平面向量

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目录

用基底表示向量

1.在  \triangle A B C  中, 点  D, E  分别在边  B C, A C  上, 且  \overrightarrow{B D}=2 \overrightarrow{D C}, \overrightarrow{C E}=3 \overrightarrow{E A} , 若  \overrightarrow{A B}= \overrightarrow{a}, ~~~~~~~\\\overrightarrow{A C}= \overrightarrow{b} , 用 \overrightarrow{a},  \overrightarrow{b} 表示  \overrightarrow{D E} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:-\frac{1}{3} \overrightarrow{a}-\frac{5}{12} \overrightarrow{b} ;提示:\overrightarrow{D E} =\frac{1}{3} \overrightarrow{B C}+\frac{3}{4} \overrightarrow{C A} ~~~~~~~~~\\
=\frac{1}{3}(\overrightarrow{A C}-\overrightarrow{A B})-\frac{3}{4} \overrightarrow{A C} 
=-\frac{1}{3} \overrightarrow{a}-\frac{5}{12} \overrightarrow{b} .~~~~~~~~~~~~~~~~~~~~~

2.在矩形  A B C D  中,  E  为  A B  边的中点, 线段  A C  和  D E  交于点  F , 则  \overrightarrow{B F}=(\quad).~~~~~~~~~~~~~~~~~~\\
A.  -\frac{1}{3} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{A D}\quad
B.  \frac{1}{3} \overrightarrow{A B}-\frac{2}{3} \overrightarrow{A D}\quad
C.  \frac{2}{3} \overrightarrow{A B}-\frac{1}{3} \overrightarrow{A D} \quad
D.  -\frac{2}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{A D} ~~~~~~~~~~~~~~~
答案 

答案:D;提示: \triangle A EF\sim  \triangle CDF \Rightarrow |AF|:|CF|=1:2,~\\
  \overrightarrow{BF}=  \overrightarrow{AF}-  \overrightarrow{AB}=\frac{1}{3}\overrightarrow{AC}-  \overrightarrow{AB}=-\frac{2}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{A D}.~~~~~~~~~~~~~
3.如图, 等腰梯形  A B C D  中,  A B=B C=C D=3 A D , 点  E  为线段  C D  上靠近  C  的三等~~~\\分点, 点  F  为线段  B C  的中点, 则  \overrightarrow{F E}=(\quad).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A.  -\frac{11}{18} \overrightarrow{A B}+\frac{5}{18} \overrightarrow{A C} \quad
B.  -\frac{11}{18} \overrightarrow{A B}+\frac{11}{9} \overrightarrow{A C} C.  -\frac{11}{18} \overrightarrow{A B}+\frac{4}{9} \overrightarrow{A C} \quad
D.  -\frac{1}{2} \overrightarrow{A B}+\frac{5}{6} \overrightarrow{A C} ~
答案 

答案: A;
提示:

\overrightarrow{F E}=\overrightarrow{F C}+\overrightarrow{C E}=\frac{1}{2} \overrightarrow{B C}+\frac{1}{3} \overrightarrow{C D}=~~~\\\frac{1}{2}(\overrightarrow{A C}-\overrightarrow{A B})+\frac{1}{3}\left(\overrightarrow{AD}-\overrightarrow{AC}\right) =-\frac{11}{18} \overrightarrow{A B}+\frac{5}{18} \overrightarrow{A C}.~

4. 在平行四边形  A B C D  中,  A C  与  B D  交于点  O, F  是线段  D C  上的点.若  D C=   3 D F , ~~~~~~~~~\\设  \overrightarrow{A C}=\overrightarrow{a}, \overrightarrow{B D}=\overrightarrow{b} , 则  \overrightarrow{A F}=(\quad) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A.  \frac{1}{4} \overrightarrow{a}+\frac{1}{2} \overrightarrow{b} \quad
B.  \frac{2}{2} \overrightarrow{a}+\frac{1}{3} \overrightarrow{b} \quad
C.  \frac{1}{2} \overrightarrow{a}+\frac{1}{4} \overrightarrow{b} \quad

D.  \frac{1}{3} \overrightarrow{a}+\frac{2}{3} \overrightarrow{b}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案: B;提示:先用\overrightarrow{a},\overrightarrow{b}表示\overrightarrow{AB}=\frac{1 }{2} \left ( \overrightarrow{a}-\overrightarrow{b} \right ),~~~~~\\\overrightarrow{AD}=\frac{1 }{2} \left ( \overrightarrow{a}+\overrightarrow{b} \right ), \overrightarrow{A F}=\frac{2}{2} \overrightarrow{a}+\frac{1}{3} \overrightarrow{b} .~~~~~~~~~~~~~~~~~~~~~~~~~~~
5.在梯形  A B C D  中,  A B / / C D, A B=2 C D, M, N  分别为  C D, B C  的中点.若  \overrightarrow{A B}=   ~~~~~~~~~~~\\\lambda \overrightarrow{A M}+\mu \overrightarrow{A N} , 则  \lambda+\mu  等于(\quad )~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A.  \frac{1}{5} \quad
B.  \frac{2}{5} \quad
C.  \frac{3}{5} \quad
D.  \frac{4}{5} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案: D;
提示: 用\overrightarrow{A B},\overrightarrow{A D}表示向量\overrightarrow{A M},\overrightarrow{A N},再解方程组~~~~~\\得到\overrightarrow{A B}.先求 \overrightarrow{A N}，\overrightarrow{A N}=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\\\overrightarrow{A B}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{A B}+\frac{1}{2}\left ( -\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{DC} \right ) 
=~~~~~~~~~~~~~~~~~~~~~~~\\\frac{1}{2}\overrightarrow{A D}+\frac{3}{4}  \overrightarrow{A B},\Rightarrow 
\left\{\begin{matrix}
 \overrightarrow{A M}=\overrightarrow{A D}+\displaystyle \frac{1}{4} \overrightarrow{A B} & \\
  \overrightarrow{A N}=\displaystyle \frac{1}{2}\overrightarrow{A D} +\displaystyle \frac{3}{4} \overrightarrow{A B} &
\end{matrix}\right.,解方程组得:~~~~~~\\\overrightarrow{A B}= \frac{8}{5} \overrightarrow{AN} - \frac{4}{5} \overrightarrow{A M} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

向量共线定理

1.已知A(x_1,y_1),B(x_2,y_2)是平面中的两个不同点，\overrightarrow{AC} =\lambda\overrightarrow{CB} ,求点C的坐标.~~~~~~~~~~~~~~~
答案 

答案 : \left ( \frac{x_1+\lambda x_2}{1+\lambda } ,\frac{y_1+\lambda y_2}{1+\lambda } \right ) ;提示: C 点坐标等于~~~~~~~~~~~~~\\\overrightarrow {OC}的坐标,即\overrightarrow {OC}=\overrightarrow {OA}+\overrightarrow {AC}=\overrightarrow {OA}+\frac{\lambda }{1+\lambda } \overrightarrow {AB}=~~~~~~~~~~~\\\left ( \frac{x_1+\lambda x_2}{1+\lambda } ,\frac{y_1+\lambda y_2}{1+\lambda } \right ).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2.(1)设 \boldsymbol{a}  与\boldsymbol{b}  是两个不共线向量, 且向量  \boldsymbol{a}+\lambda \boldsymbol{b}  与  -(\boldsymbol{b}-2\boldsymbol{a})  共线, 求  \lambda.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
(2)设\boldsymbol{  e_{1} } 与 \boldsymbol{  e_{2} }  是两个不共线向量,  \overrightarrow{A B}=3 \boldsymbol{  e_{1} }+2 \boldsymbol{  e_{2} } , \overrightarrow{C B}=k \boldsymbol{  e_{1} }+\boldsymbol{  e_{2} } , \overrightarrow{C D}=3 \boldsymbol{  e_{1} }-2 k \boldsymbol{  e_{2} }  ,~~~\\若  A, B, D  三点共线, 求  k  的值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案: (1)-\frac{1}{2} ;提示: 设  \boldsymbol{a}+\lambda \boldsymbol{b}=k(2 \boldsymbol{a}   -\boldsymbol{b}) , 得  k=\frac{1}{2}, \lambda=-\frac{1}{2} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\
(2)-\frac{9}{4} ;
提示:由题意，  A, B, D 三 点共线，故必存在一个实数  \lambda  ，使得  \overrightarrow{A B}=\lambda \overrightarrow{B D} .~~~~~\\所以3 e_{1}+2 e_{2}=\lambda(3-k) e_{1}-\lambda(2 k+1) e_{2} ,所以  \left\{\begin{array}{l}3=\lambda(3-k), \\ 2=-\lambda(2 k+1)\end{array}\right. , 解得  k=-\frac{9}{4} .~~~~~~

3.如图所示, 在  \triangle A B C  中, 点  O  是  B C  的中点, 过点  O  的直线分别交  A B, A C  所在直线于不\\同的两点  M, N , 若  \overrightarrow{A B}=m \overrightarrow{A M}, \overrightarrow{A C}   =n \overrightarrow{A N} , 求  m+n  的值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案: 2;
提示:连接  A O , 则  \overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=~~~~~~~~~~~~~~~~~~~\\\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} , 因为  M, O, N  点共线,所以  \frac{m}{2}+\frac{n}{2}=1 ,~~~~~~~~~~~ \\所以  m+n=2 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4. 已知点M  为  \triangle A B C  中  B C  边上的中点, 点  N  满足  \overrightarrow{A N}=\frac{1}{5} \overrightarrow{A M} , 过点N  的直线与  A B, ~~~~~\\A C  分别交于  P, Q  两点, 且设  \overrightarrow{A P}=x \overrightarrow{A B}, \overrightarrow{A Q}=y \overrightarrow{A C} , 求  \frac{1}{x}+\frac{1}{y}  的值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:10;解析:
 \overrightarrow{A N}=\frac{1}{5} \overrightarrow{A M}=\frac{1}{10}(\overrightarrow{A B}+\overrightarrow{A C})=~~~~~~~~~~~\\\frac{1}{10}\left(\frac{1}{x} \overrightarrow{A P}+\frac{1}{y} \overrightarrow{A Q}\right),\because P, N, Q 三点共线，~~~~~~~~~~~~~~~~~~~~~\\  \therefore \frac{1}{10 x}+\frac{1}{10 y}=1 , 即  \frac{1}{x}+\frac{1}{y}=10 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
5.如图,  \triangle A B C  中, 点  M  是  B C  的中点, 点  N  满足  \overrightarrow{A N}=\frac{2}{3} \overrightarrow{A B}, A M  与  C N 交于点  D, \overrightarrow{A D}=\\\lambda \overrightarrow{A M} , 则  \lambda  等于 (\quad )~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A.  \frac{2}{3} \quad
B.  \frac{3}{4} \quad
C.  \frac{4}{5} \quad
D.  \frac{5}{6} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案: C;
提示:\overrightarrow{A D}=\lambda \overrightarrow{A M}=\frac{\lambda}{2} \overrightarrow{A B}+\frac{\lambda}{2} \overrightarrow{A C}=~~~~~~~~~~~~~~~~~~~~~~\\\frac{3 \lambda}{4} \overrightarrow{A N}+\frac{\lambda}{2} \overrightarrow{A C} ,

因为点  C, D, N  共线, 则有  \frac{3 \lambda}{4}+\frac{\lambda}{2}=1 , ~~~~~~~~~~~\\解得  \lambda=\frac{4}{5} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
6.在  \triangle A B C  中, 点  P  是  A B  上一点, 且  \overrightarrow{C P}=\frac{2}{3} \overrightarrow{C A}+\frac{1}{3} \overrightarrow{C B}, Q  是  B C  的中点,  A Q 与  C P  的~~\\交点为  M , 又  \overrightarrow{C M}=t \overrightarrow{C P} , 求  t  的值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案  :\frac{3}{4}; 提示: A, M, Q  三点共线,\Rightarrow
 \overrightarrow{C M}=t \overrightarrow{C P}=~~~~~~~~~\\t\left(\frac{2}{3} \overrightarrow{C A}+\frac{1}{3} \overrightarrow{C B}\right)=\frac{2 t}{3} \overrightarrow{C A}+\frac{2 t}{3} \overrightarrow{C Q} , 得  t=\frac{3}{4} . ~~~~~~~~~~~~~~~~~~~

平面向量数量积的应用

1.已知向量  \boldsymbol{a}, \boldsymbol{b}  满足  |\boldsymbol{a}|=1,|\boldsymbol{b}|=2, \boldsymbol{a}-\boldsymbol{b}=(\sqrt{3}, \sqrt{2}) , 则  |2 \boldsymbol{a}-\boldsymbol{b}|  等于  (\quad) ~~~~~~~~~~~~~~~~~~~~~~~\\
A.  2 \sqrt{2} \quad
B.  \sqrt{17} \quad
C.  \sqrt{15} \quad
D.  2 \sqrt{5} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:A;提示: \boldsymbol{a} \cdot \boldsymbol{b} =0，(2 \boldsymbol{a}-\boldsymbol{b})^2=8.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2.已知非零向量  \boldsymbol{a}, \boldsymbol{b}  满足  |\boldsymbol{a}|=2|\boldsymbol{b}| , 且  (\boldsymbol{a}-\boldsymbol{b}) \perp \boldsymbol{b} , 则  \boldsymbol{a}  与  \boldsymbol{b}  的夹角为( \quad)~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A.  \frac{\pi}{6} \quad
B.  \frac{\pi}{3} \quad
C.  \frac{2 \pi}{3} \quad
D.  \frac{5 \pi}{6} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:B;提示:  (\boldsymbol{a}-\boldsymbol{b}) \cdot \boldsymbol{b}=0, \therefore \boldsymbol{a} \cdot \boldsymbol{b}=\boldsymbol{b}^{2} . \because|\boldsymbol{a}|=2|\boldsymbol{b}|, \therefore \cos \langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| \cdot|\boldsymbol{b}|}   =~~~~~~~\\\frac{\boldsymbol{b}^{2}}{2 \boldsymbol{b}^{2}}=\frac{1}{2} \because 0 \leqslant\langle \boldsymbol{a}, \boldsymbol{b}\rangle \leqslant \pi, \therefore \boldsymbol{a}  与  \boldsymbol{b}  的夹角为  \frac{\pi}{3} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
3.已知向量  \boldsymbol{a}, \boldsymbol{b}  满足  |\boldsymbol{a}|=5,|\boldsymbol{b}|=6, \boldsymbol{a} \cdot \boldsymbol{b}=-6 , 则  \cos \langle\boldsymbol{a}, \boldsymbol{a}+\boldsymbol{b}\rangle=(\quad)~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\
A.  -\frac{31}{35} \quad
B.  -\frac{19}{35} \quad
C.  \frac{17}{35} \quad
D.  \frac{19}{35} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:D;提示:  |\boldsymbol{a}+\boldsymbol{b}|^{2}=(\boldsymbol{a}+\boldsymbol{b})^{2}=49, \therefore|\boldsymbol{a}+\boldsymbol{b}|=7, \therefore \cos \langle\boldsymbol{a}, \boldsymbol{a}+\boldsymbol{b}\rangle=\frac{19}{35} .~~~~~~~~~~~~~~

4.如图, 在边长为 2 的等边  \triangle A B C  中, 点  E  为中线  B D  的三等分点(靠近点  B  ),点  F  为  B C  ~~~~\\的中点, 则  \overrightarrow{F E} \cdot \overrightarrow{F C}=(\quad) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A.  -\frac{\sqrt{3}}{4} \quad
B.  -\frac{1}{2} \quad
C.  \frac{3}{4} \quad
D.  \frac{1}{2} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:B;提示:
法一、 以F为原点建立坐标系，~~~~~~~~~~~~~~~~~\\  \overrightarrow{F E}=\left(\frac{1}{2}, \frac{\sqrt{3}}{6}\right), \overrightarrow{F C}=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) ,
 \therefore \overrightarrow{F E} \cdot \overrightarrow{F C}=-\frac{1}{2} .\\
法二 、转化为基底的运算，设\overrightarrow{BC} =\overrightarrow{m}, \overrightarrow{BA} =\overrightarrow{n}, ~~~~~~~~~~~~~~\\\overrightarrow{F E} \cdot \overrightarrow{F C}=\frac{1}{2} \vec{m} \cdot\left(\frac{1}{6} \vec{n}-\frac{1}{3} \vec{m}\right)=-\frac{1}{2} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
★★数量积的运算，通常转化为基底运算或坐标运算.~~~~~~
5. 已知向量  \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}  满足  |\boldsymbol{a}|=|\boldsymbol{b}|=1,|\boldsymbol{c}|=\sqrt{2} , 且  \boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}=\mathbf{0} , 求  \cos \langle\boldsymbol{a}-\boldsymbol{c}, \boldsymbol{b}   -\boldsymbol{c}\rangle.~~~~~~
答案 

答案: \frac{4}{5} 

;提示: -\boldsymbol{c}=\boldsymbol{a}+\boldsymbol{b},  \boldsymbol{c}^{2}=\boldsymbol{a}^{2}+\boldsymbol{b}^{2}+2 \boldsymbol{a} \cdot \boldsymbol{b} , 解得~~~~~~~~~~\\  \boldsymbol{a} \cdot \boldsymbol{b}=0 . 建系如图,  \therefore \boldsymbol{a}-\boldsymbol{c}=(2,1), \boldsymbol{b}-\boldsymbol{c}=(1,2) ,则 ~~~~~~~~~~~~ \\\cos \langle\boldsymbol{a}-\boldsymbol{c}, \boldsymbol{b}-\boldsymbol{c}\rangle=\frac{4}{5} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
6. 在平面直角坐标系中,  O  为原点,  A(-1,0), B(0, \sqrt{3}), C(3,0) , 动点  D  满足  |\overrightarrow{C D}|=1 , 求 ~\\ |\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O D}|  的最大值是.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案 : 1+\sqrt{7}; 
提示: 设  D(x, y) ,   |\overrightarrow{C D}|=1 \Rightarrow  (x-3)^{2}+y^{2}=1 ,
\\则  \overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O D}=(x-1, y+\sqrt{3}) ,
故  |\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O D}|=~~\\\sqrt{(x-1)^{2}+(y+\sqrt{3})^{2}} , 几何意义为点  (1,-\sqrt{3}) 到圆 ~~~~~~~~~~~~~~~~~~~~ \\(x-3)^{2}+y^{2}=1  的距离, 最大值为  |E C|+1=\sqrt{7}+1 .~~~~~~~~~~~~~~~~
7.已知  \boldsymbol{a}, \boldsymbol{b}  是单位向量,  \boldsymbol{a} \cdot \boldsymbol{b}=0 . 若向量  \boldsymbol{c}  满足  |\boldsymbol{c}-\boldsymbol{a}-\boldsymbol{b}|=1 , 求  |\boldsymbol{c}|  的最大值是.~~~~~~~~~~~~~~
答案 

答案: \sqrt{2}+1 ;
提示: 由  \boldsymbol{a} \cdot \boldsymbol{b}=0 , 得  \boldsymbol{a} \perp \boldsymbol{b} ,建立坐标系, ~~~~~~~~~~\\则  \overrightarrow{O A}=\boldsymbol{a}=(1,0), \overrightarrow{O B}=\boldsymbol{b}=(0,1) .

设  \boldsymbol{c}=\overrightarrow{O C}=(x, y) , ~~~~~~~~\\由  |\boldsymbol{c}-\boldsymbol{a}-\boldsymbol{b}|=1 ,
得  (x-1)^{2}+(y-1)^{2}=1 ,
所以点  C  在以~~~~ \\ (1,1)  为圆心, 1 为半径的圆上.
所以  |c|_{\text {max }}=\sqrt{2}+1 .~~~~~~~~~~~~~~~~~~
8.如图, 在  \triangle A B C  中,  \cos \angle B A C=\frac{1}{4} , 点,  D  在线段  B C  上, 且  B D=3 D C, A D=   \frac{\sqrt{15}}{2} , ~~~~\\求  \triangle A B C  的面积的最大值.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案 : \sqrt{15} 提示:
\overrightarrow{A D}=\frac{1}{4} \overrightarrow{A B}+\frac{3}{4} \overrightarrow{A C},   \Rightarrow~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\\overrightarrow{A D}=\left(\frac{1}{4} \overrightarrow{A B}+\frac{3}{4} \overrightarrow{A C}\right)^{2} \Rightarrow \frac{15}{4}=\frac{1}{16} c^{2}+\frac{9}{16} b^{2}+\frac{3}{32} b c \geqslant \frac{15}{32} b c,\\


当且仅当  c=3 b  时，等号成立。
所以  b c \leqslant 8 , 又~~~~~~~~~~~~~~~~~~~~~~~~~\\  \sin \angle B A C=\frac{\sqrt{15}}{4} , 所以  S_{\triangle A B C}=\frac{1}{2} b c \sin \angle B A C \leqslant \sqrt{15} .~~~~~~~~

9.在  \triangle A B C  中,  A C=9, \angle A=60^{\circ}, D  点满足  \overrightarrow{C D}=2 \overrightarrow{D B}, A D=\sqrt{37} , 则  B C  的长为 (\quad )\\
A.  3 \sqrt{7} \quad 
B.  3 \sqrt{6} \quad 
C.  3 \sqrt{3} \quad 
D. 6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案 :A;提示:   \overrightarrow{A D}=\overrightarrow{A B}+\frac{1}{3} \overrightarrow{B C}=\overrightarrow{A B}+\frac{1}{3}(\overrightarrow{A C}-\overrightarrow{A B})=~\\\frac{2}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{A C} ,
设  A B=x , 则  \overrightarrow{A D}{ }^{2}=\left(\frac{2}{3} \overrightarrow{A B}+\frac{1}{3} \overrightarrow{A C}\right)^{2} , ~~~~~~~~~~\\即  2 x^{2}+9 x-126=0 ,
 x=6 , 即  A B=6 , 所以  |\overrightarrow{B C}|==3 \sqrt{7} .~~
10.已知点  A, B, C  均在半径为  \sqrt{2}  的圆上, 若  |A B|=2 , 求  \overrightarrow{A C} \cdot \overrightarrow{B C}  的取值范围.~~~~~~~~~~~~~~~~~
答案 

答案:[-1,3];提示:A(\sqrt{2},0 ),B(0,\sqrt{2} ),设C(\cos \theta ,\sin \theta ),~~~~~~\\\overrightarrow{AC}\cdot \overrightarrow{BC}
=(\cos\theta -\sqrt{2},\sin\theta  ) (\cos\theta,\sin\theta-\sqrt{2} )=~~~~~~~~~~~~~~~~~~\\1-2\sin(\theta+\frac{\pi }{4} )
\in [-1,3].~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
11.在  \triangle A B C  中,  A C=3, B C=4, \angle C=90^{\circ} ,P  为  \triangle A B C  所在平面内的动点，且  PC=1  ，\\
求  \overrightarrow{PA} \cdot \overrightarrow{PB}  的取值范围.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
答案 

答案:[-4,6];提示:设A(0,3 ),B(4,0 ),P(\cos \theta ,\sin \theta ),~~~~~~~~\\\overrightarrow{PA}\cdot \overrightarrow{PB}
=(-\cos\theta ,3-\sin\theta  ) (4-\cos\theta,-\sin\theta)=~~~~~~~~~~~~~\\1-5\sin(\theta+\varphi   )
\in [-4,6].~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
12.在\triangle ABC中,BC = 2\sqrt{3},A = 60^{\circ},则\overrightarrow{BA} \cdot \overrightarrow{BC}的最大值为(\quad)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
A. 6\quad
B. 3 + 2\sqrt{3}\quad
C. 12\quad
D. 6 + 4\sqrt{3}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad~
参考图象

答案 

答案:D;提示:
法一、先求\overrightarrow{BA}在\overrightarrow{BC}上的投影,最大值为BD + r=\sqrt{3}+2,~~~~~~~~~~~~~~~~\\
\therefore\overrightarrow{BA} \cdot \overrightarrow{BC}最大值为(\sqrt{3}+2)\cdot2\sqrt{3}=6 + 4\sqrt{3}.
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
法二、 \overrightarrow{BA} \cdot \overrightarrow{BC}=|\overrightarrow{BA}||\overrightarrow{BC}|\cos B = 2\sqrt{3}|\overrightarrow{BA}|\cos B,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
由正弦定理得：\frac{BC}{\sin A}=\frac{BA}{\sin C},\therefore|\overrightarrow{BA}| = 4\sin C,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
\therefore\overrightarrow{BA} \cdot \overrightarrow{BC}=8\sqrt{3}\sin C\cdot\cos B = 8\sqrt{3}\sin(\frac{2\pi}{3}-B)\cdot\cos B
=4\sqrt{3}\sin(2B + \frac{\pi}{3})+6~\\\leqslant6 + 4\sqrt{3}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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