2023届一轮复习北师大版 平面向量的概念及线性运算 作业

1、 课时分层作业(三十二)平面向量的概念及线性运算 一、选择题1如图所示，在正六边形ABCDEF中，eq o(BA,sup8()eq o(CD,sup8()eq o(EF,sup8()等于()A0Beq o(BE,sup8()Ceq o(AD,sup8()Deq o(CF,sup8()D由图知，eq o(BA,sup8()eq o(CD,sup8()eq o(EF,sup8()eq o(CD,sup8()eq o(DE,sup8()eq o(EF,sup8()eq o(CF,sup8()，故选D2(2021天水月考)已知a，b是两个非零向量，且|ab|a|b|，则下列说法正确的是()Aab0Ba

2、bCa与b共线反向D存在正实数，使abD由a，b是非零向量且|ab|a|b|知，向量a，b是方向相同的共线向量，故选D3设D，E，F分别为ABC的三边BC，CA，AB的中点，则eq o(EB,sup8()eq o(FC,sup8()()Aeq o(AD,sup8()Beq f(1,2)eq o(AD,sup8()Ceq f(1,2)eq o(BC,sup8()Deq o(BC,sup8()A由题意得eq o(EB,sup8()eq o(FC,sup8()eq f(1,2)(eq o(AB,sup8()eq o(CB,sup8()eq f(1,2)(eq o(AC,sup8()eq o(BC,s

3、up8()eq f(1,2)(eq o(AB,sup8()eq o(AC,sup8()eq o(AD,sup8()4已知向量eq o(AB,sup8()a3b，eq o(BC,sup8()5a3b，eq o(CD,sup8()3a3b，则()AA，B，C三点共线BA，B，D三点共线CA，C，D三点共线DB，C，D三点共线Beq o(BC,sup8()eq o(CD,sup8()2a6b2(a3b)2eq o(AB,sup8()，eq o(BD,sup8()2eq o(AB,sup8()，A，B，D三点共线，故选B5在ABC中，eq o(AN,sup8()eq f(1,4)eq o(NC,sup

4、8()，P是直线BN上一点，若eq o(AP,sup8()meq o(AB,sup8()eq f(2,5)eq o(AC,sup8()，则实数m的值为()A4B1C1D4Beq o(AN,sup8()eq f(1,4)eq o(NC,sup8()，eq o(AC,sup8()5eq o(AN,sup8()又eq o(AP,sup8()meq o(AB,sup8()eq f(2,5)eq o(AC,sup8()，eq o(AP,sup8()meq o(AB,sup8()2eq o(AN,sup8()，由B，P，N三点共线可知，m21，m16(2021南昌模拟)如图，在ABC中，点D在BC边上，且

5、CD2DB，点E在AD边上，且AD3AE，则用向量eq o(AB,sup8()，eq o(AC,sup8()表示eq o(CE,sup8()为()Aeq f(2,9)eq o(AB,sup8()eq f(8,9)eq o(AC,sup8()Beq f(2,9)eq o(AB,sup8()eq f(8,9)eq o(AC,sup8()Ceq f(2,9)eq o(AB,sup8()eq f(7,9)eq o(AC,sup8()Deq f(2,9)eq o(AB,sup8()eq f(7,9)eq o(AC,sup8()B由平面向量的三角形法则及向量共线的性质可得eq o(CE,sup8()eq

6、o(AE,sup8()eq o(AC,sup8()eq f(1,3)eq o(AD,sup8()eq o(AC,sup8()eq f(1,3)(eq o(AB,sup8()eq f(1,3)eq o(BC,sup8()eq o(AC,sup8()eq f(1,3)eq blcrc(avs4alco1(o(AB,sup8()f(1,3)o(AC,sup8()o(AB,sup8()eq o(AC,sup8()eq f(2,9)eq o(AB,sup8()eq f(8,9)eq o(AC,sup8()7已知平面内一点P及ABC，若eq o(PA,sup8()eq o(PB,sup8()eq o(PC

7、,sup8()eq o(AB,sup8()，则点P与ABC的位置关系是()A点P在线段AB上B点P在线段BC上C点P在线段AC上D点P在ABC外部Ceq o(PA,sup8()eq o(PB,sup8()eq o(PC,sup8()eq o(AB,sup8()eq o(AP,sup8()eq o(PB,sup8()，eq o(PC,sup8()2eq o(PA,sup8()，因此点P在线段AC上，故选C8设O在ABC的内部，D为AB的中点，且eq o(OA,sup8()eq o(OB,sup8()2eq o(OC,sup8()0，则ABC的面积与AOC的面积的比值为()A3B4C5D6B如图，

8、D为AB的中点，则eq o(OD,sup8()eq f(1,2)(eq o(OA,sup8()eq o(OB,sup8()，又eq o(OA,sup8()eq o(OB,sup8()2eq o(OC,sup8()0，eq o(OD,sup8()eq o(OC,sup8()，O为CD的中点，又D为AB中点，SAOCeq f(1,2)SADCeq f(1,4)SABC，则eq f(SABC,SAOC)4二、填空题9若|eq o(AB,sup8()|eq o(AC,sup8()|eq o(AB,sup8()eq o(AC,sup8()|2，则|eq o(AB,sup8()eq o(AC,sup8()

9、|_2eq r(3)由|eq o(AB,sup8()|eq o(AC,sup8()|eq o(AB,sup8()eq o(AC,sup8()|2知ABC是等边三角形，设BC的中点为D，则|eq o(AB,sup8()eq o(AC,sup8()|2|eq o(AD,sup8()|2eq r(3)10已知向量a，b不共线，且cab，da(21)b，若c与d共线反向，则实数的值为_eq f(1,2)由于c与d共线反向，则存在实数k使ckd(k0)，于是abka(21)b整理得abka(2kk)b由于a，b不共线，所以有eq blcrc (avs4alco1(k，,2kk1，)整理得2210，解得1

10、或eq f(1,2)又因为k0，所以0，故eq f(1,2)11在等腰梯形ABCD中， eq o(AB,sup8()2eq o(DC,sup8()，点E是线段BC的中点，若eq o(AE,sup8()eq o(AB,sup8()eq o(AD,sup8()，则_，_eq f(3,4)eq f(1,2)取AB的中点F，连接CF(图略)，则由题可得CFAD，且CFADeq o(AE,sup8()eq o(AB,sup8()eq o(BE,sup8()eq o(AB,sup8()eq f(1,2)eq o(BC,sup8()eq o(AB,sup8()eq f(1,2)(eq o(FC,sup8()

11、eq o(FB,sup8()eq o(AB,sup8()eq f(1,2)eq blc(rc)(avs4alco1(o(AD,sup8()f(1,2)o(AB,sup8()eq f(3,4)eq o(AB,sup8()eq f(1,2)eq o(AD,sup8()，eq f(3,4)，eq f(1,2)12已知ABC和点M满足eq o(MA,sup8()eq o(MB,sup8()eq o(MC,sup8()0，若存在实数m使得eq o(AB,sup8()eq o(AC,sup8()meq o(AM,sup8()成立，则m_3由已知条件得eq o(MB,sup8()eq o(MC,sup8()

12、eq o(MA,sup8()，M为ABC的重心，eq o(AM,sup8()eq f(1,3)(eq o(AB,sup8()eq o(AC,sup8()，即eq o(AB,sup8()eq o(AC,sup8()3eq o(AM,sup8()，则m31已知等边三角形ABC内接于O，D为线段OA的中点，则eq o(BD,sup8()()Aeq f(2,3)eq o(BA,sup8()eq f(1,6)eq o(BC,sup8()Beq f(4,3)eq o(BA,sup8()eq f(1,6)eq o(BC,sup8()Ceq o(BA,sup8()eq f(2,3)eq o(BC,sup8()

13、Deq f(2,3)eq o(BA,sup8()eq f(1,3)eq o(BC,sup8()A设BC的中点为E，由题意知，点O为等边ABC的重心，则eq o(AD,sup8()eq f(1,3)eq o(AE,sup8()，eq o(BD,sup8()eq o(BA,sup8()eq o(AD,sup8()eq o(BA,sup8()eq f(1,3)eq o(AE,sup8()eq o(BA,sup8()eq f(1,3)(eq o(AB,sup8()eq o(BE,sup8()eq f(2,3)eq o(BA,sup8()eq f(1,6)eq o(BC,sup8()故选A2如图，在AB

14、C中，eq o(AN,sup8()eq f(2,3)eq o(NC,sup8()，P是BN上一点，若eq o(AP,sup8()teq o(AB,sup8()eq f(1,3)eq o(AC,sup8()，则实数t的值为()Aeq f(2,3)Beq f(2,5)Ceq f(1,6)Deq f(3,4)C法一：因为eq o(AN,sup8()eq f(2,3)eq o(NC,sup8()，所以eq o(AN,sup8()eq f(2,5)eq o(AC,sup8()设eq o(NP,sup8()eq o(NB,sup8()，则eq o(AP,sup8()eq o(AN,sup8()eq o(N

15、P,sup8()eq f(2,5)eq o(AC,sup8()eq o(NB,sup8()eq f(2,5)eq o(AC,sup8()(eq o(NA,sup8()eq o(AB,sup8()eq f(2,5)eq o(AC,sup8()eq blc(rc)(avs4alco1(f(2,5)o(AC,sup8()o(AB,sup8()eq o(AB,sup8()eq f(2,5)(1)eq o(AC,sup8()又eq o(AP,sup8()teq o(AB,sup8()eq f(1,3)eq o(AC,sup8()，所以teq o(AB,sup8()eq f(1,3)eq o(AC,sup8()eq o(AB,sup8()eq f(2,5)(1)eq o(AC,sup8()，得eq blcrc (avs4alco1(t，,f(2,5)1f(