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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].~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
