2021-2022学年高中数学人教版B版必修第二册6.1.3向量的减法练习题-【含答案】

1、向量的减法一、选择题1(多选题)在ABC中，向量eq o(BC,sup7(）可表示为()Aeq o(AB,sup7(）eq o(AC,sup7(）Beq o(AC,sup7(）eq o(AB,sup7(）Ceq o(BA,sup7(）eq o(AC,sup7(）Deq o(BA,sup7(）eq o(CA,sup7(）2下列各式结果是eq o(AB,sup7(）的是()Aeq o(AM,sup7(）eq o(MN,sup7(）eq o(MB,sup7(）Beq o(AC,sup7(）eq o(BF,sup7(）eq o(CF,sup7(）Ceq o(AB,sup7(）eq o(DC,sup7(

2、）eq o(CB,sup7(）Deq o(AB,sup7(）eq o(FC,sup7(）eq o(BC,sup7(）3在ABC中，|eq o(AB,sup7(）|eq o(BC,sup7(）|eq o(CA,sup7(）|1，则|eq o(BC,sup7(）eq o(AC,sup7(）|的值为()A0B1 Ceq r(3)D24下列各式不能化简为eq o(PQ,sup7(）的是()Aeq o(AB,sup7(）(eq o(PA,sup7(）eq o(BQ,sup7(）)B(eq o(AB,sup7(）eq o(PC,sup7(）)(eq o(BA,sup7(）eq o(QC,sup7(）)Ce

3、q o(QC,sup7(）eq o(QP,sup7(）eq o(CQ,sup7(）Deq o(PA,sup7(）eq o(AB,sup7(）eq o(BQ,sup7(）5已知平面内M，N，P三点满足eq o(MN,sup7(）eq o(PN,sup7(）eq o(PM,sup7(）0，则下列说法正确的是()AM，N，P是一个三角形的三个顶点BM，N，P是一条直线上的三个点CM，N，P是平面内的任意三个点D以上都不对二、填空题6.eq o(AC,sup7(）eq o(BD,sup7(）eq o(CD,sup7(）eq o(AB,sup7(）的化简结果为_7已知两向量a和b，如果a的方向与b的方向

4、垂直，那么|ab|_|ab|.(填“”“”或“”)8已知|a|7，|b|2，若ab，则|ab|_.三、解答题9如图，解答下列各题：(1)用a，d，e表示eq o(DB,sup7(）；(2)用b，c表示eq o(DB,sup7(）；(3)用a，b，e表示eq o(EC,sup7(）；(4)用d，c表示eq o(EC,sup7(）.10已知ABC是等腰直角三角形，ACB90，M是斜边AB的中点，eq o(CM,sup7(）a，eq o(CA,sup7(）b，求证：(1)|ab|a|；(2)|a(ab)|b|.素养达标11如图，P，Q是ABC的边BC上的两点，且eq o(BP,sup7(）eq o(

5、QC,sup7(），则eq o(AB,sup7(）eq o(AC,sup7(）eq o(AP,sup7(）eq o(AQ,sup7(）的结果为()A0Beq o(BP,sup7(）Ceq o(PQ,sup7(）Deq o(PC,sup7(）12(多选题)已知D，E，F分别是ABC的边AB，BC，CA的中点，则()Aeq o(AD,sup7(）eq o(BE,sup7(）eq o(CF,sup7(）0Beq o(BD,sup7(）eq o(CF,sup7(）eq o(DF,sup7(）eq o(BA,sup7(）Ceq o(AD,sup7(）eq o(CE,sup7(）eq o(CF,sup7(

6、）eq o(AB,sup7(）Deq o(BD,sup7(）eq o(BE,sup7(）eq o(FC,sup7(）013如图所示，已知一点O到平行四边形ABCD的三个顶点A，B，C的向量分别为r1，r2，r3，则eq o(OD,sup7(）_.14已知如图，在正六边形ABCDEF中，与eq o(OA,sup7(）eq o(OC,sup7(）eq o(CD,sup7(）相等的向量有_(填序号)eq o(CF,sup7(）；eq o(AD,sup7(）；eq o(BE,sup7(）；eq o(DE,sup7(）eq o(FE,sup7(）eq o(CD,sup7(）；eq o(CE,sup7(）

7、eq o(BC,sup7(）；eq o(CA,sup7(）eq o(CD,sup7(）；eq o(AB,sup7(）eq o(AE,sup7(）.15已知OAB中，eq o(OA,sup7(）a，eq o(OB,sup7(）b，满足|a|b|ab|2，求|ab|与OAB的面积一、选择题1(多选题)在ABC中，向量eq o(BC,sup7(）可表示为()Aeq o(AB,sup7(）eq o(AC,sup7(）Beq o(AC,sup7(）eq o(AB,sup7(）Ceq o(BA,sup7(）eq o(AC,sup7(）Deq o(BA,sup7(）eq o(CA,sup7(）BCD由向量的

8、减法与加法可知B、C、D正确2下列各式结果是eq o(AB,sup7(）的是()Aeq o(AM,sup7(）eq o(MN,sup7(）eq o(MB,sup7(）Beq o(AC,sup7(）eq o(BF,sup7(）eq o(CF,sup7(）Ceq o(AB,sup7(）eq o(DC,sup7(）eq o(CB,sup7(）Deq o(AB,sup7(）eq o(FC,sup7(）eq o(BC,sup7(）Beq o(AC,sup7(）eq o(BF,sup7(）eq o(CF,sup7(）eq o(AC,sup7(）eq o(CF,sup7(）eq o(BF,sup7(）eq

9、o(AF,sup7(）eq o(FB,sup7(）eq o(AB,sup7(）.3在ABC中，|eq o(AB,sup7(）|eq o(BC,sup7(）|eq o(CA,sup7(）|1，则|eq o(BC,sup7(）eq o(AC,sup7(）|的值为()A0B1 Ceq r(3)D2B|eq o(BC,sup7(）eq o(AC,sup7(）|eq o(BC,sup7(）eq o(CA,sup7(）|eq o(BA,sup7(）|14下列各式不能化简为eq o(PQ,sup7(）的是()Aeq o(AB,sup7(）(eq o(PA,sup7(）eq o(BQ,sup7(）)B(eq

10、o(AB,sup7(）eq o(PC,sup7(）)(eq o(BA,sup7(）eq o(QC,sup7(）)Ceq o(QC,sup7(）eq o(QP,sup7(）eq o(CQ,sup7(）Deq o(PA,sup7(）eq o(AB,sup7(）eq o(BQ,sup7(）DA项中，原式eq o(AB,sup7(）eq o(PA,sup7(）eq o(BQ,sup7(）eq o(PA,sup7(）eq o(AB,sup7(）eq o(BQ,sup7(）eq o(PQ,sup7(）；B项中，原式(eq o(AB,sup7(）eq o(BA,sup7(）)(eq o(PC,sup7(）e

11、q o(QC,sup7(）)0eq o(PC,sup7(）eq o(CQ,sup7(）eq o(PQ,sup7(）；C项中，原式eq o(QC,sup7(）eq o(CQ,sup7(）eq o(QP,sup7(）0eq o(PQ,sup7(）eq o(PQ,sup7(）；D项中，原式eq o(PB,sup7(）eq o(BQ,sup7(）eq o(PB,sup7(）eq o(QB,sup7(）eq o(PQ,sup7(）.5已知平面内M，N，P三点满足eq o(MN,sup7(）eq o(PN,sup7(）eq o(PM,sup7(）0，则下列说法正确的是()AM，N，P是一个三角形的三个顶点

12、BM，N，P是一条直线上的三个点CM，N，P是平面内的任意三个点D以上都不对C因为eq o(MN,sup7(）eq o(PN,sup7(）eq o(PM,sup7(）eq o(MN,sup7(）eq o(NP,sup7(）eq o(PM,sup7(）eq o(MP,sup7(）eq o(PM,sup7(）0，eq o(MN,sup7(）eq o(NP,sup7(）eq o(PM,sup7(）0对任意情况是恒成立的故M，N，P是平面内的任意三个点故选C二、填空题6.eq o(AC,sup7(）eq o(BD,sup7(）eq o(CD,sup7(）eq o(AB,sup7(）的化简结果为_0原式

13、(eq o(AC,sup7(）eq o(CD,sup7(）)(eq o(AB,sup7(）eq o(BD,sup7(）)eq o(AD,sup7(）eq o(AD,sup7(）0.7已知两向量a和b，如果a的方向与b的方向垂直，那么|ab|_|ab|.(填“”“”或“”)以a，b为邻边的平行四边形是矩形，矩形的对角线相等由加减法的几何意义知|ab|ab|.8已知|a|7，|b|2，若ab，则|ab|_.5或9ab，当a与b同向时，|ab|72|5，当a与b反向时，|ab|72|9.三、解答题9如图，解答下列各题：(1)用a，d，e表示eq o(DB,sup7(）；(2)用b，c表示eq o(D

14、B,sup7(）；(3)用a，b，e表示eq o(EC,sup7(）；(4)用d，c表示eq o(EC,sup7(）.解eq o(AB,sup7(）a，eq o(BC,sup7(）b，eq o(CD,sup7(）c，eq o(DE,sup7(）d，eq o(EA,sup7(）e.(1)eq o(DB,sup7(）eq o(DE,sup7(）eq o(EA,sup7(）eq o(AB,sup7(）dea；(2)eq o(DB,sup7(）eq o(CB,sup7(）eq o(CD,sup7(）eq o(BC,sup7(）eq o(CD,sup7(）bc；(3)eq o(EC,sup7(）eq o

15、(EA,sup7(）eq o(AB,sup7(）eq o(BC,sup7(）eab；(4)eq o(EC,sup7(）eq o(CE,sup7(）(eq o(CD,sup7(）eq o(DE,sup7(）)cd.10已知ABC是等腰直角三角形，ACB90，M是斜边AB的中点，eq o(CM,sup7(）a，eq o(CA,sup7(）b，求证：(1)|ab|a|；(2)|a(ab)|b|.证明如图，在等腰RtABC中，由M是斜边AB的中点，得|eq o(CM,sup7(）|eq o(AM,sup7(）|，|eq o(CA,sup7(）|eq o(CB,sup7(）|.(1)在ACM中，eq o

16、(AM,sup7(）eq o(CM,sup7(）eq o(CA,sup7(）ab.于是由|eq o(AM,sup7(）|eq o(CM,sup7(）|，得|ab|a|.(2)在MCB中，eq o(MB,sup7(）eq o(AM,sup7(）ab，所以eq o(CB,sup7(）eq o(MB,sup7(）eq o(MC,sup7(）abaa(ab)从而由|eq o(CB,sup7(）|eq o(CA,sup7(）|，得|a(ab)|b|.素养达标11如图，P，Q是ABC的边BC上的两点，且eq o(BP,sup7(）eq o(QC,sup7(），则eq o(AB,sup7(）eq o(AC,

17、sup7(）eq o(AP,sup7(）eq o(AQ,sup7(）的结果为()A0Beq o(BP,sup7(）Ceq o(PQ,sup7(）Deq o(PC,sup7(）A因为eq o(BP,sup7(）eq o(QC,sup7(），所以eq o(AB,sup7(）eq o(AC,sup7(）eq o(AP,sup7(）eq o(AQ,sup7(）(eq o(AB,sup7(）eq o(AP,sup7(）)(eq o(AC,sup7(）eq o(AQ,sup7(）)eq o(PB,sup7(）eq o(QC,sup7(）eq o(BP,sup7(）eq o(QC,sup7(）0.12(多选

18、题)已知D，E，F分别是ABC的边AB，BC，CA的中点，则()Aeq o(AD,sup7(）eq o(BE,sup7(）eq o(CF,sup7(）0Beq o(BD,sup7(）eq o(CF,sup7(）eq o(DF,sup7(）eq o(BA,sup7(）Ceq o(AD,sup7(）eq o(CE,sup7(）eq o(CF,sup7(）eq o(AB,sup7(）Deq o(BD,sup7(）eq o(BE,sup7(）eq o(FC,sup7(）0AC因为D，E，F分别是ABC的边AB，BC，CA的中点，所以eq o(AD,sup7(）eq o(DB,sup7(），eq o(C

19、F,sup7(）eq o(ED,sup7(），eq o(FC,sup7(）eq o(DE,sup7(），eq o(FE,sup7(）eq o(DB,sup7(），所以eq o(AD,sup7(）eq o(BE,sup7(）eq o(CF,sup7(）eq o(DB,sup7(）eq o(BE,sup7(）eq o(ED,sup7(）0，故A项成立eq o(BD,sup7(）eq o(CF,sup7(）eq o(DF,sup7(）eq o(BD,sup7(）eq o(DF,sup7(）eq o(CF,sup7(）eq o(BF,sup7(）eq o(FC,sup7(）eq o(BC,sup7(）

20、，故B项不成立eq o(AD,sup7(）eq o(CE,sup7(）eq o(CF,sup7(）eq o(AD,sup7(）eq o(FE,sup7(）eq o(AD,sup7(）eq o(DB,sup7(）eq o(AB,sup7(），故C项成立eq o(BD,sup7(）eq o(BE,sup7(）eq o(FC,sup7(）eq o(ED,sup7(）eq o(DE,sup7(）eq o(ED,sup7(）eq o(ED,sup7(）0，故D项不成立13如图所示，已知一点O到平行四边形ABCD的三个顶点A，B，C的向量分别为r1，r2，r3，则eq o(OD,sup7(）_.r3r1r

21、2因为eq o(OD,sup7(）eq o(OC,sup7(）eq o(CD,sup7(），eq o(CD,sup7(）eq o(BA,sup7(）eq o(OA,sup7(）eq o(OB,sup7(），所以eq o(OD,sup7(）eq o(OC,sup7(）eq o(OA,sup7(）eq o(OB,sup7(）r3r1r2.14已知如图，在正六边形ABCDEF中，与eq o(OA,sup7(）eq o(OC,sup7(）eq o(CD,sup7(）相等的向量有_(填序号)eq o(CF,sup7(）；eq o(AD,sup7(）；eq o(BE,sup7(）；eq o(DE,sup7(）eq o(FE,sup7(）eq o(CD,sup7(）；eq o(CE,sup7(）eq o(BC,sup7(）；eq o(CA,sup7(）eq o(CD,sup7(）；eq o(AB,sup7(）eq o(AE,