高三數(shù)學(xué)一輪復(fù)習(xí) （教材回扣+考點(diǎn)分類+課堂內(nèi)外+限時(shí)訓(xùn)練）專講專練 5.1　平面向量的概念及其線性運(yùn)算

1、PAGE PAGE 5高三數(shù)學(xué)一輪復(fù)習(xí)專講專練（教材回扣+考點(diǎn)分類+課堂內(nèi)外+限時(shí)訓(xùn)練）：5.1平面向量的概念及其線性運(yùn)算一、選擇題1若O、E、F是不共線的任意三點(diǎn)，則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：由減法的三角形法則知eq

2、o(EF,sup15()eq o(OF,sup15()eq o(OE,sup15().答案：B2(2013德州調(diào)研)在ABC中，M為邊BC上任意一點(diǎn)，N為AM中點(diǎn)，eq o(AN,sup15()eq o(AB,sup15()eq o(AC,sup15()，則的值為()A.eq f(1,2) B.eq f(1,3)C.eq f(1,4) D1解析：M為邊BC上任意一點(diǎn)，可設(shè)eq o(AM,sup15()xeq o(AB,sup15()yeq o(AC,sup15()(xy1)N為AM中點(diǎn)，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設(shè)P是ABC所在平面內(nèi)的一點(diǎn)，eq o(BC,sup15()eq o(BA,sup15()eq o(2BP,sup15()，則()AP、A、B三點(diǎn)共線 BP、A、C三點(diǎn)共線CP、B、C三點(diǎn)共線 D以上均不正確解析：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三點(diǎn)共線答案：B4已知點(diǎn)O，N在ABC所在平面內(nèi)，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，則點(diǎn)O，N依次是ABC的()A重心外心 B重心內(nèi)心C外心重心 D外心內(nèi)心解析：由|eq o(OA,sup15()|eq o(OB,sup15()|eq o(OC,sup15()|知，O為ABC的外心；eq o(NA,sup15()eq o(NB,sup15()e

5、q o(NC,sup15()0，知，N為ABC的重心答案：C5如圖，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()，則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，點(diǎn)D是BC的中點(diǎn)，過點(diǎn)D的直線分別交直線AB、AC于E、F兩點(diǎn)，若eq o(AB,sup15()eq o(AE,sup15()(0)，eq o(AC,sup15()eq o(AF,sup15()(0)，則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：由題意得，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在同一條直線上，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)，當(dāng)且僅當(dāng)2時(shí)取等號答案：D二、填空題7設(shè)向量a，b滿足|a|2eq r(5)，b(2,1)，且a與b的方向相反，則a的坐標(biāo)為_解析：設(shè)a(x，y)，x0，y0，則x2y0且x2y220，解得x4，y2(舍去)，或者x4，y2，即a(4，2)答案：(4，2)8(2013無錫質(zhì)檢)設(shè)a，b是兩個(gè)不共線的非零向量，若8akb與ka2b共線，則實(shí)數(shù)k_.解析：因?yàn)?akb與ka2b共線，所以存在實(shí)數(shù)，使8akb(ka2b)，即(8k)a(k2)b0.又a，b是兩個(gè)不共線的非零向量，故eq blcrc (avs4alco1(8k0，,k20，)解得k4.答案：49如圖所示，平面內(nèi)的兩條相交

9、直線OP1和OP2將該平面分割成四個(gè)部分、(不包括邊界)，若eq o(OP,sup15()aeq o(OP1,sup15()beq o(OP2,sup15()，且點(diǎn)P落在第部分，則實(shí)數(shù)a，b滿足a_0，b_0(用“”，“”或“”填空)解析：由于點(diǎn)P落在第部分，且eq o(OP,sup15()aeq o(OP,sup15()1beq o(OP,sup15()2，則根據(jù)實(shí)數(shù)與向量的積的定義及平行四邊形法則知a0，b0.答案：三、解答題10(2013東莞階段檢測)如圖所示，在ABC中，點(diǎn)D、F分別是BC、AC的中點(diǎn)，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)求證：B，E，F(xiàn)三點(diǎn)共線解析：(1)延長AD到G，使eq o(AD,sup15()eq f(1,2)eq o(AG,sup15()，連接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(xiàn)三點(diǎn)共線11(2013臨沂模擬)

12、若a，b是兩個(gè)不共線的非零向量，a與b起點(diǎn)相同，則當(dāng)t為何值時(shí)，a，tb，eq f(1,3)(ab)三向量的終點(diǎn)在同一條直線上？解析：設(shè)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三點(diǎn)共線，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).)當(dāng)teq f(1,2)時(shí)，三向量終點(diǎn)在同一直線上12已知ABC中，eq o(AB,sup15()a，eq o(AC,sup15()b，對于平面ABC上任意一點(diǎn)O，動(dòng)點(diǎn)P滿足eq o(OP,sup15()eq o(OA,sup15()ab，則動(dòng)點(diǎn)P的軌跡是什么？其軌跡是否過定點(diǎn)，并說明理由解析：依題意，由eq o(OP,sup15()eq