平面向量

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目录

用基底表示向量

1.在△ABC中,点D,E分别在边BC,AC上,且BD→=2DC→,CE→=3EA→,若AB→=a→,       AC→=b→,用a→,b→表示DE→.                                                                                                       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} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:−13a→−512b→;提示:DE→=13BC→+34CA→         =13(AC→−AB→)−34AC→=−13a→−512b→.                     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ABCD中,E为AB边的中点,线段AC和DE交于点F,则BF→=().                  A.−13AB→+23AD→B.13AB→−23AD→C.23AB→−13AD→D.−23AB→+13AD→               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} ~~~~~~~~~~~~~~~
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答案:D;提示:△AEF∼△CDF⇒∣AF∣:∣CF∣=1:2, BF→=AF→−AB→=13AC→−AB→=−23AB→+13AD→.             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ABCD中,AB=BC=CD=3AD,点E为线段CD上靠近C的三等   分点,点F为线段BC的中点,则FE→=().                                                                            A.−1118AB→+518AC→B.−1118AB→+119AC→C.−1118AB→+49AC→D.−12AB→+56AC→ 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} ~
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答案:A;提示:FE→=FC→+CE→=12BC→+13CD→=   12(AC→−AB→)+13(AD→−AC→)=−1118AB→+518AC→. 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ABCD中,AC与BD交于点O,F是线段DC上的点.若DC=3DF,         设AC→=a→,BD→=b→,则AF→=()                                                                                        A.14a→+12b→B.22a→+13b→C.12a→+14b→D.13a→+23b→                                               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}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:B;提示:先用a→,b→表示AB→=12(a→−b→),     AD→=12(a→+b→),AF→=22a→+13b→.                           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ABCD中,AB//CD,AB=2CD,M,N分别为CD,BC的中点.若AB→=           λAM→+μAN→,则λ+μ等于()                                                                                              A.15B.25C.35D.45                                                                                                           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} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:D;提示:用AB→,AD→表示向量AM→,AN→,再解方程组     得到AB→.先求AN→，AN→=                                                            AB→+12BC→=AB→+12(−AB→+AD→+DC→)=                       12AD→+34AB→,⇒{AM→=AD→+14AB→AN→=12AD→+34AB→,解方程组得:      AB→=85AN→−45AM→.                                                                   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(x1,y1),B(x2,y2)是平面中的两个不同点，AC→=λCB→,求点C的坐标.               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.~~~~~~~~~~~~~~~
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答案:(x1+λx21+λ,y1+λy21+λ);提示:C点坐标等于             OC→的坐标,即OC→=OA→+AC→=OA→+λ1+λAB→=           (x1+λx21+λ,y1+λy21+λ).                                                            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a与b是两个不共线向量,且向量a+λb与−(b−2a)共线,求λ.                             (2)设e1与e2是两个不共线向量,AB→=3e1+2e2,CB→=ke1+e2,CD→=3e1−2ke2,   若A,B,D三点共线,求k的值.                                                                                                      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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:(1)−12;提示:设a+λb=k(2a−b),得k=12,λ=−12.                                        (2)−94;提示:由题意，A,B,D三点共线，故必存在一个实数λ，使得AB→=λBD→.     所以3e1+2e2=λ(3−k)e1−λ(2k+1)e2,所以{3=λ(3−k),2=−λ(2k+1),解得k=−94.      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△ABC中,点O是BC的中点,过点O的直线分别交AB,AC所在直线于不同的两点M,N,若AB→=mAM→,AC→=nAN→,求m+n的值.                                              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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:2;提示:连接AO,则AO→=12(AB→+AC→)=                   m2AM→+n2AN→,因为M,O,N点共线,所以m2+n2=1,           所以m+n=2.                                                                             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△ABC中BC边上的中点,点N满足AN→=15AM→,过点N的直线与AB,     AC分别交于P,Q两点,且设AP→=xAB→,AQ→=yAC→,求1x+1y的值.                                 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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:10;解析:AN→=15AM→=110(AB→+AC→)=           110(1xAP→+1yAQ→),∵P,N,Q三点共线，                     ∴110x+110y=1,即1x+1y=10.                                       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,△ABC中,点M是BC的中点,点N满足AN→=23AB→,AM与CN交于点D,AD→=λAM→,则λ等于()                                                                                                                   A.23B.34C.45D.56                                                                                                           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} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:C;提示:AD→=λAM→=λ2AB→+λ2AC→=                      3λ4AN→+λ2AC→,因为点C,D,N共线,则有3λ4+λ2=1,           解得λ=45.                                                                                     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△ABC中,点P是AB上一点,且CP→=23CA→+13CB→,Q是BC的中点,AQ与CP的  交点为M,又CM→=tCP→,求t的值.                                                                                          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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:34;提示:A,M,Q三点共线,⇒CM→=tCP→=         t(23CA→+13CB→)=2t3CA→+2t3CQ→,得t=34.                   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a,b满足∣a∣=1,∣b∣=2,a−b=(3,2),则∣2a−b∣等于()                       A.22B.17C.15D.25                                                                                         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} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:A;提示:a⋅b=0，(2a−b)2=8.                                                                             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a,b满足∣a∣=2∣b∣,且(a−b)⊥b,则a与b的夹角为()                           A.π6B.π3C.2π3D.5π6                                                                                                     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} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:B;提示:(a−b)⋅b=0,∴a⋅b=b2.∵∣a∣=2∣b∣,∴cos⁡⟨a,b⟩=a⋅b∣a∣⋅∣b∣=       b22b2=12∵0⩽⟨a,b⟩⩽π,∴a与b的夹角为π3.                                                                     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a,b满足∣a∣=5,∣b∣=6,a⋅b=−6,则cos⁡⟨a,a+b⟩=()                            A.−3135B.−1935C.1735D.1935                                                                                         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} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:D;提示:∣a+b∣2=(a+b)2=49,∴∣a+b∣=7,∴cos⁡⟨a,a+b⟩=1935.              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△ABC中,点E为中线BD的三等分点(靠近点B),点F为BC    的中点,则FE→⋅FC→=()                                                                                                        A.−34B.−12C.34D.12                                                                                               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} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:B;提示:法一、以F为原点建立坐标系，                 FE→=(12,36),FC→=(−12,−32),∴FE→⋅FC→=−12.法二、转化为基底的运算，设BC→=m→,BA→=n→,              FE→⋅FC→=12m⃗⋅(16n⃗−13m⃗)=−12.                                   ★★数量积的运算，通常转化为基底运算或坐标运算.      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a,b,c满足∣a∣=∣b∣=1,∣c∣=2,且a+b+c=0,求cos⁡⟨a−c,b−c⟩.      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.~~~~~~
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答案:45;提示:−c=a+b,c2=a2+b2+2a⋅b,解得          a⋅b=0.建系如图,∴a−c=(2,1),b−c=(1,2),则            cos⁡⟨a−c,b−c⟩=45.                                                                  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(0,3),C(3,0),动点D满足∣CD→∣=1,求 ∣OA→+OB→+OD→∣的最大值是.                                                                                                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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:1+7;提示:设D(x,y),∣CD→∣=1⇒(x−3)2+y2=1,则OA→+OB→+OD→=(x−1,y+3),故∣OA→+OB→+OD→∣=  (x−1)2+(y+3)2,几何意义为点(1,−3)到圆                    (x−3)2+y2=1的距离,最大值为∣EC∣+1=7+1.                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a,b是单位向量,a⋅b=0.若向量c满足∣c−a−b∣=1,求∣c∣的最大值是.              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.~~~~~~~~~~~~~~
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答案:2+1;提示:由a⋅b=0,得a⊥b,建立坐标系,          则OA→=a=(1,0),OB→=b=(0,1).设c=OC→=(x,y),        由∣c−a−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∣max =2+1.                  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△ABC中,cos⁡∠BAC=14,点,D在线段BC上,且BD=3DC,AD=152,    求△ABC的面积的最大值.                                                                                                      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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:15提示:AD→=14AB→+34AC→,⇒                                 AD→=(14AB→+34AC→)2⇒154=116c2+916b2+332bc⩾1532bc,当且仅当c=3b时，等号成立。所以bc⩽8,又                         sin⁡∠BAC=154,所以S△ABC=12bcsin⁡∠BAC⩽15.        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△ABC中,AC=9,∠A=60∘,D点满足CD→=2DB→,AD=37,则BC的长为()A.37B.36C.33D.6                                                                                               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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:A;提示:AD→=AB→+13BC→=AB→+13(AC→−AB→)= 23AB→+13AC→,设AB=x,则AD→2=(23AB→+13AC→)2,          即2x2+9x−126=0,x=6,即AB=6,所以∣BC→∣==37.  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2的圆上,若∣AB∣=2,求AC→⋅BC→的取值范围.                 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.~~~~~~~~~~~~~~~~~
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答案:[−1,3];提示:A(2,0),B(0,2),设C(cos⁡θ,sin⁡θ),      AC→⋅BC→=(cos⁡θ−2,sin⁡θ)(cos⁡θ,sin⁡θ−2)=                  1−2sin⁡(θ+π4)∈[−1,3].                                                            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△ABC中,AC=3,BC=4,∠C=90∘,P为△ABC所在平面内的动点，且PC=1，求PA→⋅PB→的取值范围.                                                                                                            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.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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答案:[−4,6];提示:设A(0,3),B(4,0),P(cos⁡θ,sin⁡θ),        PA→⋅PB→=(−cos⁡θ,3−sin⁡θ)(4−cos⁡θ,−sin⁡θ)=             1−5sin⁡(θ+φ)∈[−4,6].                                                       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△ABC中,BC=23,A=60∘,则BA→⋅BC→的最大值为()                                   A.6B.3+23C.12D.6+43 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~
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答案:D;提示:法一、先求BA→在BC→上的投影,最大值为BD+r=3+2,                ∴BA→⋅BC→最大值为(3+2)⋅23=6+43.                                                              法二、BA→⋅BC→=∣BA→∣∣BC→∣cos⁡B=23∣BA→∣cos⁡B,                                                     由正弦定理得：BCsin⁡A=BAsin⁡C,∴∣BA→∣=4sin⁡C,                                                            ∴BA→⋅BC→=83sin⁡C⋅cos⁡B=83sin⁡(2π3−B)⋅cos⁡B=43sin⁡(2B+π3)+6 ⩽6+43.                                                                                                                            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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