\(\displaystyle \lambda\)的数乘。

当$\displaystyle \lambda\neq 0$且$\displaystyle \vv{a}\neq \vv{0}$时，$\displaystyle \lambda\vv{a}$的模长为$\displaystyle |\lambda||\vv{a}|$，当$\displaystyle \lambda > 0$时，与$\displaystyle \vv{a}$的方向相同；当$\displaystyle \lambda < 0$时，与$\displaystyle \vv{a}$的方向相反；

当$\displaystyle \lambda = 0$或$\displaystyle \vv{a}=\vv{0}$时，$\displaystyle \lambda\vv{a}=\vv{0}$。

向量的加法、数乘运算统称为向量的线性运算。对于向量\(\displaystyle \vv{a},\vv{b}\)与实数\(\displaystyle \lambda,\mu\)，称\(\displaystyle \lambda \vv{a}+\mu \vv{b}\)为向量\(\displaystyle \vv{a}\)和\(\displaystyle \vv{b}\)的线性组合，若\(\displaystyle c=\lambda_1\vv{e}_1+\lambda_2\vv{e}_2\)，则称\(\displaystyle \vv{a}\)可以由\(\displaystyle \vv{e}_1,\vv{e}_2\)线性表出。

定理：平面向量基本定理

平面内取定不共线的两个向量\(\displaystyle \vv{e}_1,\vv{e}_2\)，则该平面内的任意一个向量\(\displaystyle \vv{a}\)可以唯一地表示成\(\displaystyle \vv{e}_1,\vv{e}_2\)的线性组合\(\displaystyle \lambda\vv{e}_1+\mu\vv{e}_2\)

称平面内不共线的两个向量组成的集合\(\displaystyle \left \{ \vv{e}_1,\vv{e}_2\right \}\)为该平面上向量的一组基底。如果平面向量的基底\(\displaystyle \left \{ \vv{e}_1,\vv{e}_2\right \}\)中\(\displaystyle \vv{e}_1\perp \vv{e}_2\)，则称这组基底为正交基底。

向量的直角坐标表示

以单位正交基底\(\displaystyle \left \{ \vv{e}_1,\vv{e}_2\right \}\)中\(\displaystyle \vv{e_1}\)的方向为\(\displaystyle x\)轴正方向，\(\displaystyle \vv{e_2}\)的方向为\(\displaystyle y\)轴正方向，建立平面直角坐标系。由定理1.1.4,平面内任意一个向量\(\displaystyle \vv{a}\)都可以唯一地表示为\(\displaystyle \vv{a}=x\vv{e}_1+y\vv{e}_2\)，其中\(\displaystyle x,y\)是实数，称\(\displaystyle (x,y)\)为向量\(\displaystyle \vv{a}\)在基底\(\displaystyle \left \{ \vv{e}_1,\vv{e}_2\right \}\)下的直角坐标。

向量的数量积

定义 1.1.5（向量的夹角）

设\(\displaystyle \vv{a},\vv{b}\)是两个非零向量，平移这两条向量，使得它们的始点重合，记为\(\displaystyle O\)，此时这两个向量的终点分别记为\(\displaystyle A,B\)，定义向量\(\displaystyle \vv{a},\vv{b}\)的夹角\(\displaystyle \theta\)为\(\displaystyle \angle AOB\)的大小，其中\(\displaystyle \angle AOB\)取不超过平角大小的情况。

定义 1.1.6（向量的数量积）

设\(\displaystyle \vv{a},\vv{b}\)是两个向量，实数\(\displaystyle \vv{a}\cdot \vv{b}=|\vv{a}|\cdot |\vv{b}|\cos\theta\)称为\(\displaystyle \vv{a}\)与\(\displaystyle \vv{b}\)的数量积（内积）。

由上述定义可以得出：两个非零向量互相垂直当且仅当\(\displaystyle \vv{a}\cdot \vv{b}=0\)。

问题

解答下述问题：

（1）在平面直角坐标系终，已知向量\(\displaystyle \vv{a}=(x_1,y_1)\)，\(\displaystyle \vv{b}=(x_2,y_2)\)，证明：\(\displaystyle \vv{a}\cdot \vv{b}=x_1x_2+y_1y_2\)

（2）已知向量\(\displaystyle \vv{a},\vv{b}\),设向量\(\displaystyle \vv{a}\)在向量\(\displaystyle \vv{b}\)方向上的投影向量为\(\displaystyle \vv{c}\)。证明：当\(\displaystyle \vv{c}\)与\(\displaystyle \vv{a}\)方向相同时，\(\displaystyle \vv{a}\cdot \vv{b}=|a|\cdot |c|\),当\(\displaystyle \vv{c}\)与\(\displaystyle \vv{a}\)方向相反时，\(\displaystyle \vv{a}\cdot \vv{b}=-|a|\cdot |c|\)。

（3）已知在\(\displaystyle \Delta ABC\)中，记\(\displaystyle BC\)的中点为\(\displaystyle D\)，连接\(\displaystyle AD\)，证明：\(\displaystyle \vv{AB}\cdot \vv{AC}=AD^2-BD^2\)

A 组习题

习\(\displaystyle \quad\)题

A组

  1. 设\(\displaystyle \vv{a},\vv{b}\)是非零平面向量,则\(\displaystyle \vv{a}\)在\(\displaystyle \vv{b}\)上的投影向量是

    • \(\displaystyle \frac{\vv{a}\cdot \vv{b}}{|\vv{b}|^2}\vv{b}\)
    • \(\displaystyle \frac{\vv{a}\cdot \vv{b}}{|\vv{a}|^2}\vv{a}\)
    • \(\displaystyle \frac{\vv{a}\cdot \vv{b}}{|\vv{b}|}\vv{b}\)
    • \(\displaystyle \frac{\vv{a}\cdot \vv{b}}{|\vv{a}|}\vv{a}\)
    答案

    A.

    设向量$\displaystyle \vv{a}$在向量$\displaystyle \vv{b}$上的投影向量$\displaystyle \vv{p}=t\vv{b}$.由投影的数量关系$\displaystyle t|\vv{b}|=|\vv{a}|\cos\theta=\frac{\vv{a}\cdot\vv{b}}{|\vv{b}|}$可得$\displaystyle t=\frac{\vv{a}\cdot\vv{b}}{|\vv{b}|^2}$.因此投影向量为$\displaystyle \frac{\vv{a}\cdot\vv{b}}{|\vv{b}|^2}\vv{b}$,选A.
    1. 设\(\displaystyle V\)是全体平面向量构成的集合,\(\displaystyle \vv{a}\)为非零平面向量,定义集合$\(\displaystyle A=\left \{ \vv{x}\in V\mid \vv{a,x} \text{方向相反}\right \} ,\quad B=\left \{\vv{x}\in V\mid |\vv{x}|=|\vv{a}| \right \},\quad C=\left \{ \vv{x}\in V\mid |\vv{x}|=|\vv{a}|\text{且}\vv{a}\parallel \vv{x} \right \}\)$则下列命题中正确的是
    • \(\displaystyle C\subseteq A\)
    • \(\displaystyle \left \{ \vv{a}\right \} \subset A\cap B\)
    • \(\displaystyle C\subseteq B\)
    • \(\displaystyle A\cup C=B\cup C\)
    答案

    C.

    由题意,$\displaystyle A$中的向量与$\displaystyle \vv{a}$方向相反,所以$\displaystyle A$是以$\displaystyle -\vv{a}$为端点、沿$\displaystyle -\vv{a}$方向延伸的射线;集合$\displaystyle B$是以原点为圆心、半径$\displaystyle |\vv{a}|$的圆;集合$\displaystyle C$中的向量既与$\displaystyle \vv{a}$共线,又与$\displaystyle \vv{a}$等长,所以$\displaystyle C=\{\vv{a},-\vv{a}\}$.
因为$\displaystyle C$中的每个向量模长都等于$\displaystyle |\vv{a}|$,所以$\displaystyle C\subseteq B$,选项C正确.而$\displaystyle \vv{a}$的方向与$\displaystyle \vv{a}$相同,故$\displaystyle \vv{a}\notin A$,选项B错误;因为$\displaystyle \vv{a}\in C$且$\displaystyle \vv{a}\notin A$,故$\displaystyle C\not\subseteq A$,选项A错误.$\displaystyle A\cup C$只包含与$\displaystyle \vv{a}$共线的向量,$\displaystyle B\cup C$还包含其他方向的向量,二者不相等,选项D错误.
    1. 【2008琼宁卷理8】 平面向量 \(\displaystyle \vv{a},\vv{b}\) 共线的充要条件是
    • \(\displaystyle \vv{a},\vv{b}\) 方向相同
    • \(\displaystyle \vv{a},\vv{b}\) 中至少有一个为零向量
    • \(\displaystyle \exists \lambda \in \mathbb{R}, \vv{b} = \lambda\vv{a}\)
    • 存在不全为零的实数 \(\displaystyle \lambda_1,\lambda_2\)使得\(\displaystyle \lambda_1\vv{a} + \lambda_2\vv{b} = \vv{0}\)
    答案

    D.

    两个向量共线,则它们的方向相同或相反.因此选项A错误。两个向量共线时，它们可能全不为零向量，因此选项B错误。当\(\displaystyle \vv{a}\)为零向量,\(\displaystyle \vv{b}\)不为零向量时，\(\displaystyle \vv{a},\vv{b}\)共线，对任意实数\(\displaystyle \lambda\)，\(\displaystyle \lambda \vv{a}=\vv{0}\neq \vv{b}\)，因此选项C错误。对于D选项，若\(\displaystyle \vv{a},\vv{b}\)共线,可以验证一定存在不全为零的实数\(\displaystyle \lambda,\mu\)使得\(\displaystyle \lambda\vv{a}+\mu\vv{b}=\vv{0}\).反过来， 若存在不全为零的实数\(\displaystyle \lambda_1,\lambda_2\),使得 \(\displaystyle \lambda_1\vv{a}+\lambda_2\vv{b}=\vv{0},\)则当\(\displaystyle \lambda_2\neq0\)时有\(\displaystyle \vv{b}=-\frac{\lambda_1}{\lambda_2}\vv{a}\);当\(\displaystyle \lambda_2=0\)时必有\(\displaystyle \lambda_1\neq0\),从而\(\displaystyle \vv{a}=\vv{0}\),此时\(\displaystyle \vv{a}\)与任意向量共线.因此该条件必然推出\(\displaystyle \vv{a},\vv{b}\)共线.因此D正确

    1. 【2024上海15】集合\(\displaystyle \Omega\subseteq \mathbb{R}^3\),且满足对任意\(\displaystyle P_1,P_2,P_3\in\Omega\),存在不全为0的实数\(\displaystyle \lambda_1,\lambda_2,\lambda_3\),使得\(\displaystyle \lambda_1 \vv{OP_1}+\lambda_2 \vv{OP_2}+\lambda_3 \vv{OP_3}= \vv{0}\).已知\(\displaystyle (1,0,0)\in\Omega\),则\(\displaystyle (0,0,1)\notin\Omega\)的充分条件是
    • \(\displaystyle (0,0,0)\in\Omega\)
    • \(\displaystyle (-1,0,0)\in\Omega\)
    • \(\displaystyle (0,1,0)\in\Omega\)
    • \(\displaystyle (0,0,-1)\in\Omega\)
    答案

    C.

    如果$\displaystyle (0,1,0)\in\Omega$，下面证明$\displaystyle (0,0,1)\notin \Omega$,采取反证法。假设$\displaystyle (0,0,1)\in \Omega$，则$\displaystyle (1,0,0),(0,1,0),(0,0,1)$构成空间中向量的一组基底，若存在实数$\displaystyle \lambda_1,\lambda_2,\lambda_3$使得$\displaystyle \lambda_1\vv{e}_1+\lambda_2\vv{e}_2+\lambda_3\vv{e}_3=\vv{0}$，则只能有$\displaystyle \lambda_1=\lambda_2=\lambda_3=0$,与题设矛盾，假设错误。故$\displaystyle (0,0,1)\notin \Omega$，选项C正确。可以验证其余选项均不正确。

注：已知向量$\displaystyle \vv{a}_1,\vv{a}_2,\cdots,\vv{a}_n$，如果存在不全为零的实数$\displaystyle \lambda_1,\lambda_2,\cdots,\lambda_n$,使得$\displaystyle \lambda_1 \vv{a}_1+\lambda_2 \vv{a}_2+\cdots +\lambda_n \vv{a}_n= \vv{0}$，则称向量$\displaystyle \vv{a}_1,\vv{a}_2,\cdots,\vv{a}_n$是线性相关的，也就是说，其中存在某个向量可以被表示为其余向量的线性组合。若由$\displaystyle \lambda_1 \vv{a}_1+\lambda_2 \vv{a}_2+\cdots +\lambda_n \vv{a}_n= \vv{0}$只能解得$\displaystyle \lambda_1=\lambda_2=\cdots \lambda_n=0$，则称向量$\displaystyle \vv{a}_1,\vv{a}_2,\cdots,\vv{a}_n$是线性无关的。

 新答案（来源：4.4 多个量词的问题（3）向量、几何.md）：

    C

    【解题思路】设\(\displaystyle O\left(0,0,0\right)\)，\(\displaystyle A\left(1,0,0\right)\)，\(\displaystyle B\left(0,0,1\right)\)．

    思路1：若\(\displaystyle O\left(0,0,0\right)\in\Omega\)，取\(\displaystyle \lambda_1=1\)，\(\displaystyle \lambda_2=0\)，\(\displaystyle \lambda_3=0\)，则\(\displaystyle \lambda_1\vv{OO}+\lambda_2\vv{OA}+\lambda_3\vv{OB}=\vv{0}\)，所以\(\displaystyle O\in\Omega\)不能推出\(\displaystyle B\notin\Omega\)．不合题意．

    若\(\displaystyle Q_1\left(-1,0,0\right)\in\Omega\)，取\(\displaystyle \lambda_1=1\)，\(\displaystyle \lambda_2=-1\)，\(\displaystyle \lambda_3=0\)，则\(\displaystyle \lambda_1\vv{OQ_1}+\lambda_2\vv{OA}+\lambda_3\vv{OB}=\vv{0}\)，所以\(\displaystyle Q_1\in\Omega\)不能推出\(\displaystyle B\notin\Omega\)，不合题意．

    若\(\displaystyle Q_2\left(0,0,-1\right)\in\Omega\)，取\(\displaystyle \lambda_1=1\)，\(\displaystyle \lambda_2=-1\)，\(\displaystyle \lambda_3=0\)，则\(\displaystyle \lambda_1\vv{OB}+\lambda_2\vv{OQ_2}+\lambda_3\vv{OA}=\vv{0}\)，所以\(\displaystyle Q_2\in\Omega\)不能推出\(\displaystyle B\notin\Omega\)，不合题意．

    排除\(\displaystyle A\)、\(\displaystyle B\)、\(\displaystyle D\)选项，选\(\displaystyle C\)．

    思路2："存在不全为零的实数\(\displaystyle \lambda_1\)、\(\displaystyle \lambda_2\)、\(\displaystyle \lambda_3\)满足\(\displaystyle \lambda_1\vv{OP_1}+\lambda_2\vv{OP_2}+\lambda_3\vv{OP_3}=\vv{0}\)"等价于"\(\displaystyle O, P_1, P_2, P_3\)不在同一平面上"．

    而四个选项中，\(\displaystyle D_1\left(0,0,0\right)\)在平面\(\displaystyle OAB\)上，\(\displaystyle D_2\left(-1,0,0\right)\)在平面\(\displaystyle OAB\)上，\(\displaystyle D_3\left(0,-1,0\right)\)不在平面\(\displaystyle OAB\)上，\(\displaystyle D_4\left(0,0,-1\right)\)在平面\(\displaystyle OAB\)上，故选\(\displaystyle C\)．

    1. 【2006陕西理9】已知非零向量 \(\displaystyle \vv{AB},\vv{AC}\) 满足 \(\displaystyle \left(\frac{\vv{AB}}{|\vv{AB}|} + \frac{\vv{AC}}{|\vv{AC}|}\right) \cdot \vv{BC} = 0\) 且 \(\displaystyle \frac{\vv{AB}}{|\vv{AB}|} \cdot \frac{\vv{AC}}{|\vv{AC}|} = \frac{1}{2}\), 则 \(\displaystyle \triangle ABC\) 为

    2. 三边均不相等的三角形

    3. 直角三角形
    4. 等腰非等边三角形
    5. 等边三角形
答案

D.

设\(\displaystyle \vv{u}=\frac{\vv{AB}}{|\vv{AB}|}\),\(\displaystyle \vv{v}=\frac{\vv{AC}}{|\vv{AC}|}\).由\(\displaystyle \vv{u}\cdot\vv{v}=\frac12\)及\(\displaystyle |\vv{u}|=|\vv{v}|=1\),可知\(\displaystyle \angle BAC=60^\circ\).

向量\(\displaystyle \frac{\vv{AB}}{|\vv{AB}|}+\frac{\vv{AC}}{|\vv{AC}|}\)的方向是\(\displaystyle \angle BAC\)的角平分线方向，由\(\displaystyle \left(\frac{\vv{AB}}{|\vv{AB}|} + \frac{\vv{AC}}{|\vv{AC}|}\right) \cdot \vv{BC} = 0\)可知，该角平分线垂直于对边，因此它也是垂线，根据三线合一可推得\(\displaystyle \triangle ABC\)为等腰三角形，进而得出\(\displaystyle \triangle ABC\)为等边三角形,选D.

  1. 【2010辽宁理8】 平面上 \(\displaystyle O,A,B\) 三点不共线, 设 \(\displaystyle \vv{OA}=\vv{a},\vv{OB}=\vv{b}\), 则 \(\displaystyle \triangle OAB\) 的面积等于
  • \(\displaystyle \sqrt{|\vv{a}|^2|\vv{b}|^2-(\vv{a} \cdot \vv{b})^2}\)
  • \(\displaystyle \sqrt{|\vv{a}|^2|\vv{b}|^2+(\vv{a} \cdot \vv{b})^2}\)
  • \(\displaystyle \frac{1}{2}\sqrt{|\vv{a}|^2|\vv{b}|^2-(\vv{a} \cdot \vv{b})^2}\)
  • \(\displaystyle \frac{1}{2}\sqrt{|\vv{a}|^2|\vv{b}|^2+(\vv{a} \cdot \vv{b})^2}\)
答案

C.

$\displaystyle \triangle OAB$的面积为
$\displaystyle S_{\triangle OAB}=\frac12|\vv{a}|\,|\vv{b}|\sin\theta.$
而$\displaystyle |\vv{a}|^2|\vv{b}|^2-(\vv{a}\cdot\vv{b})^2=|\vv{a}|^2|\vv{b}|^2\sin^2\theta.$
因此$\displaystyle \triangle OAB$的面积可表示为$\displaystyle \frac12\sqrt{|\vv{a}|^2|\vv{b}|^2-(\vv{a}\cdot\vv{b})^2}$,选C.

  1. 【2014浙江理8】已知 \(\displaystyle \vv{a},\vv{b}\) 为平面向量, 则

  2. \(\displaystyle \min\{|\vv{a}+\vv{b}|,|\vv{a}-\vv{b}|\}\leqslant \min\{|\vv{a}|,|\vv{b}|\}\)

  3. \(\displaystyle \min\{|\vv{a}+\vv{b}|,|\vv{a}-\vv{b}|\}\geqslant \min\{|\vv{a}|,|\vv{b}|\}\)
  4. \(\displaystyle \min\{|\vv{a}+\vv{b}|^2,|\vv{a}-\vv{b}|^2\leqslant |\vv{a}|^2+|\vv{b}|^2\}\)
  5. \(\displaystyle \min\{|\vv{a}+\vv{b}|^2,|\vv{a}-\vv{b}|^2\geqslant |\vv{a}|^2+|\vv{b}|^2\}\)

8. 【2014安徽10】在平面直角坐标系中,单位向量\(\displaystyle \vv{a},\vv{b}\)互相垂直,点\(\displaystyle Q\)满足\(\displaystyle \vv{OQ} =\sqrt{2}(\vv{a}+\vv{b})\),曲线\(\displaystyle C=\left \{ P\mid \vv{OP} =\vv{a}\cos\theta+\vv{b}\sin\theta,0\leqslant\theta<2\pi \right \}\),区域\(\displaystyle \Omega=\left \{ P\mid 0<r\leqslant|\vv{PQ}|\leqslant R,r<R\right \}\),若\(\displaystyle C\cap\Omega\)为两段分离的曲线,则

<div class="choices choices--4" markdown>
答案

A.

因为\(\displaystyle \vv{a},\vv{b}\)是互相垂直的单位向量,所以\(\displaystyle \sqrt{2}(\vv{a}+\vv{b})\)的模长为\(\displaystyle 2\).区域\(\displaystyle \Omega\)是以\(\displaystyle Q\)为圆心、内外半径分别为\(\displaystyle r,R\)的圆环，作示意图如下。 \centering 查看图形文件 \bookcaption{第8题示意图} 圆\(\displaystyle C\)与圆环的交集由条件\(\displaystyle r\leqslant|\vv{PQ}|\leqslant R\)给出.要使交集为两段分离的曲线,内圆必须真正截过圆\(\displaystyle C\),故\(\displaystyle r>1\);外圆也必须截过圆\(\displaystyle C\),故\(\displaystyle R<3\).结合题设\(\displaystyle r<R\),得到\(\displaystyle 1<r<R<3\),选A.

  1. 【2026“fiddie”模拟考8】已知 \(\displaystyle \vv{x}, \vv{y}\) 是平面向量,\(\displaystyle |\vv{x}| = 1,|\vv{y}| \neq 0\).设甲：\(\displaystyle |\vv{x} \cdot \vv{y}| \leqslant 1\)；乙：存在平面向量 \(\displaystyle \vv{z}\),使得 \(\displaystyle |\vv{y} - \vv{z}| \leqslant 1\) 且 \(\displaystyle (\vv{x} \cdot \vv{y})(\vv{x} \cdot \vv{z}) \leqslant 0\).则

  2. 甲是乙的充分不必要条件

  3. 甲是乙的必要不充分条件
  4. 甲是乙的充要条件
  5. 甲是乙的既不充分又不必要条件
答案

C.

取单位向量\(\displaystyle \vv{x}\)为横轴正方向,设\(\displaystyle \vv{y}=(s,t)\),其中\(\displaystyle s=\vv{x}\cdot\vv{y}\).若\(\displaystyle s>0\),条件乙要求存在\(\displaystyle \vv{z}=(u,v)\)满足\(\displaystyle |\vv{y}-\vv{z}|\leqslant1\)且\(\displaystyle u\leqslant0\);若\(\displaystyle s<0\),则要求存在\(\displaystyle \vv{z}\)满足\(\displaystyle |\vv{y}-\vv{z}|\leqslant1\)且\(\displaystyle u\geqslant0\). 当\(\displaystyle s>0\)时,点\(\displaystyle \vv{y}\)到半平面\(\displaystyle u\leqslant0\)的最短距离为\(\displaystyle s\),所以存在这样的\(\displaystyle \vv{z}\)当且仅当\(\displaystyle s\leqslant1\).当\(\displaystyle s<0\)时同理,存在这样的\(\displaystyle \vv{z}\)当且仅当\(\displaystyle -s\leqslant1\).当\(\displaystyle s=0\)时取\(\displaystyle \vv{z}=\vv{y}\)即可满足乙.综上,乙成立当且仅当\(\displaystyle |s|\leqslant1\),即当且仅当\(\displaystyle |\vv{x}\cdot\vv{y}|\leqslant1\),这正是甲. 因此甲是乙的充要条件,选C.

  1. 【2013上海理18】在边长为 1 的正六边形 \(\displaystyle ABCDEF\) 中, 记以 \(\displaystyle A\) 为起点, 其余顶点为终点的向量分别为 \(\displaystyle \vv{a}_1,\vv{a}_2,\vv{a}_3,\vv{a}_4,\vv{a}_5\); 以 \(\displaystyle D\) 为起点, 其余顶点为终点的向量分别为 \(\displaystyle \vv{d}_1,\vv{d}_2,\vv{d}_3,\vv{d}_4,\vv{d}_5\). 若 \(\displaystyle m,M\) 分别为 \(\displaystyle (\vv{a}_i+\vv{a}_j+\vv{a}_k) \cdot (\vv{d}_r+\vv{d}_s+\vv{d}_t), \{i,j,k\} \subseteq \{1,2,3,4,5\},\{r,s,t\} \subseteq \{1,2,3,4,5\}\) 的最小值、最大值,则
答案

新答案（来源：1.8 三角恒等变换.md）： A

【解题思路】若\(\displaystyle \sin x=0\)，则\(\displaystyle \sin2x=2\sin x\cos x=0\)，所以\(\displaystyle E\subset F\)．但反之不对，\(\displaystyle x=\frac{\pi}{2}\)满足\(\displaystyle \sin2x=0\)但\(\displaystyle \sin x\ne0\)，所以\(\displaystyle E\subsetneq F\)．

  1. 【2014浙江文9改编】（多选）设\(\displaystyle \theta\)为非零向量\(\displaystyle \vv{a},\vv{b}\)的夹角,已知对任意实数\(\displaystyle t\),\(\displaystyle |\vv{b}+t\vv{a}|\)的最小值为\(\displaystyle 1\),则下列命题中错误的是

  2. 若\(\displaystyle \theta\)确定,则\(\displaystyle |\vv{a}|\)唯一确定

  3. 若\(\displaystyle \theta\)确定,则\(\displaystyle |\vv{b}|\)唯一确定
  4. \(\displaystyle |\vv{a}|\)确定,则\(\displaystyle \theta\)唯一确定
  5. 若\(\displaystyle |\vv{b}|\)确定,则\(\displaystyle \theta\)唯一确定

12. 【2021新高考I10】（多选） 已知\(\displaystyle O\)为坐标原点，点\(\displaystyle P_1\left(\cos\alpha, \sin\alpha\right), P_2\left(\cos\beta, -\sin\beta\right), P_3\left(\cos\left(\alpha + \beta\right), \sin\left(\alpha + \beta\right)\right), A\left(1, 0\right),\)则（多选题）\(\displaystyle (\text{ })\)

13. 已知向量\(\displaystyle \vv{a},\vv{b}\)均为单位向量,求\(\displaystyle \frac{|\vv{a}-2\vv{b}|}{\sqrt{1-(\vv{a}\cdot\vv{b})^2}}\)的最小值. 14. 若向量 \(\displaystyle \vv{a}\) 与 \(\displaystyle \vv{b}\) 不共线, \(\displaystyle \vv{a}\cdot\vv{b}\neq 0\), 且 \(\displaystyle \vv{c}=\vv{a}-\left(\frac{\vv{a}\cdot\vv{a}}{\vv{a}\cdot\vv{b}}\right)\vv{b}\), 求向量 \(\displaystyle \vv{a}\) 与 \(\displaystyle \vv{c}\) 的夹角.

答案

方法一：由\(\displaystyle \vv{c}=\vv{a}-\frac{\vv{a}\cdot\vv{a}}{\vv{a}\cdot\vv{b}}\vv{b}\),有 \(\displaystyle \vv{a}\cdot\vv{c}=\vv{a}\cdot\vv{a}-\frac{\vv{a}\cdot\vv{a}}{\vv{a}\cdot\vv{b}}(\vv{a}\cdot\vv{b})=0.\) 又因为\(\displaystyle \vv{a},\vv{b}\)不共线,所以\(\displaystyle \vv{c}\neq\vv{0}\).因此\(\displaystyle \vv{a}\perp\vv{c}\),它们的夹角为\(\displaystyle 90^\circ\).

方法二：记\(\displaystyle \theta\)为向量\(\displaystyle \vv{a},\vv{b}\)的夹角，有\(\displaystyle \frac{\vv{a}\cdot\vv{a}}{\vv{a}\cdot\vv{b}}\vv{b}=\frac{|\vv{a}|}{|\vv{b}|\cos\theta}\vv{b}\)，

  1. 【2011上海理17】设 \(\displaystyle A_1,A_2,A_3,A_4,A_5\) 是平面上给定的 5 个不同点, 求使得 \(\displaystyle \vv{MA_1}+\vv{MA_2}+\vv{MA_3}+\vv{MA_4}+\vv{MA_5}=\vv{0}\) 成立的点 \(\displaystyle M\) 的个数.
答案

在平面内取定一个点\(\displaystyle O\)， 由\(\displaystyle \vv{MA_i}=\vv{OA_i}-\vv{OM}\)可得 $\(\displaystyle \vv{MA_1}+\vv{MA_2}+\vv{MA_3}+\vv{MA_4}+\vv{MA_5}=\vv{OA_1}+\vv{OA_2}+\vv{OA_3}+\vv{OA_4}+\vv{OA_5}-5\vv{OM}.\)$ 令其等于零,得到 $\(\displaystyle \vv{OM}=\frac15(\vv{OA_1}+\vv{OA_2}+\vv{OA_3}+\vv{OA_4}+\vv{OA_5}).\)$ 于是\(\displaystyle \vv{OM}\)是唯一确定的,所以满足条件的点\(\displaystyle M\)有\(\displaystyle 1\)个.

评：这是求解三角形重心的其中一种方法的延申。设平面直角坐标系中\(\displaystyle \triangle ABC\)的三个顶点\(\displaystyle A,B,C\)的坐标分别为\(\displaystyle (x_1,y_1),(x_2,y_2),(x_3,y_3)\)，则该三角形的重心为\(\displaystyle G(\frac{1}{3}(x_1+x_2+x_3),\frac{1}{3}(y_1+y_2+y_3))\)。一种证明方法利用率\(\displaystyle \vv{GA}+\vv{GB}+\vv{GC}=\vv{0}\)，再复用该题采用的上述方法即可。

  1. 【2025上海12】已知 \(\displaystyle f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}\) \(\displaystyle ，\) \(\displaystyle \vv{a}, \vv{b}, \vv{c}\) 是平面内三个不同的单位向量．若 \(\displaystyle f(\vv{a} \cdot \vv{b}) + f(\vv{b} \cdot \vv{c}) + f(\vv{c} \cdot \vv{a}) = 0\)，则 \(\displaystyle |\vv{a} + \vv{b} + \vv{c}|\) 的取值范围是\(\displaystyle (\triangle)\)
  2. 【2020浙江17】单位向量\(\displaystyle \vv{e}_1, \vv{e}_2\) 满足 \(\displaystyle |2\vv{e}_1 - \vv{e}_2| \leqslant \sqrt{2}\),已知\(\displaystyle \vv{a} = \vv{e}_1 + \vv{e}_2, \vv{b} = 3\vv{e}_1 + \vv{e}_2\),设 \(\displaystyle \vv{a}, \vv{b}\) 的夹角为 \(\displaystyle \theta\),求 \(\displaystyle \cos^2 \theta\) 的最小值.
答案

设\(\displaystyle \vv{e}_1\cdot\vv{e}_2=t,-1\leqslant t\leqslant 1\).对\(\displaystyle |2\vv{e}_1-\vv{e}_2|\leqslant\sqrt2\)两边平方，结合单位向量条件,化简得\(\displaystyle 5-4t\leqslant2,\)所以\(\displaystyle t\geqslant\frac34\).综合得到\(\displaystyle t\)的取值范围为\(\displaystyle \frac34\leqslant t\leqslant 1\).

计算得\(\displaystyle \vv{a}\cdot\vv{b}=4+4t\),\(\displaystyle |\vv{a}|^2=2+2t\),\(\displaystyle |\vv{b}|^2=10+6t\),故 $\(\displaystyle \cos^2\theta=\frac{(\vv{a}\cdot\vv{b})^2}{|\vv{a}|^2|\vv{b}|^2}=\frac{(4+4t)^2}{(2+2t)(10+6t)}=\frac{4(1+t)}{5+3t}.\)$ 设\(\displaystyle f(t)=\frac{4(1+t)}{5+3t}\),因为\(\displaystyle f'(t)=\frac{8}{(5+3t)^2}>0\)，所以\(\displaystyle f(t)\)在\(\displaystyle \frac34\leqslant t\leqslant 1\)时单调递增,最小值在\(\displaystyle t=\frac34\)时取得,带入得到\(\displaystyle \cos^2\theta_{\min}=\frac{28}{29}.\)

注：求\(\displaystyle f(t)=\frac{4(1+t)}{5+3t}\)也可分离常数得\(\displaystyle f(t)=\frac{4}{3}-\frac{8}{3(5+3t)}\)，据此判断\(\displaystyle f(t)\)在\(\displaystyle \frac34\leqslant t\leqslant 1\)时单调递增。

  1. 【2007重庆理10】 如图, 在四边形 \(\displaystyle ABCD\) 中, \(\displaystyle |\vv{AB}| + |\vv{BD}| + |\vv{DC}| = 4\), \(\displaystyle |\vv{AB}| \cdot |\vv{BD}| + |\vv{BD}| \cdot |\vv{DC}| = 4\), \(\displaystyle \vv{AB} \cdot \vv{BD} = \vv{BD} \cdot \vv{DC} = 0\), 求 \(\displaystyle (\vv{AB} + \vv{DC}) \cdot \vv{AC}\) 的值.
  2. 【2006辽宁12】 设 \(\displaystyle O(0,0)\), \(\displaystyle A(1,0)\), \(\displaystyle B(0,1)\), 点 \(\displaystyle P\) 是线段 \(\displaystyle AB\) 上的一个动点, \(\displaystyle \vv{AP} = \lambda\vv{AB}\), 若 \(\displaystyle \vv{OP} \cdot \vv{AB} \geqslant \vv{PA} \cdot \vv{PB}\), 求实数 \(\displaystyle \lambda\) 的取值范围.
  3. 【2017上海10】在平面直角坐标系 \(\displaystyle xOy\) 中, 已知椭圆 \(\displaystyle C_1\): \(\displaystyle \frac{x^2}{36}+\frac{y^2}{4}=1\) 和 \(\displaystyle C_2\): \(\displaystyle x^2+\frac{y^2}{9}=1\). \(\displaystyle P\) 为 \(\displaystyle C_1\) 上的动点, \(\displaystyle Q\) 为 \(\displaystyle C_2\) 上的动点, \(\displaystyle w\) 是 \(\displaystyle \vv{OP}\cdot\vv{OQ}\) 的最大值. 记 $\(\displaystyle \Omega=\{(P,Q)\mid P \text{在} C_1 \text{上}, Q \text{在} C_2 上, \text{且} \vv{OP}\cdot\vv{OQ}=w\}\)$ 求 \(\displaystyle |\Omega|\).
  4. 【2018天津理8】在平面四边形 \(\displaystyle ABCD\) 中,\(\displaystyle AB \perp BC, AD \perp CD, \angle BAD = 120^\circ, AB = AD = 1\). 若点 \(\displaystyle E\) 为边 \(\displaystyle CD\) 上的动点,求 \(\displaystyle \vv{AE} \cdot \vv{BE}\) 的最小值.
  5. 【2017全国II卷理12】已知 \(\displaystyle \triangle ABC\) 是边长为 \(\displaystyle 2\) 的等边三角形,\(\displaystyle P\) 为平面 \(\displaystyle ABC\) 内一点,求 \(\displaystyle \vv{PA} \cdot (\vv{PB} + \vv{PC})\) 的最小值.
答案

设\(\displaystyle BC\)的中点为\(\displaystyle D\)，\(\displaystyle AD\)的中点为\(\displaystyle O\),则$\(\displaystyle \vv{PA}\cdot(\vv{PB}+\vv{PC})=2\vv{PA}\cdot \vv{PD}=2(PO^2-OD^2)\geqslant -OD^2=-\frac32.\)$ 故最小值为\(\displaystyle -\frac32\).

  1. 【2023全国乙卷理12】已知 \(\displaystyle \odot O\) 的半径为 \(\displaystyle 1\),直线 \(\displaystyle PA\) 与 \(\displaystyle \odot O\) 相切于点 \(\displaystyle A\),直线 \(\displaystyle PB\) 与 \(\displaystyle \odot O\) 交于 \(\displaystyle B, C\) 两点,\(\displaystyle D\) 为 \(\displaystyle BC\) 的中点. 若 \(\displaystyle |PO| = \sqrt{2}\),求 \(\displaystyle \vv{PA} \cdot \vv{PD}\) 的最大值.
  2. 【2013安徽9】 平面直角坐标系中,\(\displaystyle O\)是坐标原点,定点\(\displaystyle A,B\)满足\(\displaystyle {\left | \vv {OA} \right |}={\left | \vv{OB} \right |}= \vv {OA} \cdot \vv {OB}=2\),求点集\(\displaystyle \left \{ P\mid \vv {OP}=\lambda\vv {OA}+ \mu\vv {OB},\left |\lambda \right |+\left | \mu \right |\leqslant 1,\lambda,\mu \in \mathbb{R}\right \}\)所表示区域的面积.
  3. 【2025绍兴一模8改编】在平面直角坐标系中,已知\(\displaystyle A(-1,0),B(0,3),C(2,0),\) \(\displaystyle D(0,-6),P\)是平面内任意一点,记\(\displaystyle \vv{PA}=\vv{a}, \vv{PB}=\vv{b}, \vv{PC}=\vv{c}, \vv{PD}=\vv{d}\),分别求 $\(\displaystyle (a).|\vv{a}+\vv{b}+\vv{c}+\vv{d}|,\quad (b).(\vv{a}+\vv{b})\cdot(\vv{c}+\vv{d})\quad (c).|\vv{a}+\vv{b}|+|\vv{c}+\vv{d}|\quad (d).(\vv{a}+\vv{b})\times(\vv{c}+\vv{d})\)$ 的最小值
  4. 在\(\displaystyle \Delta ABC\)中,\(\displaystyle \vv{CE}=2\vv{EA},\vv{CD}=\vv{DB}\),\(\displaystyle F\)为\(\displaystyle AD\)与\(\displaystyle BE\)的交点,且\(\displaystyle \vv{CA}\cdot \vv{CB}=3\),当向量\(\displaystyle \vv{CF}\)在\(\displaystyle \vv{CB}\)上的投影向量的模取最小值时,求\(\displaystyle |\vv{CB}|\).
  5. 【2021八省联考变式6】已知\(\displaystyle A,B,C\)是单位圆上的三个动点,求\(\displaystyle \vv{AB}\cdot \vv{AC}\)的最小值.
  6. 【上海11】已知实数\(\displaystyle \lambda>0\),向量\(\displaystyle \vv{a},\vv{b},\vv{c}\)满足\(\displaystyle |\vv{a}|=|\vv{b}|=|\vv{c}|=\lambda\),且\(\displaystyle \vv{a}\cdot \vv{b}=0,\vv{b}\cdot\vv{c}=1,\vv{c}\cdot\vv{a}=2\),求\(\displaystyle \lambda\).
  7. 【2017浙江15】已知向量\(\displaystyle \vv{a},\vv{b}\)满足\(\displaystyle |\vv{a}|=1,|\vv{b}|=2\),求\(\displaystyle |\vv{a}+\vv{b}|+|\vv{a}-\vv{b}|\)的最小值与最大值.
答案

设\(\displaystyle \vv{a}\cdot\vv{b}=t\),其中\(\displaystyle -2\leqslant t\leqslant2\).令\(\displaystyle x=|\vv{a}+\vv{b}|\), \(\displaystyle y=|\vv{a}-\vv{b}|\),则 $\(\displaystyle x^2+y^2&=2|\vv{a}|^2+2|\vv{b}|^2=10 x^2y^2&=(|\vv{a}+\vv{b}|^2+|\vv{a}-\vv{b}|)^2 &=(|\vv{a}|^2+|\vv{b}|^2+2\vv{a}\cdot \vv{b})(|\vv{a}|^2+|\vv{b}|^2-2\vv{a}\cdot \vv{b}) &=(|\vv{a}|^2+|\vv{b}|^2)^2-4(\vv{a}\cdot\vv{b})^2 &=25-4t^2.\)$ 所以\(\displaystyle (x+y)^2=10+2\sqrt{25-4t^2}.\)当\(\displaystyle |t|=2\)时取得最小值\(\displaystyle 16\),当\(\displaystyle t=0\)时取得最大值\(\displaystyle 20\),故所求最小值为\(\displaystyle 4\),最大值为\(\displaystyle 2\sqrt5\).

  1. 【2019江苏12】如图, 在 \(\displaystyle \triangle ABC\) 中, \(\displaystyle D\) 是 \(\displaystyle BC\) 的中点, \(\displaystyle E\) 在边 \(\displaystyle AB\) 上, \(\displaystyle BE=2EA\), \(\displaystyle AD\) 与 \(\displaystyle CE\) 交于点 \(\displaystyle O\). 若 \(\displaystyle \vv{AB}\cdot\vv{AC}=6\vv{AO}\cdot\vv{EC}\), 求 \(\displaystyle \frac{AB}{AC}\) 的值.
  2. 【2007陕西理15】如图, 平面内有三个向量 \(\displaystyle \vv{OA},\vv{OB},\vv{OC}\), 其中 \(\displaystyle \vv{OA}\) 与 \(\displaystyle \vv{OB}\) 的夹角为 \(\displaystyle 120^\circ\), \(\displaystyle \vv{OA}\) 与 \(\displaystyle \vv{OC}\) 的夹角为 \(\displaystyle 30^\circ\), 且 \(\displaystyle |\vv{OA}| = |\vv{OB}| = 1\), \(\displaystyle |\vv{OC}| = 2\sqrt{3}\). 若 \(\displaystyle \vv{OC} = \lambda\vv{OA} + \mu\vv{OB}\) (\(\displaystyle \lambda,\mu \in \mathbb{R}\)), 求 \(\displaystyle \lambda + \mu\) 的值.
  3. 【2020江苏13】在 \(\displaystyle \triangle ABC\) 中, \(\displaystyle AB=4\), \(\displaystyle AC=3\), \(\displaystyle \angle BAC=90^\circ\), \(\displaystyle D\) 在边 \(\displaystyle BC\) 上, 延长 \(\displaystyle AD\) 到 \(\displaystyle P\), 使得 \(\displaystyle AP=9\), 若 \(\displaystyle \vv{PA}=m\vv{PB}+\left(\frac{3}{2}-m\right)\vv{PC}\) (\(\displaystyle m\) 为常数), 求 \(\displaystyle CD\) 的长度.（只考虑\(\displaystyle CD>0\)的情况）
  4. 【2017江苏12】如图,在同一个平面内,向量\(\displaystyle \vv{OA},\vv{OB},\vv{OC}\)的模分别为\(\displaystyle 1,1,\sqrt{2}\),\(\displaystyle \vv{OA}\)与\(\displaystyle \vv{OC}\)的夹角为\(\displaystyle \alpha\),且\(\displaystyle \tan\alpha=7\),\(\displaystyle \vv{OB}\)与\(\displaystyle \vv{OC}\)的夹角为\(\displaystyle 45^{\circ}\).若\(\displaystyle \vv{OC}=m\vv{OA}+n\vv{OB}\),求\(\displaystyle m+n\).
答案

新答案（来源：2024上海，15（立体几何、向量）.md）： \(\displaystyle \frac{5}{28}\)

【解题思路】\(\displaystyle \vv{OP}+\vv{OA_i}+\vv{OA_j}=\vv{0}\)等价于\(\displaystyle \vv{OP}=-\left(\vv{OA_i}+\vv{OA_j}\right)\)，\(\displaystyle P\)点由取定的\(\displaystyle A_i\)，\(\displaystyle A_j\)唯一确定．在八个点中任取不同的两点的取法数为\(\displaystyle \mathrm{C}_8^2\)即\(\displaystyle 28\)．其中，使得\(\displaystyle \vv{OA_i}+\vv{OA_j}=\vv{0}\)即\(\displaystyle P\)落在原点的取法有\(\displaystyle 4\)种，分别是\(\displaystyle A_1\)与\(\displaystyle A_5\)、\(\displaystyle A_2\)与\(\displaystyle A_6\)、\(\displaystyle A_3\)与\(\displaystyle A_7\)、\(\displaystyle A_4\)与\(\displaystyle A_8\)；使得\(\displaystyle \vv{OA_i}+\vv{OA_j}\)是与坐标轴平行的向量即\(\displaystyle P\)落在除原点外的坐标轴上的取法也有\(\displaystyle 4\)种，分别是\(\displaystyle A_2\)与\(\displaystyle A_8\)、\(\displaystyle A_4\)与\(\displaystyle A_6\)、\(\displaystyle A_2\)与\(\displaystyle A_4\)、\(\displaystyle A_6\)与\(\displaystyle A_8\)．因此，使得\(\displaystyle P\)落在四个象限的取法有\(\displaystyle 20\)种，而\(\displaystyle P\)落在每个象限的机会均等，从而使得\(\displaystyle P\)落在第一象限的取法有\(\displaystyle 5\)种，由此求得\(\displaystyle P\)落在第一象限的概率是\(\displaystyle \frac{5}{28}\)．

【实测数据】本题理科难度为\(\displaystyle 0.425\)，区分度为\(\displaystyle 0.50\)．

 新答案（来源：1.18 空间直角坐标系与空间向量.md）：

\(\displaystyle 1\)；\(\displaystyle 2\)；\(\displaystyle 2\sqrt{2}\)

【解题思路】因为\(\displaystyle \left|\vv{b}-\left(x\vv{e_1}+y\vv{e_2}\right)\right|\geqslant\left|\vv{b}-\left(x_0\vv{e_1}+y_0\vv{e_2}\right)\right|=1\left(x_0,y_0\in\mathbb{R}\right)\)，

两边平方，得

\[\displaystyle \left|\vv{b}\right|^2-2\vv{b}\cdot\left(x\vv{e_1}+y\vv{e_2}\right)+\left|x\vv{e_1}+y\vv{e_2}\right|^2\geqslant\left|\vv{b}\right|^2-2\vv{b}\cdot\left(x_0\vv{e_1}+y_0\vv{e_2}\right)+\left|x_0\vv{e_1}+y_0\vv{e_2}\right|^2\]

由于\(\displaystyle \vv{b}\cdot\vv{e_1}=2\)，\(\displaystyle \vv{b}\cdot\vv{e_2}=\frac{5}{2}\)，且

\[\displaystyle \left|x\vv{e_1}+y\vv{e_2}\right|^2=x^2\left|\vv{e_1}\right|^2+y^2\left|\vv{e_2}\right|^2+2xy\vv{e_1}\cdot\vv{e_2}=x^2+y^2+xy\]

所以

\[\displaystyle \left|\vv{b}\right|^2-4x-5y+x^2+y^2+xy\geqslant\left|\vv{b}\right|^2-4x_0-5y_0+x_0^2+y_0^2+x_0y_0=1\]

考虑函数\(\displaystyle g\left(x,y\right)=\left|\vv{b}\right|^2-4x-5y+x^2+y^2+xy\)．由题意，\(\displaystyle g\left(x,y\right)\)在\(\displaystyle \left(x,y\right)=\left(x_0,y_0\right)\)取到最小值，且最小值为\(\displaystyle 1\)．

将\(\displaystyle g\left(x,y\right)\)视作\(\displaystyle x\)的函数，然后再配方，可得

\[\displaystyle g\left(x,y\right)=x^2+\left(y-4\right)x+y^2-5y+\left|\vv{b}\right|^2=\left(x+\frac{y-4}{2}\right)^2-\frac{\left(y-4\right)^2}{4}+y^2-5y+\left|\vv{b}\right|^2=\left(x+\frac{y-4}{2}\right)^2+\frac{3y^2}{4}-3y+\left|\vv{b}\right|^2-4=\left(x+\frac{y-4}{2}\right)^2+\frac{3}{4}\left(y-2\right)^2+\left|\vv{b}\right|^2-7\]

因此，当\(\displaystyle x+\frac{y-4}{2}=0\)，且\(\displaystyle y-2=0\)时\(\displaystyle g\left(x,y\right)\)取到最小值，所以\(\displaystyle x_0+\frac{y_0-4}{2}=0\)且\(\displaystyle y_0=2\)，故\(\displaystyle x_0=1\)．

最小值\(\displaystyle \left|\vv{b}\right|^2-7=1\)，所以\(\displaystyle \left|\vv{b}\right|=2\sqrt{2}\)．

  1. 【2015上海理14】在锐角三角形 \(\displaystyle ABC\) 中, \(\displaystyle \tan A=\frac{1}{2}\), \(\displaystyle D\) 为边 \(\displaystyle BC\) 上的点, \(\displaystyle \triangle ABD\) 与 \(\displaystyle \triangle ACD\) 的面积分别为 \(\displaystyle 2\) 和 \(\displaystyle 4\). 过 \(\displaystyle D\) 作 \(\displaystyle DE\perp AB\) 于 \(\displaystyle E\), \(\displaystyle DF\perp AC\) 于 \(\displaystyle F\), 求 \(\displaystyle \vv{DE}\cdot\vv{DF}\).
  2. 【2026上海春10】在\(\displaystyle \Delta ABC\)中,\(\displaystyle D,E\)在\(\displaystyle BC\)上,且\(\displaystyle \vv{BD}=\vv{DE}=\vv{EC},|\vv{AD}|=1,<\vv{AD},\vv{AE}>=\frac{\pi}{3}\),求\(\displaystyle \vv{AB}\cdot \vv{AC}\)的最大值.
  3. 【2011天津理14】已知直角梯形 \(\displaystyle ABCD\) 中, \(\displaystyle AD \parallel BC,\angle ADC=90^\circ,AD=2,BC=1\), \(\displaystyle P\) 是腰 \(\displaystyle DC\) 上的动点, 求 \(\displaystyle |\vv{PA}+3\vv{PB}|\) 的最小值.

  4. 【2006湖南理15】如图, \(\displaystyle OM \parallel AB\), 点 \(\displaystyle P\) 在由射线 \(\displaystyle OM\)、线段 \(\displaystyle OB\) 及 \(\displaystyle AB\) 的延长线围成的区域内 (不含边界) 运动, 设 \(\displaystyle \vv{OP} = x\vv{OA} + y\vv{OB}\).

    1. 求 \(\displaystyle x\) 的取值范围；
      1. 当 \(\displaystyle x = -\frac{1}{2}\) 时, 求\(\displaystyle y\) 的取值范围.
    2. 【2009天津理15】 四边形 \(\displaystyle ABCD\) 中, \(\displaystyle \vv{AB} = \vv{DC} = (1,1)\), \(\displaystyle \frac{1}{|\vv{BA}|}\vv{BA} + \frac{1}{|\vv{BC}|}\vv{BC} = \frac{\sqrt{3}}{|\vv{BD}|}\vv{BD}\), 求四边形\(\displaystyle ABCD\)的面积.
    3. 【2021天津15】在边长为 \(\displaystyle 1\) 的等边三角形 \(\displaystyle ABC\) 中，\(\displaystyle D\) 为线段 \(\displaystyle BC\) 上的动点，\(\displaystyle DE \perp AB\) 且交 \(\displaystyle AB\) 于点 \(\displaystyle E\)，\(\displaystyle DF \parallel AB\) 且交 \(\displaystyle AC\) 于点 \(\displaystyle F\)，求 \(\displaystyle \left|2\vv{BE} + \vv{DF}\right|\) 的值与\(\displaystyle \left(\vv{DE} + \vv{DF}\right) \cdot \vv{DA}\) 的最小值。
    4. 设\(\displaystyle A,B\)是平面直角坐标系中关于\(\displaystyle y\)轴对称的两点,且\(\displaystyle |\vv{OA}|=2\),若存在\(\displaystyle m,n\in\mathbb{R}\),使得\(\displaystyle m\vv{AB}+\vv{OA}\)与\(\displaystyle n\vv{AB}+\vv{OB}\)垂直,且\(\displaystyle |(m\vv{AB}+\vv{OA})-(n\vv{AB}+\vv{OB})|=2\),求\(\displaystyle |\vv{AB}|\)的最小值.
    5. 【2019浙江17】已知正方形 \(\displaystyle ABCD\) 的边长为 \(\displaystyle 1\), 求$\(\displaystyle |\lambda_1\vv{AB}+\lambda_2\vv{BC}+\lambda_3\vv{CD}+\lambda_4\vv{DA}+\lambda_5\vv{AC}+\lambda_6\vv{BD}|,\quad \lambda_i\in\{-1,1\}(i=1,2,3,4,5,6)\)$ 的最小值与最大值.
    6. 【2016上海理14】 如图, 在平面直角坐标系 \(\displaystyle xOy\) 中, \(\displaystyle O\) 为正八边形 \(\displaystyle A_1A_2\cdots A_8\) 的中心, \(\displaystyle A_1(1,0)\), 任取不同的两点 \(\displaystyle A_i,A_j\), 点 \(\displaystyle P\) 满足 \(\displaystyle \vv{OP}+\vv{OA_i}+\vv{OA_j}=\vv{0}\), 求点 \(\displaystyle P\) 落在第一象限的概率.
    7. 【2010浙江理16】已知平面向量 \(\displaystyle \vv{\alpha},\vv{\beta}\) (\(\displaystyle \vv{\alpha} \neq 0,\vv{\alpha} \neq \vv{\beta}\)) 满足 \(\displaystyle |\vv{\beta}|=1,<\vv{\alpha},\vv{\beta}-\vv{\alpha}>=\frac{2\pi}{3}\), 求 \(\displaystyle |\vv{\alpha}|\) 的取值范围.
    8. 【2013浙江理17】设 \(\displaystyle \vv{e}_1,\vv{e}_2\) 为单位向量, 非零向量 \(\displaystyle \vv{b}=x\vv{e}_1+y\vv{e}_2, x,y \in \mathbb{R}\). 若 \(\displaystyle <\vv{e}_1,\vv{e}_2>=\frac{\pi}{6}\), 求 \(\displaystyle \frac{|x|}{|\vv{b}|}\) 的最大值.
    9. 【2015浙江15】已知\(\displaystyle \vv{e}_1,\vv{e}_2\)是空间单位向量.若空间向量\(\displaystyle \vv{b}\)满足\(\displaystyle \vv{b}\cdot \vv{e}_1=2,\vv{b}\cdot \vv{e}_2=\frac{5}{2}\),且对于任意\(\displaystyle x,y\in\mathbb{R}\),$\(\displaystyle |\vv{b}-(x\vv{e}_1+y\vv{e}_2)|\geqslant |\vv{b}-(x_0\mathrm{e}_1+y_0\mathrm{e}_2)|=1(x_0,y_0\in\mathbb{R})\)\(求\)\displaystyle x_0,y_0,|\vv{b}|$.
    10. 【2026闵行二模16】已知平面上存在13个向量 \(\displaystyle \vv{a_1},\vv{a_2},\dots,\vv{a_{13}}\),其中 \(\displaystyle |\vv{a_1}|=1\).且对任意 \(\displaystyle n \in \mathbb{N},1 \leqslant n \leqslant 12\),有 \(\displaystyle |\vv{a_{n+1}}|=\sqrt{2}|\vv{a_n}|,\vv{a_n} \cdot \vv{a_{n+1}}=0\),求 \(\displaystyle |\vv{a_1}+\vv{a_2}+\dots+\vv{a_{13}}|\) 的最小值. ??? answer "答案"

    相邻向量互相垂直,故奇数项方向均在同一直线上,偶数项方向均在与其垂直的直线上.奇数项模长为\(\displaystyle 1,2,4,\ldots,64\),其带符号和的模最小为\(\displaystyle 64-(1+2+4+8+16+32)=1\);偶数项模长为\(\displaystyle \sqrt2,2\sqrt2,\ldots,32\sqrt2\),其带符号和的模最小为\(\displaystyle 32\sqrt2-(\sqrt2+2\sqrt2+\cdots+16\sqrt2)=\sqrt2\).两方向垂直,故总和模长的最小值为\(\displaystyle \sqrt{1^2+(\sqrt2)^2}=\sqrt3\). 47. 【2021浙江17】已知平面向量\(\displaystyle \vv{a},\vv{b},\vv{c}(\vv{c}\neq 0)\)满足\(\displaystyle |\vv{a}|=1,|\vv{b}|=2,\vv{a}\cdot\vv{b}=0,(\vv{a}-\vv{b})\cdot \vv{c}=0\),记平面向量\(\displaystyle \vv{d}\)在\(\displaystyle \vv{a},\vv{b}\)方向上的投影分别为\(\displaystyle \vv{x},\vv{y}\),\(\displaystyle \vv{d}-\vv{a}\)在\(\displaystyle \vv{c}\)方向上的投影为\(\displaystyle \vv{z}\),求\(\displaystyle \vv{x}^2+\vv{y}^2+\vv{z}^2\)的最小值. 48. 【2022浙江17】设点\(\displaystyle P\)在单位圆的内接正八边形\(\displaystyle A_1A_2\cdots A_8\)的边\(\displaystyle A_1A_2\)上,求\(\displaystyle \vv{PA_1}^2+\vv{PA_2}^2+\cdots +\vv{PA_8}^2\)的取值范围. 49. 【2026海淀一模10】已知平面上互不重合的点\(\displaystyle A_1,A_2,A_3,A_4,A_5\)满足\(\displaystyle \angle A_1A_2A_3=\angle A_2A_3A_4=\angle A_3A_4A_5=120^\circ\)且\(\displaystyle \vv{A_1A_2}\cdot\vv{A_2A_3}=\vv{A_2A_3}\cdot\vv{A_3A_4}=\vv{A_3A_4}\cdot\vv{A_4A_5}\),下列命题中一定不成立的是

  5. \(\displaystyle \vv{A_1A_5}=\vv{A_2A_4}\)

  6. \(\displaystyle \vv{A_1A_5}=\dfrac{3}{2}\vv{A_2A_5}\)
  7. \(\displaystyle \vv{A_1A_5}=2\vv{A_2A_4}\)
  8. \(\displaystyle \vv{A_1A_5}=\dfrac{5}{2}\vv{A_4A_5}\)

50. 在\(\displaystyle \Delta ABC\)中,\(\displaystyle P\)为\(\displaystyle AB\)中点,\(\displaystyle O\)在边\(\displaystyle AC\)上,\(\displaystyle BO\)交\(\displaystyle CP\)于\(\displaystyle R\),且\(\displaystyle |\vv{AO} |=2|\vv{AC} |\),设\(\displaystyle |\vv{AB}|=2, |\vv{AC}|=1 ,<\vv{AB}, \vv{AC}>=\theta\). 1. 若\(\displaystyle \theta = 60^{\circ }\),求\(\displaystyle \cos \angle ARB\)； 2. 若\(\displaystyle H\)在线段\(\displaystyle BC\)上,且\(\displaystyle RH\perp BC,\theta\in[\frac{\pi}{3},\frac{2\pi}{3}]\),设,求\(\displaystyle \frac{|CH|}{|CB|}\)的取值范围.

B 组习题

C 组习题

杂\(\displaystyle \quad\)题

    1. **【2026“数海漫游一模”（网络联考）5】**已知向量$\displaystyle \vv{a},\vv{b}$满足$\displaystyle \vv{a}+\vv{b}=(1,2)$，求$\displaystyle \vv{a}\cdot \vv{b}$的最大值。
  1. 【2024宁波二模9改编】若平面向量\(\displaystyle \vv{a},\vv{b},\vv{c}\)满足\(\displaystyle |\vv{a}| = 1\)，\(\displaystyle |\vv{b}| = 1\)，\(\displaystyle |\vv{c}| = 3\)且\(\displaystyle \vv{a} \cdot \vv{c} = \vv{b} \cdot \vv{c}\)，分别求\(\displaystyle |\vv{a} + \vv{b} + \vv{c}|,|\vv{a} - \vv{b} + \vv{c}|\)的最大值与最小值。
  2. 已知\(\displaystyle O\)是\(\displaystyle \triangle ABC\)所在平面内一点,且\(\displaystyle |\vv{AB}|=2\),\(\displaystyle \vv{OA}\cdot\vv{AC}=-1\),\(\displaystyle \vv{OC}\cdot\vv{AC}=1\),求\(\displaystyle \angle ABC\)的最大值。
  3. 【2021浙江十校联考17】已知空间向量\(\displaystyle \vec{a},\vec{b},\vec{c}\)两两夹角均为\(\displaystyle 60^{\circ}\)，且\(\displaystyle |\vec{a}|=|\vec{b}|=2,|\vec{c}|=6\)，若向量\(\displaystyle \vec{x},\vec{y}\)分别满足\(\displaystyle \vec{x}\cdot(\vec{x}+\vec{a}-\vec{b})=0\)与\(\displaystyle \vec{y}\cdot\vec{c}=8\)，求\(\displaystyle |\vec{x}-\vec{y}|\)的最小值。

D 组习题