高三数学一轮复习 （教材回扣+考点分类+课堂内外+限时训练）专讲专练 5.1　平面向量的概念及其线性运算

1、PAGE PAGE 5高三数学一轮复习专讲专练（教材回扣+考点分类+课堂内外+限时训练）：5.1平面向量的概念及其线性运算一、选择题1若O、E、F是不共线的任意三点，则以下各式中成立的是()A.eq o(EF,sup15()eq o(OF,sup15()eq o(OE,sup15()B.eq o(EF,sup15()eq o(OF,sup15()eq o(OE,sup15()C.eq o(EF,sup15()eq o(OF,sup15()eq o(OE,sup15() D.eq o(EF,sup15()eq o(OF,sup15()eq o(OE,sup15()解析：由减法的三角形法则知eq

2、o(EF,sup15()eq o(OF,sup15()eq o(OE,sup15().答案：B2(2013德州调研)在ABC中，M为边BC上任意一点，N为AM中点，eq o(AN,sup15()eq o(AB,sup15()eq o(AC,sup15()，则的值为()A.eq f(1,2) B.eq f(1,3)C.eq f(1,4) D1解析：M为边BC上任意一点，可设eq o(AM,sup15()xeq o(AB,sup15()yeq o(AC,sup15()(xy1)N为AM中点，eq o(AN,sup15()eq f(1,2)eq o(AM,sup15()eq f(1,2)xeq o(

3、AB,sup15()eq f(1,2)yeq o(AC,sup15()eq o(AB,sup15()eq o(AC,sup15().eq f(1,2)(xy)eq f(1,2).答案：A3设P是ABC所在平面内的一点，eq o(BC,sup15()eq o(BA,sup15()eq o(2BP,sup15()，则()AP、A、B三点共线 BP、A、C三点共线CP、B、C三点共线 D以上均不正确解析：eq o(BC,sup15()eq o(BA,sup15()2eq o(BP,sup15()，eq o(BC,sup15()eq o(BP,sup15()eq o(BP,sup15()eq o(BA

4、,sup15()，即eq o(PC,sup15()eq o(AP,sup15()，P、A、C三点共线答案：B4已知点O，N在ABC所在平面内，且|eq o(OA,sup15()|eq o(OB,sup15()|eq o(OC,sup15()|，eq o(NA,sup15()eq o(NB,sup15()eq o(NC,sup15()0，则点O，N依次是ABC的()A重心外心 B重心内心C外心重心 D外心内心解析：由|eq o(OA,sup15()|eq o(OB,sup15()|eq o(OC,sup15()|知，O为ABC的外心；eq o(NA,sup15()eq o(NB,sup15()e

5、q o(NC,sup15()0，知，N为ABC的重心答案：C5如图，已知eq o(AB,sup15()a，eq o(AC,sup15()b，eq o(BD,sup15()3eq o(DC,sup15()，用a，b表示eq o(AD,sup15()，则eq o(AD,sup15()()Aaeq f(3,4)bB.eq f(1,4)aeq f(3,4)bC.eq f(1,4)aeq f(1,4)bD.eq f(3,4)aeq f(1,4)b解析：eq o(CB,sup15()eq o(AB,sup15()eq o(AC,sup15()ab，又eq o(BD,sup15()3eq o(DC,sup1

6、5()，eq o(CD,sup15()eq f(1,4)eq o(CB,sup15()eq f(1,4)(ab)，eq o(AD,sup15()eq o(AC,sup15()eq o(CD,sup15()beq f(1,4)(ab)eq f(1,4)aeq f(3,4)b.答案：B6已知ABC中，点D是BC的中点，过点D的直线分别交直线AB、AC于E、F两点，若eq o(AB,sup15()eq o(AE,sup15()(0)，eq o(AC,sup15()eq o(AF,sup15()(0)，则eq f(1,)eq f(4,)的最小值是()A9 B.eq f(7,2)C5 D.eq f(9,

7、2)解析：由题意得，eq o(AB,sup15()eq o(AC,sup15()2eq o(AD,sup15()eq o(AE,sup15()eq o(AF,sup15()eq o(AD,sup15()eq f(,2)eq o(AE,sup15()eq f(,2)eq o(AF,sup15()，又D、E、F在同一条直线上，可得eq f(,2)eq f(,2)1.所以eq f(1,)eq f(4,)eq blc(rc)(avs4alco1(f(,2)f(,2)eq blc(rc)(avs4alco1(f(1,)f(4,)eq f(5,2)eq f(2,)eq f(,2)eq f(5,2)2eq

8、f(9,2)，当且仅当2时取等号答案：D二、填空题7设向量a，b满足|a|2eq r(5)，b(2,1)，且a与b的方向相反，则a的坐标为_解析：设a(x，y)，x0，y0，则x2y0且x2y220，解得x4，y2(舍去)，或者x4，y2，即a(4，2)答案：(4，2)8(2013无锡质检)设a，b是两个不共线的非零向量，若8akb与ka2b共线，则实数k_.解析：因为8akb与ka2b共线，所以存在实数，使8akb(ka2b)，即(8k)a(k2)b0.又a，b是两个不共线的非零向量，故eq blcrc (avs4alco1(8k0，,k20，)解得k4.答案：49如图所示，平面内的两条相交

9、直线OP1和OP2将该平面分割成四个部分、(不包括边界)，若eq o(OP,sup15()aeq o(OP1,sup15()beq o(OP2,sup15()，且点P落在第部分，则实数a，b满足a_0，b_0(用“”，“”或“”填空)解析：由于点P落在第部分，且eq o(OP,sup15()aeq o(OP,sup15()1beq o(OP,sup15()2，则根据实数与向量的积的定义及平行四边形法则知a0，b0.答案：三、解答题10(2013东莞阶段检测)如图所示，在ABC中，点D、F分别是BC、AC的中点，eq o(AE,sup15()eq f(2,3)eq o(AD,sup15()，eq

10、 o(AB,sup15()a，eq o(AC,sup15()b.(1)用a，b表示向量eq o(AD,sup15()、eq o(AE,sup15()、eq o(AF,sup15()、eq o(BE,sup15()、eq o(BF,sup15()；(2)求证：B，E，F三点共线解析：(1)延长AD到G，使eq o(AD,sup15()eq f(1,2)eq o(AG,sup15()，连接BG、CG，得到ABGC，所以eq o(AG,sup15()ab，eq o(AD,sup15()eq f(1,2)eq o(AG,sup15()eq f(1,2)(ab)，eq o(AE,sup15()eq f(

11、2,3)eq o(AD,sup15()eq f(1,3)(ab)，eq o(AF,sup15()eq f(1,2)eq o(AC,sup15()eq f(1,2)b，eq o(BE,sup15()eq o(AE,sup15()eq o(AB,sup15()eq f(1,3)(ab)aeq f(1,3)(b2a)，eq o(BF,sup15()eq o(AF,sup15()eq o(AB,sup15()eq f(1,2)baeq f(1,2)(b2a)(2)由(1)可知，eq o(BE,sup15()eq f(2,3)eq o(BF,sup15()，所以B，E，F三点共线11(2013临沂模拟)

12、若a，b是两个不共线的非零向量，a与b起点相同，则当t为何值时，a，tb，eq f(1,3)(ab)三向量的终点在同一条直线上？解析：设eq o(OA,sup15()a，eq o(OB,sup15()tb，eq o(OC,sup15()eq f(1,3)(ab)，eq o(AC,sup15()eq o(OC,sup15()eq o(OA,sup15()eq f(2,3)aeq f(1,3)b，eq o(AB,sup15()eq o(OB,sup15()eq o(OA,sup15()tba.要使A，B，C三点共线，只需eq o(AC,sup15()eq o(AB,sup15().即eq f(2,3)aeq f(1,3)btba.eq blcrc (avs4alco1(f(2,3)，,f(1,3)t，)解得eq blcrc (avs4alco1(f(2,3)，,tf(1,2).)当teq f(1,2)时，三向量终点在同一直线上12已知ABC中，eq o(AB,sup15()a，eq o(AC,sup15()b，对于平面ABC上任意一点O，动点P满足eq o(OP,sup15()eq o(OA,sup15()ab，则动点P的轨迹是什么？其轨迹是否过定点，并说明理由解析：依题意，由eq o(OP,sup15()eq