6.2.4 向量的数量积
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基础知识
概念
如果两个非零向量\(\vec{a}\), \(\vec{b}\),它们的夹角为\(θ\),我们把数量\(|\vec{a}||\vec{b}| \cos \theta\)叫做\(\vec{a}\)与\(\vec{b}\)的数量积(或内积),
记作:\(\vec{a} \cdot \vec{b}\),即 \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\).
解释
(1) 如果一个物体在力\(\vec{F}\)的作用下产生位移\(\vec{s}\),那么力所做的功\(W=\vec{F} \cdot \vec{s}=|\vec{F}||\vec{s}| \cos \theta\);
(2) 规定:零向量与任一向量的数量积是\(0\);
(3) 若\(\theta=\dfrac{\pi}{2}\),那我们说\(\vec{a}\)与\(\vec{b}\)垂直,记作\(\vec{a} \perp \vec{b}\);
(4) 数量积是一个实数,不再是一个向量;
(5) 注意确定向量的夹角\(θ\),其中\(\vec{a}\)与\(\vec{b}\)的夹角\(θ=40°\),而\(\vec{a}\)与\(\vec{c}\)的夹角\(θ=180°-40°=140°\).
【例】若\(∆ABC\)是边长为\(2\)的等边三角形,求数量积\(\overrightarrow{A B} \cdot \overrightarrow{C A}\).
解 \(\overrightarrow{A B} \cdot \overrightarrow{C A}=|\overrightarrow{A B}||\overrightarrow{C A}| \cos 120^{\circ}=2 \times 2 \times\left(-\dfrac{\sqrt{3}}{2}\right)=-2 \sqrt{3}\).
投影
\(\vec{a}\) ,\(\vec{b}\)是非零向量,向量\(\vec{a}\)在向量\(\vec{b}\)上的投影: \(|\vec{a}| \cos \theta\),它是一个实数,但不一定大于\(0\).
数量积的性质
设\(\vec{a}\),\(\vec{b}\)是非零向量,它们的夹角是\(θ\),\(\vec{e}\)是与\(\vec{b}\)方向相同的单位向量,则
(1) \(\vec{a} \cdot \vec{e}=\vec{e} \cdot \vec{a}=|\vec{a}| \cos \theta\),
(2) \(\vec{a} \perp \vec{b} \Leftrightarrow \vec{a} \cdot \vec{b}=0\),
(3) 当\(\vec{a}\)与\(\vec{b}\)同向时, \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\);当\(\vec{a}\)与\(\vec{b}\)反向时, \(\vec{a} \cdot \vec{b}=-|\vec{a}||\vec{b}|\).
特殊地, \(\vec{a}^2=|\vec{a}|^2\) 或 \(|\vec{a}|=\sqrt{\vec{a}^2}\).
(4) \(|\vec{a} \cdot \vec{b}| \leqslant|\vec{a}||\vec{b}|\).
运算法则
对于向量 \(\vec{a}\) ,\(\vec{b}\) ,\(\vec{c}\)和实数\(λ\),有
(1) \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) ; (2) \((\lambda \vec{a}) \cdot \vec{b}=\lambda(\vec{a} \cdot \vec{b})=\vec{a} \cdot(\lambda \vec{b})\) ; (3) \((\vec{a}+\vec{b}) \cdot \vec{c}=\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{c}\).
但是\((\vec{a} \cdot \vec{b}) \vec{c}=\vec{a}(\vec{b} \cdot \vec{c})\)不一定成立.
(当向量\(\vec{a}\),\(\vec{c}\)不共线时,向量\(\vec{a}(\vec{b} \cdot \vec{c})\)与向量\((\vec{a} \cdot \vec{b}) \vec{c}\)肯定不共线,那怎么可能相等呢)
即向量的数量积满足交换律,分配率,但不满足结合律.
基本方法
【题型1】 求数量积
【典题1】 在边长为\(2\)的菱形\(ABCD\)中,若\(AC=2\),则 \(\overrightarrow{A B} \cdot \overrightarrow{C A}=\)\(\underline{\quad \quad}\) .
解析 \(\overrightarrow{A B} \cdot \overrightarrow{C A}=-|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cos \langle\overrightarrow{A B}, \overrightarrow{A C}\rangle=-\dfrac{1}{2}|\overrightarrow{A C}|^2=-2\).
点拨 根据数量积的定义\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)求解.
【典题2】 知平行四边形\(ABCD\)中, \(|\overrightarrow{A B}|=|\overrightarrow{A D}|=\overrightarrow{A B} \cdot \overrightarrow{A D}=2\),点\(E\)为边\(BC\)的中点,则\(\overrightarrow{A E} \cdot \overrightarrow{A C}\)的值为 \(\underline{\quad \quad}\) .
解析 \(\because|\overrightarrow{A B}|=|\overrightarrow{A D}|=\overrightarrow{A B} \cdot \overrightarrow{A D}=2\), \(\therefore \overrightarrow{A B} \cdot \overrightarrow{A D}=4 \cos \angle B A D=2\),
\(\therefore \cos \angle B A D=\dfrac{1}{2}\), \(\therefore \angle B A D=\dfrac{\pi}{3}\),
\(∵\)点\(E\)为边\(BC\)的中点,
\(\therefore \overrightarrow{A E} \cdot \overrightarrow{A C}=(\overrightarrow{A B}+\overrightarrow{B E}) \cdot(\overrightarrow{A B}+\overrightarrow{A D})=\left(\overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{B C}\right) \cdot(\overrightarrow{A B}+\overrightarrow{A D})\)
\(=\left(\overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}\right) \cdot(\overrightarrow{A B}+\overrightarrow{A D})=\overrightarrow{A B}^2+\dfrac{3}{2} \overrightarrow{A B} \cdot \overrightarrow{A D}+\dfrac{1}{2} \overrightarrow{A D}^2\)
\(=4+\dfrac{3}{2} \times 2 \times 2 \cos \dfrac{\pi}{3}+\dfrac{1}{2} \times 4=9\).
点拨 直接用数量积的定义求\(\overrightarrow{A E} \cdot \overrightarrow{A C}\)显然不好求,则把\(\overrightarrow{A E} \cdot \overrightarrow{A C}\)转化为向量\(\overrightarrow{A B}\),\(\overrightarrow{A D}\)的数量积问题,因为已知条件中均与\(\overrightarrow{A B}\),\(\overrightarrow{A D}\)有关.
【典题3】 已知\(|\vec{a}|=4\), \(|\vec{b}|=8\),\(\vec{a}\)与\(\vec{b}\)的夹角是\(120^∘\).
(1)计算\(|\vec{a}+\vec{b}|\); (2)当\(k\)为何值时, \((\vec{a}+2 \vec{b}) \perp(k \vec{a}+\vec{b})\).
解析 (1) \(|\vec{a}+\vec{b}|=\sqrt{(\vec{a}+\vec{b})^2}=\sqrt{\vec{a}^2+\vec{b}^2+2 \vec{a} \cdot \vec{b}}=\sqrt{16+64-2 \times 4 \times 8 \times \dfrac{1}{2}}=4 \sqrt{3}\).
(2) \(\because(\vec{a}+2 \vec{b}) \perp(k \vec{a}+\vec{b})\), \(\therefore(\vec{a}+2 \vec{b}) \cdot(k \vec{a}+\vec{b})=0\),
\(\therefore k \vec{a}^2+(2 k-1) \vec{a} \cdot \vec{b}-2 \vec{b}^2=0\),
即\(16 k-16(2 k-1)-2 \times 64=0\),解得\(k=-7\),
故当\(k=-7\)时,\((\vec{a}+2 \vec{b}) \perp(k \vec{a}+\vec{b})\)成立.
点拨 利用数量积的性质\(|\vec{a}|^2=\vec{a}^2\)求向量的模.
【巩固练习】
1.在\(△ABC\)中,\(AB=4\),\(AC=3\),\(∠A=60^∘\),则 \(\overrightarrow{B C} \cdot \overrightarrow{A C}=\)\(\underline{\quad \quad}\).
2.四边形\(ABCD\)中,\(AB//DC\),\(AB=3\),\(DC=2\),\(∠ABC=90^∘\),则 \((\overrightarrow{A C}+\overrightarrow{B D}) \cdot \overrightarrow{A B}=\)\(\underline{\quad \quad}\).
3.已知向量\(\vec{a}\),\(\vec{b}\)满足\(|\vec{a}|=|\vec{b}|=2\),\(|\vec{a}-\vec{b}|=2 \sqrt{3}\),则 \(\vec{a} \cdot \vec{b}=\) \(\underline{\quad \quad}\) .
4.已知向量\(\vec{a}\),\(\vec{b}\)满足\(|\vec{a}+\vec{b}|=|\vec{b}|\),且\(|\vec{a}|=2\),则\(\vec{a} \cdot \vec{b}=\)\(\underline{\quad \quad}\) .
5.已知平面向量\(\vec{a}\),\(\vec{b}\)的夹角为\(\dfrac{\pi}{3}\),且\(\vec{a} \cdot \vec{b}=1\),则 \(|\vec{a}+\vec{b}|\)的最小值为\(\underline{\quad \quad}\).
参考答案
-
答案 \(3\)
解析 在\(△ABC\)中,\(AB=4\),\(AC=3\),\(∠A=60^∘\),
由 \(\overrightarrow{B C} \cdot \overrightarrow{A C}=(\overrightarrow{A C}-\overrightarrow{A B}) \cdot \overrightarrow{A C}=\overrightarrow{A C}^2-\overrightarrow{A B} \cdot \overrightarrow{A C}=3^2-4 \times 3 \cos 60^{\circ}=3\). -
答案 \(3\)
解析 如图所示,
四边形\(ABCD\)中,\(AB=3\),\(DC=2\),\(∠ABC=90^∘\),
所以 \((\overrightarrow{A C}+\overrightarrow{B D}) \cdot \overrightarrow{A B}=(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{B C}+\overrightarrow{C D}) \cdot \overrightarrow{A B}\)
\(=\overrightarrow{A B}^2+2 \overrightarrow{B C} \cdot \overrightarrow{A B}+\overrightarrow{C D} \cdot \overrightarrow{A B}=3^2+0-2 \times 3=3\)
-
答案 \(-2\)
解析 因为\(|\vec{a}|=|\vec{b}|=2\),\(|\vec{a}-\vec{b}|=2 \sqrt{3}\),
所以\((\vec{a}-\vec{b})^2=\vec{a}^2-2 \vec{a} \cdot \vec{b}+\vec{b}^2=4-2 \vec{a} \cdot \vec{b}+4=12\),
解得\(\vec{a} \cdot \vec{b}=-2\). -
答案 \(-2\)
解析 因为\(|\vec{a}+\vec{b}|=|\vec{b}|\),即有\(|\vec{a}+\vec{b}|^2=|\vec{b}|^2\),
所以 \(\vec{a}^2+2 \vec{a} \cdot \vec{b}+\vec{b}^2=\vec{b}^2\),则 \(2 \vec{a} \cdot \vec{b}=-\vec{a}^2=-4\),
所以\(\vec{a} \cdot \vec{b}=-2\). -
答案 \(\sqrt{6}\)
解析 \(∵\)平面向量\(\vec{a}\),\(\vec{b}\)的夹角为\(\dfrac{\pi}{3}\),
\(\therefore \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos 60^{\circ}=\dfrac{1}{2}|\vec{a}| \cdot|\vec{b}|=1\),
\(\therefore|\vec{a}| \cdot|\vec{b}|=2\),
则\(|\vec{a}+\vec{b}|=\sqrt{(\vec{a}+\vec{b})^2}=\sqrt{\overrightarrow{a^2}+2 \vec{a} \cdot \vec{b}+\vec{b}^2} \geq \sqrt{2+2|\vec{a}||\vec{b}|}=\sqrt{6}\),
当且仅当 \(|\vec{a}|=|\vec{b}|=\sqrt{2}\)时取等号,
故\(|\vec{a}+\vec{b}|\)的最小值为\(\sqrt{6}\).
【题型2】 求向量夹角
【典题1】 已知\(|\vec{a}|=1\), \(|\vec{b}|=2\), \(|2 \vec{a}-\vec{b}|=2 \sqrt{3}\).
(1)求\(\vec{a}\)与\(\vec{b}\)的夹角及\(|2 \vec{a}+\vec{b}|\);
(2)当\(k \vec{a}-\vec{b}\)与 \(\vec{a}+2 \vec{b}\)的夹角为钝角时,求实数\(k\)的取值范围.
解析 (1)根据题意,设\(\vec{a}\)与\(\vec{b}\)的夹角为\(θ\),
若\(|2 \vec{a}-\vec{b}|=2 \sqrt{3}\),则有 \((2 \vec{a}-\vec{b})^2=4 \vec{a}^2+\vec{b}^2-4 \vec{a} \cdot \vec{b}=8-8 \cos \theta=12\),
解可得 \(\cos \theta=-\dfrac{1}{2}\),
又由\(0⩽θ⩽π\),则 \(\theta=\dfrac{2 \pi}{3}\);
\((2 \vec{a}+\vec{b})^2=4 \vec{a}^2+\vec{b}^2+4 \vec{a} \cdot \vec{b}=8-4=4\),则 \(|2 \vec{a}+\vec{b}|=2\);
(2)根据题意,当\(k \vec{a}-\vec{b}\)与 \(\vec{a}+2 \vec{b}\)的夹角为钝角时,
有 \((k \vec{a}-\vec{b}) \cdot(\vec{a}+2 \vec{b})<0\)且\(k \vec{a}-\vec{b}\)与 \(\vec{a}+2 \vec{b}\)不共线,
则有 \((k \vec{a}-\vec{b}) \cdot(\vec{a}+2 \vec{b})=k \vec{a}^2-2 \vec{b}^2+(2 k-1) \vec{a} \cdot \vec{b}=-k-7<0\),解可得\(k>-7\),
若\(k \vec{a}-\vec{b}\)与 \(\vec{a}+2 \vec{b}\)共线,设\(k \vec{a}-\vec{b}=t(\vec{a}+2 \vec{b})\),则有 \(\left\{\begin{array}{l}
k=t \\
-1=2 t
\end{array}\right.\),则有 \(k=-\dfrac{1}{2}\),
若\(k \vec{a}-\vec{b}\)与 \(\vec{a}+2 \vec{b}\)不共线,必有 \(k \neq-\dfrac{1}{2}\),
故\(k\)的取值范围为\(\left(-7,-\dfrac{1}{2}\right) \cup\left(-\dfrac{1}{2},+\infty\right)\).
点拨 \(\vec{a}\)与\(\vec{b}\)的夹角为钝角等价于\(\vec{a} \cdot \vec{b}<0\)且\(\vec{a}\)与\(\vec{b}\)不共线.
【巩固练习】
1.若向量\(\vec{a}\),\(\vec{b}\)满足 \(|\vec{a}|=|\vec{b}|=1\),且 \((\vec{a}+\vec{b}) \cdot \vec{b}=\dfrac{3}{2}\),向量\(\vec{a}\),\(\vec{b}\)的夹角为\(\underline{\quad \quad}\) .
2.已知向量\(\vec{a}\),\(\vec{b}\)是两个非零向量,且\(|\vec{a}|=|\vec{b}|=|\vec{a}-\vec{b}|,\),则\(\vec{a}\)与\(\vec{b}\)的夹角为\(\underline{\quad \quad}\) .
3.已知向量\(\vec{a}\),\(\vec{b}\)满足 \(|\vec{a}|=1\), \((\vec{a}-\vec{b}) \perp(3 \vec{a}-\vec{b})\),则\(\vec{a}\)与\(\vec{b}\)的夹角的最大值为\(\underline{\quad \quad}\) .
参考答案
- 答案 \(60^∘\)
解析 由 \((\vec{a}+\vec{b}) \cdot \vec{b}=\dfrac{3}{2}\),则 \(\vec{a} \cdot \vec{b}+\vec{b}^2=\dfrac{3}{2}\),
又 \(|\vec{a}|=|\vec{b}|=1\),则 \(\vec{a} \cdot \vec{b}=\dfrac{1}{2}\),
则 \(\cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{1}{2}\),
又 \(\langle\vec{a}, \vec{b}\rangle \in\left[0^{\circ}, 180^{\circ}\right]\),则 \(\langle\vec{a}, \vec{b}\rangle=60^{\circ}\). - 答案 \(\dfrac{\pi}{3}\)
解析 设\(<\vec{a}, \vec{b}>=\theta\),且设 \(|\vec{a}|=|\vec{b}|=|\vec{a}-\vec{b}|=1\),
所以 \((\vec{a}-\vec{b})^2=\vec{a}^2-2 \vec{a} \cdot \vec{b}+\vec{b}^2=1\),
即 \(1-2 \vec{a} \cdot \vec{b}+1=1\),得 \(\vec{a} \cdot \vec{b}=\dfrac{1}{2}\),
所以 \(\cos \theta=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{1}{2}\),
\(\because \theta \in[0, \pi ]\), \(\therefore \theta=\dfrac{\pi}{3}\). - 答案 \(30°\)
解析 \(\because|\vec{a}|=1\), \((\vec{a}-\vec{b}) \perp(3 \vec{a}-\vec{b})\),
\(\therefore(\vec{a}-\vec{b}) \cdot(3 \vec{a}-\vec{b})=3 \vec{a}^2+\vec{b}^2-4 \vec{a} \cdot \vec{b}=3+\vec{b}^2-4 \vec{a} \cdot \vec{b}=0\),
\(\therefore \vec{a} \cdot \vec{b}=\dfrac{|\vec{b}|^2+3}{4}\),
\(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{|\vec{b}|^2+3}{4|\vec{b}|}=\dfrac{|\vec{b}|+\dfrac{3}{|\vec{b}|}}{4} \geq \dfrac{\sqrt{3}}{2}\),且 \(0^{\circ} \leq\langle\vec{a}, \vec{b}\rangle \leq 180^{\circ}\),
\(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\sqrt{3}}{2}\)时,\(\vec{a}\),\(\vec{b}\)的夹角最大为\(30°\).
【题型3】数量积的综合运用
【典题1】 已知\(O\)是\(△ABC\)所在平面内一点,向量 \(\overrightarrow{O P_1}\), \(\overrightarrow{O P_2}\), \(\overrightarrow{O P_3}\)满足条件\(\overrightarrow{O P_1}+\overrightarrow{O P_2}+\overrightarrow{O P_3}=\overrightarrow{0}\),且 \(\left|\overrightarrow{O P_1}\right|=\left|\overrightarrow{O P_2}\right|=\left|\overrightarrow{O P_3}\right|=1\),则\(\triangle P_1 P_2 P_3\)是( )
A.等腰三角形 \(\qquad \qquad \qquad\) B.直角三角形 \(\qquad \qquad \qquad\) C.等腰直角三角形 \(\qquad \qquad \qquad\) D.等边三角形
解析 \(∵\overrightarrow{O P_1}+\overrightarrow{O P_2}+\overrightarrow{O P_3}=\overrightarrow{0}\), \(\therefore \overrightarrow{O P_1}+\overrightarrow{O P_2}=-\overrightarrow{O P_3}\),
\(\therefore\left(\overrightarrow{O P}_1+\overrightarrow{O P}_2\right)^2=\left(-\overrightarrow{O P}_3\right)^2\), \(\therefore \overrightarrow{O P}_1^2+2 \overrightarrow{O P}_1 \cdot \overrightarrow{O P}_2+{\overrightarrow{O P_2}}^2={\overrightarrow{O P_3}}^2\),
\(\because\left|\overrightarrow{O P_1}\right|=\left|\overrightarrow{O P_2}\right|=\left|\overrightarrow{O P_3}\right|=1\),
\(\therefore\left|\overrightarrow{O P_1}\right|^2+\left|\overrightarrow{O P_2}\right|^2=\left|\overrightarrow{O P_3}\right|^2=1\), \(\therefore \overrightarrow{O P_1} \cdot \overrightarrow{O P_2}=-\dfrac{1}{2}\),
\(\therefore\left|\overrightarrow{P_1 P_2}\right|^2=\left|\overrightarrow{O P_2}-\overrightarrow{O P_1}\right|^2={\overrightarrow{O P_2}}^2-\overrightarrow{O P}_1 \cdot \overrightarrow{O P}_2+\overrightarrow{O P}_1^2=3\),
\(\therefore\left|\overrightarrow{P_1 P_2}\right|=\sqrt{3}\),
同理 \(\left|\overrightarrow{P_1 P_3}\right|=\left|\overrightarrow{P_2 P_3}\right|=\sqrt{3}\), \(\therefore \Delta P_1 P_2 P_3\)是等边三角形,
故选:\(D\).
【典题2】 已知\(P\)为\(△ABC\)所在平面内的一点, \(\overrightarrow{B P}=2 \overrightarrow{P C}\), \(|\overrightarrow{A P}|=4\),若点\(Q\)在线段\(AP\)上运动,则 \(\overrightarrow{Q A} \cdot(\overrightarrow{Q B}+2 \overrightarrow{Q C})\)的最小值为( )
A. \(-9 \sqrt{2}\) \(\qquad \qquad \qquad \qquad\) B.\(-12\) \(\qquad \qquad \qquad \qquad\) C. \(-3 \sqrt{2}\) \(\qquad \qquad \qquad \qquad\) D.\(-4\)
解析 由题意,画图如下,
根据题意及图,可知 \(\overrightarrow{B P}=\overrightarrow{Q P}-\overrightarrow{Q B}\), \(\overrightarrow{P C}=\overrightarrow{Q C}-\overrightarrow{Q P}\),
\(\because \overrightarrow{B P}=2 \overrightarrow{P C}\), \(\therefore \overrightarrow{Q P}-\overrightarrow{Q B}=2(\overrightarrow{Q C}-\overrightarrow{Q P})\)
整理,得 \(\overrightarrow{Q B}+2 \overrightarrow{Q C}=3 \overrightarrow{Q P}\),
则 \(\overrightarrow{Q A} \cdot(\overrightarrow{Q B}+2 \overrightarrow{Q C})=\overrightarrow{Q A} \cdot 3 \overrightarrow{Q P}=-3| \overrightarrow{Q A}|\cdot| \overrightarrow{Q P}|=-3| \overrightarrow{Q A} \mid \cdot(4-|\overrightarrow{Q A}|)\)
\(=3\left(|\overrightarrow{Q A}|^2-4|\overrightarrow{Q A}|\right)\),
设\(|\overrightarrow{Q A}|=m\),很明显 \(m \in[0,4]\),
故 \(\overrightarrow{Q A} \cdot(\overrightarrow{Q B}+2 \overrightarrow{Q C})=3\left(|\overrightarrow{Q A}|^2-4|\overrightarrow{Q A}|\right)=3\left(m^2-4 m\right)=3(m-2)^2-12\),
根据二次函数的性质,可知:
当\(m=2\)时, \(\overrightarrow{Q A} \cdot(\overrightarrow{Q B}+2 \overrightarrow{Q C})\)取得最小值为\(-12\).
故选:\(B\).
【巩固练习】
1.在四边形\(ABCD\)中,若 \(\overrightarrow{A B}=\overrightarrow{D C}\),且 \(|\overrightarrow{A B}-\overrightarrow{A D}|=|\overrightarrow{A B}+\overrightarrow{A D}|\),则该四边形一定是 ( )
A.正方形 \(\qquad \qquad \qquad \qquad\) B.菱形 \(\qquad \qquad \qquad \qquad\) C.矩形 \(\qquad \qquad \qquad \qquad\) D.等腰梯形
2.\(△ABC\)所在平面内,满足 \(\overrightarrow{M A}+\overrightarrow{M B}+\overrightarrow{M C}=\overrightarrow{0}\), \(|\overrightarrow{N A}|=|\overrightarrow{N B}|=|\overrightarrow{N C}|\)且 \(\overrightarrow{P A} \cdot \overrightarrow{P B}=\overrightarrow{P B} \cdot \overrightarrow{P C}=\overrightarrow{P C} \cdot \overrightarrow{P A}\),则\(M\),\(N\),\(P\)依次是\(△ABC\)的( )
A.重心,外心,内心 \(\qquad \qquad \qquad \qquad\) B.重心,外心,垂心
C.外心,重心,内心 \(\qquad \qquad \qquad \qquad\) D.外心,重心,垂心
3.如图,已知等腰梯形\(ABCD\)中,\(AB=2DC=4\), \(A D=B C=\sqrt{3}\),\(E\)是\(DC\)的中点,\(F\)是线段\(BC\)上的动点,则 \(\overrightarrow{E F} \cdot \overrightarrow{B F}\)最小值是\(\underline{\quad \quad}\).
4.已知\(△ABC\)中,\(∠C\)是直角,\(CA=CB\),点\(D\)是\(CB\)的中点,\(E\)为\(AB\)上一点.
(1)设\(\overrightarrow{C A}=\vec{a}\), \(\overrightarrow{C D}=\vec{b}\),当 \(\overrightarrow{A E}=\dfrac{1}{2} \overrightarrow{A B}\),请用\(\vec{a}\), \(\vec{b}\)来表示 \(\overrightarrow{A B}\),\(\overrightarrow{C E}\).
(2)当\(\overrightarrow{A E}=2 \overrightarrow{E B}\)时,求证:\(AD⊥CE\).
参考答案
-
答案 \(C\)
解析 根据题意,在四边形\(ABCD\)中,若 \(\overrightarrow{A B}=\overrightarrow{D C}\),
即\(AB//DC\)且\(AB=DC\),四边形为平行四边形,
又由 \(|\overrightarrow{A B}-\overrightarrow{A D}|=|\overrightarrow{A B}+\overrightarrow{A D}|\),
则有\(|\overrightarrow{A B}-\overrightarrow{A D}|^2=|\overrightarrow{A B}+\overrightarrow{A D}|^2\),变形可得 \(\overrightarrow{A B} \cdot \overrightarrow{A D}=0\),
则有 \(\overrightarrow{A B} \perp \overrightarrow{A D}\),即\(AB\)与\(AD\)垂直,则该四边形一定是矩形,
故选:\(C\). -
答案 \(B\)
解析 \(\because \overrightarrow{M A}+\overrightarrow{M B}+\overrightarrow{M C}=\overrightarrow{0}\), \(\therefore \overrightarrow{M A}+\overrightarrow{M B}=-\overrightarrow{M C}\),
设\(AB\)的中点\(D\),则 \(\overrightarrow{M A}+\overrightarrow{M B}=2 \overrightarrow{M D}\),
\(∴C\),\(M\),\(D\)三点共线,即\(M\)为\(△ABC\)的中线\(CD\)上的点,且\(MC=2MD\).
\(∴M\)为\(△ABC\)的重心.
\(\because|\overrightarrow{N A}|=|\overrightarrow{N B}|=|\overrightarrow{N C}|\),
\(\therefore|N A|=|N B|=|N C|\),\(∴N\)为\(△ABC\)的外心;
\(\because \overrightarrow{P A} \cdot \overrightarrow{P B}=\overrightarrow{P B} \cdot \overrightarrow{P C}\), \(\therefore \overrightarrow{P B} \cdot(\overrightarrow{P A}-\overrightarrow{P C})=0\),
即 \(\overrightarrow{P B} \cdot \overrightarrow{C A}=0\),\(∴PB⊥AC\),
同理可得:\(PA⊥BC\),\(PC⊥AB\),
\(∴P\)为\(△ABC\)的垂心;
故选:\(B\).
-
答案 \(-\dfrac{4}{3}\)
解析 由等腰梯形的知识可知 \(\cos B=\dfrac{\sqrt{3}}{3}\),
设\(BF=x\),则 \(C F=\sqrt{3}-x\),
\(\therefore \overrightarrow{E F} \cdot \overrightarrow{B F}=(\overrightarrow{E C}+\overrightarrow{C F}) \cdot \overrightarrow{B F}=\overrightarrow{E C} \cdot \overrightarrow{B F}+\overrightarrow{C F} \cdot \overrightarrow{B F}\)
\(=1 \cdot x\left(-\dfrac{\sqrt{3}}{3}\right)+(\sqrt{3}-x) \cdot x \cdot(-1)=x^2-\dfrac{4}{3} \sqrt{3} x\),
\(\because 0 \leq x \leq \sqrt{3}\),
\(∴\)当\(x=\dfrac{2}{3} \sqrt{3}\)时, \(\overrightarrow{E F} \cdot \overrightarrow{B F}\)取得最小值,最小值为 \(\left(\dfrac{2}{3} \sqrt{3}\right)^2-\dfrac{2}{3} \sqrt{3} \times \dfrac{4}{3} \sqrt{3}=-\dfrac{4}{3}\).
故答案为:\(-\dfrac{4}{3}\). -
答案 (1) \(\overrightarrow{C E}=\dfrac{1}{2} \vec{a}+\vec{b}\),(2)略.
解析 (1) \(\because \overrightarrow{C A}=\vec{a}\), \(\overrightarrow{C D}=\vec{b}\),点\(D\)是\(CB\)的中点, \(\therefore \overrightarrow{C B}=2 \vec{b}\),
\(\therefore \overrightarrow{A B}=\overrightarrow{C B}-\overrightarrow{C A}=2 \vec{b}-\vec{a}\),
\(\because \overrightarrow{A E}=\dfrac{1}{2} \overrightarrow{A B}\), \(\therefore \overrightarrow{C E}=\overrightarrow{C A}+\overrightarrow{A E}=\overrightarrow{C A}+\dfrac{1}{2} \overrightarrow{A B}=\vec{a}+\dfrac{1}{2}(2 \vec{b}-\vec{a})=\dfrac{1}{2} \vec{a}+\vec{b}\).
证明:(2)设 \(\overrightarrow{C A}=\vec{a}\), \(\overrightarrow{C D}=\vec{b}\),则 \(\overrightarrow{C B}=2 \vec{b}\),
\(\because \overrightarrow{A E}=2 \overrightarrow{E B}\),
\(\therefore \overrightarrow{C E}=\overrightarrow{C A}+\overrightarrow{A E}=\overrightarrow{C A}+\dfrac{2}{3} \overrightarrow{A B}=\overrightarrow{C A}+\dfrac{2}{3}(\overrightarrow{C B}-\overrightarrow{C A})=\dfrac{1}{3}(\vec{a}+4 \vec{b})\),
\(\overrightarrow{A D}=\overrightarrow{A C}+\overrightarrow{C D}=-\vec{a}+\vec{b}\),
\(∵△ABC\)中,\(∠C\)是直角,\(CA=CB\),
\(\therefore|\vec{a}|=2 \vec{b} \mid\), \(\vec{a} \cdot \vec{b}=0\),
\(\therefore \overrightarrow{A D} \cdot \overrightarrow{C E}=\dfrac{1}{3}(\vec{a}+4 \vec{b}) \cdot(-\vec{a}+\vec{b})=\dfrac{1}{3}\left(4 \vec{b}^2-\vec{a}^2\right)=0\),
\(∴AD⊥CE\).
分层练习
【A组---基础题】
1.在四边形\(ABCD\)中,若\(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{A D}\)且 \(\overrightarrow{A C} \cdot(\overrightarrow{A B}-\overrightarrow{A D})=0\),则( )
A.四边形\(ABCD\)是矩形 \(\qquad \qquad \qquad \qquad\) B.四边形\(ABCD\)是菱形
C.四边形\(ABCD\)是正方形 \(\qquad \qquad \qquad \qquad\) D.四边形\(ABCD\)是平行四边形
2.已知向量\(\vec{a}\),\(\vec{b}\)满足 \(|\vec{a}|=1\), \(|\vec{b}|=2\), \(|\vec{a}+2 \vec{b}|=\sqrt{21}\),那么向量\(\vec{a}\)与\(\vec{b}\)的夹角为( )
A. \(\dfrac{\pi}{6}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{\pi}{4}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{\pi}{3}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{2 \pi}{3}\)
3.如图,在梯形\(ABCD\)中,\(AB∥CD\),\(AB=4\),\(AD=3\),\(CD=2\), \(\overrightarrow{A M}=2 \overrightarrow{M D}\), \(\overrightarrow{A C} \cdot \overrightarrow{B M}=-3\),则 \(\overrightarrow{A B} \cdot \overrightarrow{A D}=\)( )
A. \(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) C . \(-\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) D. \(-\dfrac{1}{2}\)
4.设\(θ\)为两个非零向量\(\vec{a}\),\(\vec{b}\)的夹角,已知当实数\(t\)变化时 \(|\vec{a}+t \vec{b}|\)的最小值为\(2\),则( )
A.若\(θ\)确定,则\(|\vec{a}|\)唯一确定 \(\qquad \qquad \qquad \qquad\) B.若\(θ\)确定,则\(|\vec{b}|\)唯一确定
C.若\(|\vec{a}|\)确定,则\(θ\)唯一确定 \(\qquad \qquad \qquad \qquad\) D.若\(|\vec{b}|\)确定,则\(θ\)唯一确定
5.(多选)设\(\vec{a}\),\(\vec{b}\)是两个非零向量.则下列命题为假命题的是( )
A.若\(|\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|\),则\(\vec{a} \perp \vec{b}\)
B.若\(\vec{a} \perp \vec{b}\),则\(|\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|\)
C.若\(|\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|\),则存在实数\(λ\),使得 \(\vec{b}=\lambda \vec{a}\)
D.若存在实数\(λ\),使得\(\vec{b}=\lambda \vec{a}\),则\(|\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|\)
6.菱形\(ABCD\)的边长为\(2\),且\(∠DAB=60^∘\),则\(\overrightarrow{A B} \cdot \overrightarrow{B C}=\) \(\underline{\quad \quad}\) .
7.已知\(|\vec{a}|=2\), \(|\vec{b}|=1\),\(\vec{a}\)与\(\vec{b}\)的夹角为\(\dfrac{\pi}{3}\),那么\(|\vec{a}-\vec{b}|=\) \(\underline{\quad \quad}\) .
8.已知正方形\(ABCD\)的边长为\(1\).当每个 \(\lambda_i(i=1,2,3,4,5,6)\)取遍\(±1\)时, \(\left|\lambda_1 \overrightarrow{A B}+\lambda_2 \overrightarrow{B C}+\lambda_3 \overrightarrow{C D}+\lambda_4 \overrightarrow{D A}+\lambda_5 \overrightarrow{A C}+\lambda_6 \overrightarrow{B D}\right|\)的最小值是\(\underline{\quad \quad}\),最大值是\(\underline{\quad \quad}\).
9.如图,在矩形\(ABCD\)中,\(AB=2BC=2\),动点\(M\)在以点\(C\)为圆心且与\(BD\)相切的圆上,则\(\overrightarrow{A M} \cdot \overrightarrow{B D}\)的最大值是\(\underline{\quad \quad}\).
10.已知非零向量\(\vec{a}\),\(\vec{b}\)满足\(\vec{a} \cdot \vec{b}=3\), \(|\vec{b}|=2\),且 \(\vec{a} \cdot(\vec{a}-3 \vec{b})=0\).
(1)求\(|\vec{a} |\);
(2)求向量\(\vec{a}\)与\(\vec{b}\)的夹角\(θ\);
(3)求\(|\vec{a}+\vec{b}|\)的值.
11.如图,在正方形\(ABCD\)中,点\(E\)是\(BC\)边上中点,点\(F\)在边\(CD\)上且满足\(\overrightarrow{C F}=x \overrightarrow{C D}\) .
(1)若\(x=\dfrac{1}{3}\),设\(\overrightarrow{E F}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A D}\),求\(λ+μ\)的值;
(2)若\(AB=2\),当 \(\overrightarrow{A E} \cdot \overrightarrow{B F}=1\)时,求\(x\)的值.
参考答案
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答案 \(B\)
解析 四边形\(ABCD\)中,若\(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{A D}\),则四边形\(ABCD\)是平行四边形;
又\(\overrightarrow{A C} \cdot(\overrightarrow{A B}-\overrightarrow{A D})=0\),即 \(\overrightarrow{A C} \cdot \overrightarrow{D B}=0\),
所以\(\overrightarrow{A C} \perp \overrightarrow{D B}\),平行四边形\(ABCD\)是菱形.
故选:\(B\). -
答案 \(C\)
解析 \(∵|\vec{a}|=1\), \(|\vec{b}|=2\), \(|\vec{a}+2 \vec{b}|=\sqrt{21}\),,
\(\therefore(\vec{a}+2 \vec{b})^2=\vec{a}^2+4 \overrightarrow{b^2}+4 \vec{a} \cdot \vec{b}=1+16+4 \vec{a} \cdot \vec{b}=21\),
\(\therefore \vec{a} \cdot \vec{b}=1\),
\(\therefore \cos \langle\vec{a}, \vec{b}\rangle=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{1}{2}\),且 \(0 \leq<\vec{a}, \vec{b}>\leq \pi\),
\(∴\vec{a}\)与\(\vec{b}\)的夹角为\(\dfrac{\pi}{3}\).
故选:\(C\). -
答案 \(B\)
解析 \(∵\)在梯形\(ABCD\)中,\(AB∥CD\),\(AB=4\),\(AD=3\),\(CD=2\), \(\overrightarrow{A M}=2 \overrightarrow{M D}\),
\(\therefore \overrightarrow{A C} \cdot \overrightarrow{B M}=(\overrightarrow{A D}+\overrightarrow{D C}) \cdot(\overrightarrow{B A}+\overrightarrow{A M})=\left(\overrightarrow{A D}+\dfrac{1}{2} \overrightarrow{A B}\right) \cdot\left(-\overrightarrow{A B}+\dfrac{2}{3} \overrightarrow{A D}\right)\)
\(=\dfrac{2}{3} \overrightarrow{A D^2}-\dfrac{1}{2} \overrightarrow{A B^2}-\dfrac{2}{3} \overrightarrow{A D} \cdot \overrightarrow{A B}=-3\),
\(\therefore \dfrac{2}{3} \times 3^2-\dfrac{1}{2} \times 4^2-\dfrac{2}{3} \overrightarrow{A B} \cdot \overrightarrow{A D}=-3\),
则 \(\overrightarrow{A B} \cdot \overrightarrow{A D}=\dfrac{3}{2}\);
故选:\(B\). -
答案 \(A\)
解析 令 \(f(t)=|\vec{a}+t \vec{b}|^2=\vec{a}^2+2 t \cdot \vec{a} \cdot \vec{b}+t^2 \cdot \vec{b}^2\)
\(\therefore \Delta=4(\vec{a} \cdot \vec{b})^2-4 \vec{a}^2 \cdot \vec{b}^2=4 \vec{a}^2 \cdot \vec{b}^2(\cos \theta-1) \leq 0\)恒成立,
当且仅当\(t=-\dfrac{2 \vec{a} \cdot \vec{b}}{2 \times \overrightarrow{b^2}}=-\dfrac{|\vec{a}|}{|\vec{b}|} \cos \theta\)时,\(f(t)\)取得最小值\(2\),
\(\therefore\left(-\dfrac{|\vec{a}|}{|\vec{b}|} \cos \theta\right)^2+\vec{b}^2+2\left(-\dfrac{|\vec{a}|}{|\vec{b}|} \cos \theta\right) \cdot \vec{a} \cdot \vec{b}+\vec{a}^2=2\),化简 \(\vec{a}^2 \sin ^2 \theta=2\).
\(∴θ\)确定,则\(|\vec{a}|\)唯一确定
故选:\(A\). -
答案 \(ABD\)
解析 对于\(A\),若\(|\vec{a}+\vec{b}|=|\vec{a}|-|\vec{b}|\),则 \(|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\),
得\(\vec{a} \cdot \vec{b}=-|\vec{a}||\vec{b}| \neq 0\),\(\vec{a}\)与\(\vec{b}\)不垂直,所以\(A\)不正确;
对于\(B\),由\(A\)解析可知, \(|\vec{a}+\vec{b}| \neq|\vec{a}|-|\vec{b}|\),所以\(B\)不正确;
对于\(C\),若 \(|\vec{a}+\vec{b}| =|\vec{a}|-|\vec{b}|\),则\(|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\),
得\(\vec{a} \cdot \vec{b}=-|\vec{a}||\vec{b}|\),则 \(\cos \theta=-1\),则\(\vec{a}\)与\(\vec{b}\)反向,
因此存在实数\(λ\),使得 \(\vec{b}=\lambda \vec{a}\),所以\(C\)正确.
对于\(D\),若存在实数\(λ\),则 \(\vec{a} \cdot \vec{b}=\lambda|\vec{a}|^2\), \(-|\vec{a}||\vec{b}|=-|\lambda||\vec{a}|^2\),由于\(λ\)不能等于\(0\),
因此\(|\vec{a} \cdot \vec{b} \neq-| \vec{a}|| \vec{b} \mid\),则\(|\vec{a}+\vec{b}| \neq|\vec{a}|-|\vec{b}|\),所以\(D\)不正确.
故选:\(ABD\). -
答案 \(2\)
解析 \(\overrightarrow{A B} \cdot \overrightarrow{B C}=|\overrightarrow{A B} \| \overrightarrow{A C}| \cos <\overrightarrow{A B}, \overrightarrow{A C}>=2 \times 2 \times \cos 60^{\circ}=4 \times \dfrac{1}{2}=2\). -
答案 \(\sqrt{3}\)
解析 由题意,可得\(\vec{a} \cdot \vec{b}=2 \times 1 \times \cos \dfrac{\pi}{3}=1\),
则 \(|\vec{a}-\vec{b}|=\sqrt{(\vec{a}-\vec{b})^2}=\sqrt{\vec{a}^2-2 \vec{a} \cdot \vec{b}+\vec{b}^2}=\sqrt{4-2+1}=\sqrt{3}\). -
答案 \(0\),\(2 \sqrt{5}\).
解析 正方形\(ABCD\)的边长为\(1\),
可得 \(\overrightarrow{A B}+\overrightarrow{A D}=\overrightarrow{A C}\),\(\overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B}\),\(\overrightarrow{A B} \cdot \overrightarrow{A D}=0\)
\(\left|\lambda_1 \overrightarrow{A B}+\lambda_2 \overrightarrow{B C}+\lambda_3 \overrightarrow{C D}+\lambda_4 \overrightarrow{D A}+\lambda_5 \overrightarrow{A C}+\lambda_6 \overrightarrow{B D}\right|\)
\(=\left|\lambda_1 \overrightarrow{A B}+\lambda_2 \overrightarrow{A D}-\lambda_3 \overrightarrow{A B}-\lambda_4 \overrightarrow{A D}+\lambda_5 \overrightarrow{A B}+\lambda_5 \overrightarrow{A D}+\lambda_6 \overrightarrow{A D}-\lambda_6 \overrightarrow{A B}\right|\)
\(=\left|\left(\lambda_1-\lambda_3+\lambda_5-\lambda_6\right) \overrightarrow{A B}+\left(\lambda_2-\lambda_4+\lambda_5+\lambda_6\right) \overrightarrow{A D}\right|\)
\(=\sqrt{\left(\lambda_1-\lambda_3+\lambda_5-\lambda_6\right)^2+\left(\lambda_2-\lambda_4+\lambda_5+\lambda_6\right)^2}\),
由于\(λ_i (i=1,2,3,4,5,6)\)取遍\(±1\),
可得 \(\lambda_1-\lambda_3+\lambda_5-\lambda_6=0\), \(\lambda_2-\lambda_4+\lambda_5+\lambda_6=0\),
可取 \(\lambda_5=\lambda_6=1\), \(\lambda_1=\lambda_3=1\), \(\lambda_2=-1\), \(\lambda_4=1\),
可得所求最小值为\(0\);
由 \(\lambda_1-\lambda_3+\lambda_5-\lambda_6\), \(\lambda_2-\lambda_4+\lambda_5+\lambda_6\)的最大值为\(4\),
可取 \(\lambda_2=1\), \(\lambda_4=-1\), \(\lambda_5=\lambda_6=1\), \(\lambda_1=1\), \(\lambda_3=-1\),
可得所求最大值为\(2 \sqrt{5}\).
故答案为:\(0\),\(2 \sqrt{5}\).
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答案 \(2\)
解析 因为在矩形\(ABCD\)中,\(AB=2BC=2\),动点\(M\)在以点\(C\)为圆心且与\(BD\)相切的圆上,
故 \(|\overrightarrow{A C}|=|\overrightarrow{B D}|=\sqrt{5}\),
设\(C\)到\(BD\)的距离为\(d\),则有 \(d=\dfrac{1 \times 2}{\sqrt{5}}=\dfrac{2 \sqrt{5}}{5}\),
故 \(\overrightarrow{A M} \cdot \overrightarrow{B D}=(\overrightarrow{A C}+\overrightarrow{C M}) \cdot \overrightarrow{B D}=\overrightarrow{A C} \cdot \overrightarrow{B D}+\overrightarrow{C M} \cdot \overrightarrow{B D}\),
其中 \(\overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B}+\overrightarrow{B C}) \cdot(\overrightarrow{B C}+\overrightarrow{C D})=-3\), \(\overrightarrow{C M} \cdot \overrightarrow{B D} \leq|\overrightarrow{C M}| \cdot|\overrightarrow{B D}|=2\),
当且仅当\(\overrightarrow{C M}\)与\(\overrightarrow{BD}\)同向时,等号成立. -
答案 (1) \(3\);(2) \(60^∘\);(3) \(\sqrt{19}\).
解析 (1)因为\(\vec{a} \cdot(\vec{a}-3 \vec{b})=0\),所以 \(\vec{a}^2-3 \vec{a} \cdot \vec{b}=0\), \(|\vec{a}|^2-3 \vec{a} \cdot \vec{b}=0\),
所以\(|\vec{a}|^2=9\),所以 \(|\vec{a}|=3\);
(2) \(\cos \theta=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\dfrac{3}{2 \times 3}=\dfrac{1}{2}\),又因为 \(0 \leqslant \theta \leqslant \pi\),所以 \(\theta=60^{\circ}\);
(3) \(|\vec{a}+\vec{b}|=\sqrt{(\vec{a}+\vec{b})^2}=\sqrt{\vec{a}^2+2 \vec{a} \cdot \vec{b}+\vec{b}^2}=\sqrt{|\vec{a}|^2+2 \vec{a} \cdot \vec{b}+|\vec{b}|^2}\)
\(=\sqrt{4+6+9}=\sqrt{19}\). -
答案 (1) \(\dfrac{1}{6}\);(2) \(\dfrac{1}{4}\).
解析 (1)由题意可得, \(\overrightarrow{E F}=\overrightarrow{E C}+\overrightarrow{C F}=\dfrac{1}{2} \overrightarrow{B C}+\dfrac{1}{3} \overrightarrow{C D}=\dfrac{1}{2} \overrightarrow{A D}-\dfrac{1}{3} \overrightarrow{A B}\),
\(\because \overrightarrow{E F}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A D}\), \(\therefore \mu=\dfrac{1}{2}\), \(\lambda=-\dfrac{1}{3}\), \(\therefore \lambda+\mu=\dfrac{1}{6}\);
(2)由题意可得, \(\overrightarrow{A E}=\overrightarrow{A B}+\overrightarrow{B E}=\overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}\),
\(\overrightarrow{B F}=\overrightarrow{C F}-\overrightarrow{C B}=x \overrightarrow{C D}-\overrightarrow{C B}=\overrightarrow{A D}-x \overrightarrow{A B}\),且\(\overrightarrow{A B} \cdot \overrightarrow{A D}=0\),
故 \(\overrightarrow{A E} \cdot \overrightarrow{B F}=\left(\overrightarrow{A B}+\dfrac{1}{2} \overrightarrow{A D}\right) \cdot(\overrightarrow{A D}-x \overrightarrow{A B})=\dfrac{1}{2} \overrightarrow{A D}^2-x \overrightarrow{A B}^2=2-4 x=1\),
解得 \(x=\dfrac{1}{4}\).
【B组---提高题】
1.已知向量\(\vec{a}\),\(\vec{b}\)满足 \(|\vec{a}|=|\vec{b}|=1\),且对任意\(t∈R\)都有 \(|\vec{a}+\vec{b}| \leq|\vec{a}-t \vec{b}|\),则\(\vec{a}\)与\(\vec{b}\)的夹角为( )
A.\(\dfrac{ \pi}{3}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{ \pi}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{2 \pi}{3}\) \(\qquad \qquad \qquad \qquad\) D.\(\pi\)
2.在\(△ABC\)中,\(AB=8\),\(AC=6\),\(∠A=60°\),\(M\)为\(△ABC\)的外心,若 \(\overrightarrow{A M}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}\),\(λ\),\(μ∈R\),则\(4λ+3μ=\)( )
A. \(\dfrac{3}{4}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{5}{3}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{7}{3}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{8}{3}\)
3.如图,在平面四边形\(ABCD\)中,已知\(AD=6\),\(BC=8\),\(E\),\(F\)为\(AB\),\(CD\)的中点,\(P\),\(Q\)为对角线\(AC\),\(BD\)的中点,则 \(\overrightarrow{P Q} \cdot \overrightarrow{E F}\)值为\(\underline{\quad \quad}\).
参考答案
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答案 \(D\)
解析 由\(|\vec{a}+\vec{b}| \leq|\vec{a}-t \vec{b}|\),可得 \((\bar{a}+\bar{b})^2 \leq(\bar{a}-t \bar{b})^2\)
即 \(\bar{a}^2+2 \bar{a} \cdot \bar{b}+\bar{b}^2 \leq \bar{a}^2-2 \bar{a} \cdot \bar{b} t+t^2 \bar{b}^2\)
设\(\vec{a}\)与\(\vec{b}\)的夹角为\(θ\),可得 \(t^2-2 \cos \theta \cdot t-2 \cos \theta-1 \geq 0\).
对任意\(t∈R\)都成立,
\(\therefore \triangle=4 \cos ^2 \theta+4(2 \cos \theta+1) \leq 0\) ,即 \(\cos \theta \leq-1\),
可得\(θ=π\).
故选:\(D\). -
答案 \(C\)
解析 如图,取线段\(AB\)的中点\(E\),连接\(ME\),则 \(\overrightarrow{A M}=\overrightarrow{A E}+\overrightarrow{E M}\),且\(EM⊥AB\),
\(\therefore \overrightarrow{A M} \cdot \overrightarrow{A B}=(\overrightarrow{A E}+\overrightarrow{E M}) \cdot \overrightarrow{A B}=\overrightarrow{A E} \cdot \overrightarrow{A B}+\overrightarrow{E M} \cdot \overrightarrow{A B}=\dfrac{1}{2} \overrightarrow{A B}^2\),
同理可得, \(\overrightarrow{A M} \cdot \overrightarrow{A C}=\dfrac{1}{2} \overrightarrow{A C}^2\),
又 \(\overrightarrow{A B} \cdot \overrightarrow{A C}=8 \times 6 \times \cos 60^{\circ}=24\),
由 \(\left\{\begin{array}{l} \overrightarrow{A M} \cdot \overrightarrow{A B}=\dfrac{1}{2} \overrightarrow{A B}^2 \\ \overrightarrow{A M} \cdot \overrightarrow{A C}=\dfrac{1}{2} \overrightarrow{A C}^2 \end{array}\right.\),可得 \(\left\{\begin{array}{l} (\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}) \cdot \overrightarrow{A B}=32 \\ (\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}) \cdot \overrightarrow{A C}=18 \end{array}\right.\),
即 \(\left\{\begin{array}{l} 64 \lambda+24 \mu=32 \\ 24 \lambda+36 \mu=18 \end{array}\right.\),解得 \(\left\{\begin{array}{l} \lambda=\dfrac{5}{12} \\ \mu=\dfrac{2}{9} \end{array}\right.\),
\(\therefore 4 \lambda+3 \mu=4 \times \dfrac{5}{12}+3 \times \dfrac{2}{9}=\dfrac{7}{3}\).
故选 \(C\). -
答案 \(-7\)
解析 如图,连接\(FP\),\(FQ\),\(EP\),\(EQ\),
\(∵E\),\(F\)为\(AB\),\(CD\)的中点,\(P\),\(Q\)为对角线\(AC\),\(BD\)的中点,
\(∴\)四边形\(EPFQ\)为平行四边形,
\(\therefore \overrightarrow{P Q}=\overrightarrow{P F}+\overrightarrow{P E}=\dfrac{1}{2}(\overrightarrow{A D}+\overrightarrow{B C}), \overrightarrow{E F}=\overrightarrow{P F}-\overrightarrow{P E}=\dfrac{1}{2}(\overrightarrow{A D}-\overrightarrow{B C})\), 且\(AD=6\),\(BC=8\),
\(\therefore \overrightarrow{P Q} \cdot \overrightarrow{E F}=\dfrac{1}{4}(\overrightarrow{A D}+\overrightarrow{B C}) \cdot(\overrightarrow{A D}-\overrightarrow{B C})=\dfrac{1}{4}\left(\overrightarrow{A D}^2-\overrightarrow{B C}^2\right)=\dfrac{1}{4} \times(36-64)=-7\).
【C组---拓展题】
1.已知 \(\vec{e}\)为单位向量,平面向量\(\vec{a}\),\(\vec{b}\)满足 \(|\vec{a}+\vec{e}|=|\vec{b}-\vec{e}|=1\), \(\vec{a} \cdot \vec{b}\)的取值范围是\(\underline{\quad \quad}\).
2.如图,在\(△ABC\)中,\(AB=2\),\(AC=1\),\(D\),\(E\)分别是直线\(AB\),\(AC\)上的点, \(\overrightarrow{A E}=2 \overrightarrow{B E}\), \(\overrightarrow{C D}=4 \overrightarrow{A C}\),且 \(\overrightarrow{B D} \cdot \overrightarrow{C E}=-2\),则\(∠BAC=\)\(\underline{\quad \quad}\).若\(P\)是线段\(DE\)上的一个动点,则 \(\overrightarrow{B P} \cdot \overrightarrow{C P}\)的最小值为\(\underline{\quad \quad}\).
3.在平面四边形\(ABCD\)中,\(AB=6\),\(CD=2\),\(AD=2\),\(P\)为\(BC\)中点,若 \(\overrightarrow{A C} \cdot\left(\overrightarrow{A D}-\dfrac{1}{3} \overrightarrow{A B}\right)=0\), \(\overrightarrow{A C} \cdot \overrightarrow{A B}=15\),则 \(\overrightarrow{A P} \cdot \overrightarrow{A D}=\)\(\underline{\quad \quad}\).
参考答案
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答案 \(\left[-4, \dfrac{1}{2}\right]\)
解析 取单位向量 \(\vec{e}=\overrightarrow{O C}\),以点\(C\)为圆心,\(1\)为半径作圆,在圆周上任取两点\(A\)、\(B\),
令 \(\vec{a}=\overrightarrow{A O}\), \(\vec{b}=\overrightarrow{O B}\),如图所示;设 \(|\vec{a}|=x\),则\(x∈[0,2]\);
作圆\(C\)的垂直于\(OA\)的切线分别交直线\(OA\)于\(B_1\),\(B_2\)两点,
易得 \(\vec{a} \cdot \vec{b} \geq \overrightarrow{A O} \cdot \overrightarrow{O B_1}=-x\left(1+\dfrac{x}{2}\right)=-\dfrac{x^2}{2}-x, x \in[0,2]\)
所以 \(\vec{a} \cdot \vec{b} \geq-4\),当且仅当\(x=2\)时等号成立;
\(\vec{a} \cdot \vec{b} \leq \overrightarrow{A O} \cdot \overrightarrow{O B}_2=x\left(1-\dfrac{x}{2}\right)=\dfrac{1}{2} x(2-x) \leq \dfrac{1}{2} \cdot\left(\dfrac{x+2-x}{2}\right)^2=\dfrac{1}{2}\),
当且仅当\(x=1\)时等号成立,即 \(\vec{a} \cdot \vec{b} \leq \dfrac{1}{2}\);
综上知, \(\vec{a} \cdot \vec{b}\)的取值范围是 \(\left[-4, \dfrac{1}{2}\right]\).
故答案为: \(\left[-4, \dfrac{1}{2}\right]\). -
答案 \(\dfrac{37}{7}\)
解析 \(\because \overrightarrow{A E}=2 \overrightarrow{B E}\), \(\overrightarrow{C D}=4 \overrightarrow{A C}\), \(\therefore \overrightarrow{A E}=2 \overrightarrow{A B}\), \(\overrightarrow{A D}=5 \overrightarrow{A C}\),
\(\because \overrightarrow{B D} \cdot \overrightarrow{C E}=-2\),
\(\therefore(\overrightarrow{A D}-\overrightarrow{A B})(\overrightarrow{A E}-\overrightarrow{A C})=(5 \overrightarrow{A C}-\overrightarrow{A B})(2 \overrightarrow{A B}-\overrightarrow{A C})=11 \overrightarrow{A B} \cdot \overrightarrow{A C}-5 \overrightarrow{A C^2}-2 \overrightarrow{A B^2}\)
\(=11 \times 2 \times 1 \times \cos \angle B A C-5 \times 1-2 \times 4=22 \cos \angle B A C-13=-2\),
解得 \(\cos \angle B A C=\dfrac{1}{2}\),
\(\because \angle B A C \in(0, \pi)\), \(\therefore \angle B A C=\dfrac{\pi}{3}\).
设 \(\overrightarrow{E P}=\lambda \overrightarrow{E D}, \lambda \in[0,1]\),
\(\therefore \overrightarrow{B P} \cdot \overrightarrow{C P}=(\overrightarrow{B E}+\overrightarrow{E P})(\overrightarrow{C D}+\overrightarrow{D P})\)\(=\left[\dfrac{1}{2} \overrightarrow{A E}+\lambda(\overrightarrow{A D}-\overrightarrow{A E})\right]\left[\dfrac{4}{5} \overrightarrow{A D}+(1-\lambda)(\overrightarrow{A E}-\overrightarrow{A D})\right]\)
\(=\left(\dfrac{1}{2}-\lambda\right)(1-\lambda) \overrightarrow{A E^2}+\lambda\left(\lambda-\dfrac{1}{5}\right) \overrightarrow{A D^2}+\left(\dfrac{17}{10} \lambda-\dfrac{1}{10}-2 \lambda^2\right) \overrightarrow{A D} \cdot \overrightarrow{A E}\)
\(=16\left(\dfrac{1}{2}-\lambda\right)(1-\lambda)+25 \lambda\left(\lambda-\dfrac{1}{5}\right)+\left(\dfrac{17}{10} \lambda-\dfrac{1}{10}-2 \lambda^2\right) \times 5 \times 4 \times \cos \dfrac{\pi}{3}\)
\(=21 \lambda^2-12 \lambda+7=21\left(\lambda-\dfrac{6}{21}\right)^2+\dfrac{37}{7}\).
\(∴\)当 \(\lambda=\dfrac{6}{21}\)时, \(\overrightarrow{B P} \cdot \overrightarrow{C P}\)有最小值,为 \(\dfrac{37}{7}\).
故答案为:\(\dfrac{37}{7}\). -
答案 \(4\)
解析 如图所示,在\(AB\)上取点\(E\),使\(\overrightarrow{A E}=\dfrac{1}{3} \overrightarrow{A B}\),连接\(DE\),
则 \(A E=\dfrac{1}{3} A B=2\),
\(\because \overrightarrow{A C} \cdot\left(\overrightarrow{A D}-\dfrac{1}{3} \overrightarrow{A B}\right)=0\), \(\therefore \overrightarrow{A C} \cdot(\overrightarrow{A D}-\overrightarrow{A E})=\overrightarrow{A C} \cdot \overrightarrow{E D}=0\),即\(AC⊥ED\),
又\(AD=CD\),\(∴DE\)垂直平分\(AC\),\(∴DE\)是\(∠ADC\)的平分线.
\(\because \overrightarrow{A C} \cdot\left(\overrightarrow{A D}-\dfrac{1}{3} \overrightarrow{A B}\right)=0\),
\(\therefore(\overrightarrow{A D}+\overrightarrow{D C}) \cdot \overrightarrow{A D}-\dfrac{1}{3} \overrightarrow{A C} \cdot \overrightarrow{A B}=0\)化简得 \(\overrightarrow{A D} \cdot \overrightarrow{D C}=1\),
而 \(\overrightarrow{A D} \cdot \overrightarrow{D C}=|\overrightarrow{A D}| \cdot|\overrightarrow{D C}| \cdot \cos (\pi-\angle A D C)\), \(\therefore \cos \angle A D C=-\dfrac{1}{4}\),
\(\therefore \cos ^2 \angle A D E=\dfrac{\cos \angle A D C+1}{2}=\dfrac{-\dfrac{1}{4}+1}{2}=\dfrac{3}{8}\),
\(\because \angle A D E=\dfrac{1}{2} \angle A D C<90^{\circ}\),
\(\therefore \cos \angle A D E=\dfrac{\sqrt{6}}{4}=\cos \angle A E D(\because A D=A E=2, \therefore \angle A D E=\angle A E D)\),
在\(△ADE\)中, \(A D^2=A E^2+D E^2-2 \cdot A E \cdot D E \cdot \cos \angle A E D\),
即 \(4=4+D E^2-2 \times 2 \times D E \times \dfrac{\sqrt{6}}{4}\), \(\therefore D E=\sqrt{6}\).
\(∵P\)为\(BC\)的中点,
\(\therefore \overrightarrow{A P} \cdot \overrightarrow{A D}=\dfrac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A E}+\overrightarrow{E D})=\dfrac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot\left(\dfrac{1}{3} \overrightarrow{A B}+\overrightarrow{E D}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3} \overrightarrow{A B^2}+\overrightarrow{A B} \cdot \overrightarrow{E D}+\dfrac{1}{3} \overrightarrow{A C} \cdot \overrightarrow{A B}+\overrightarrow{A C} \cdot \overrightarrow{E D}\right)\)
\(=\dfrac{1}{2}\left[\dfrac{1}{3} \times 6^2+6 \times \sqrt{6} \times\left(-\dfrac{\sqrt{6}}{4}\right)+\dfrac{1}{3} \times 15+0\right]=4\).