高中数学必修二 第六章 6.2 6.2.3课后课时精练

1、A级：“四基”巩固训练一、选择题1下列各式计算正确的个数是()(7)5a35a；a2b2(ab)3a；ab(ab)0.A0 B1 C2 D3答案C解析根据向量数乘的运算律可验证正确；错误，因为向量的和、差及数乘运算的结果仍为一个向量，而不是实数2如图所示，D是ABC的边AB上的中点，则向量eq o(CD,sup16()()A.eq o(BC,sup16()eq f(1,2)eq o(BA,sup16()Beq o(BC,sup16()eq f(1,2)eq o(BA,sup16()Ceq o(BC,sup16()eq f(1,2)eq o(BA,sup16()D.eq o(BC,sup16()

2、eq f(1,2)eq o(BA,sup16()答案B解析解法一：D是AB的中点，eq o(BD,sup16()eq f(1,2)eq o(BA,sup16()，eq o(CD,sup16()eq o(CB,sup16()eq o(BD,sup16()eq o(BC,sup16()eq f(1,2)eq o(BA,sup16().解法二：由eq o(CD,sup16()eq f(1,2)(eq o(CB,sup16()eq o(CA,sup16()eq f(1,2)eq o(CB,sup16()(eq o(CB,sup16()eq o(BA,sup16()eq o(CB,sup16()eq f

3、(1,2)eq o(BA,sup16()eq o(BC,sup16()eq f(1,2)eq o(BA,sup16().3设Aeq o(B,sup16()eq f(r(2),2)(a5b)，Beq o(C,sup16()2a8b，Ceq o(D,sup16()3(ab)，则共线的三点是()AA，B，D BA，B，C CB，C，D DA，C，D答案A解析Beq o(D,sup16()Beq o(C,sup16()Ceq o(D,sup16()a5b，Aeq o(B,sup16()eq f(r(2),2)Beq o(D,sup16()，A，B，D三点共线故选A.4若eq o(AB,sup16()3

4、e1，eq o(CD,sup16()5e1，且|eq o(AD,sup16()|eq o(BC,sup16()|，则四边形ABCD是()A平行四边形 B菱形C等腰梯形 D不等腰的梯形答案C解析因为eq o(AB,sup16()eq f(3,5)eq o(CD,sup16()，所以ABCD，且|eq o(AB,sup16()|eq o(CD,sup16()|.而|eq o(AD,sup16()|eq o(BC,sup16()|，所以四边形ABCD为等腰梯形5在平行四边形ABCD中，AC与BD交于点O，E是线段OD的中点，AE的延长线与CD交于点F.若eq o(AC,sup16()a，eq o(B

5、D,sup16()b，则eq o(AF,sup16()等于()A.eq f(1,4)aeq f(1,2)b B.eq f(2,3)aeq f(1,3)bC.eq f(1,2)aeq f(1,4)b D.eq f(1,3)aeq f(2,3)b答案B解析如图所示，E是OD的中点，eq o(OE,sup16()eq f(1,4)eq o(BD,sup16()eq f(1,4)b.又ABEFDE，eq f(AE,FE)eq f(BE,DE)eq f(3,1).eq o(AE,sup16()3eq o(EF,sup16()，eq o(AE,sup16()eq f(3,4)eq o(AF,sup16()

6、，在AOE中，eq o(AE,sup16()eq o(AO,sup16()eq o(OE,sup16()eq f(1,2)aeq f(1,4)b，eq o(AF,sup16()eq f(4,3)eq o(AE,sup16()eq f(2,3)aeq f(1,3)b.故选B.二、填空题6设e1，e2是两个不共线的向量，若向量ke12e2与8e1ke2方向相反，则k_.答案4解析ke12e2与8e1ke2共线，ke12e2(8e1ke2)8e1ke2.eq blcrc (avs4alco1(k8，,2k，)解得eq blcrc (avs4alco1(f(1,2)，,k4)或eq blcrc (av

7、s4alco1(f(1,2)，,k4.)ke12e2与8e1ke2反向，eq f(1,2)，k4.7若ae13e2，b4e12e2，c3e112e2，则向量a写为1b2c的形式是_答案eq f(1,18)beq f(7,27)c解析若a1b2c，则e13e21(4e12e2)2(3e112e2)，e13e2(4132)e1(21122)e2.eq blcrc (avs4alco1(41321，,211223.)解得eq blcrc (avs4alco1(1f(1,18)，,2f(7,27).)8如图，在ABC中，点O是BC的中点，过点O的直线分别交直线AB，AC于不同的两点M，N，若eq o(

8、AB,sup16()meq o(AM,sup16()，eq o(AC,sup16()neq o(AN,sup16()，则mn的值为_答案2解析解法一：因为eq o(AB,sup16()meq o(AM,sup16()，eq o(AC,sup16()neq o(AN,sup16()，所以eq o(AM,sup16()eq f(1,m)eq o(AB,sup16()，eq o(AN,sup16()eq f(1,n)eq o(AC,sup16()，则eq o(MN,sup16()eq o(AN,sup16()eq o(AM,sup16()eq f(1,n)eq o(AC,sup16()eq f(1,

9、m)eq o(AB,sup16().因为点O为BC的中点，连接AO，所以eq o(AO,sup16()eq f(1,2)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()，则eq o(MO,sup16()eq o(AO,sup16()eq o(AM,sup16()eq f(1,2)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()eq f(1,m)eq o(AB,sup16()eq blc(rc)(avs4alco1(f(1,2)f(1,m)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()，因为M，O，N三点

10、共线，所以可设eq o(MO,sup16()eq o(MN,sup16()，即eq blc(rc)(avs4alco1(f(1,2)f(1,m)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16()eq f(,n)eq o(AC,sup16()eq f(,m)eq o(AB,sup16()，则eq blc(rc)(avs4alco1(f(1,2)f(1,m)f(,m)eq o(AB,sup16()eq blc(rc)(avs4alco1(f(1,2)f(,n)eq o(AC,sup16()0，由于eq o(AB,sup16()，eq o(AC,sup16()不共线，所以

11、eq blcrc (avs4alco1(f(1,2)f(1,m)f(,m)0，,f(1,2)f(,n)0，)消去得eq f(1,2)eq f(1,m)eq f(n,2m)0，变形整理可得mn2.解法二：在ABC中，连接AO.由于O是BC的中点，因此eq o(AO,sup16()eq f(1,2)(eq o(AB,sup16()eq o(AC,sup16()eq f(1,2)eq o(AB,sup16()eq f(1,2)eq o(AC,sup16().由于eq o(AB,sup16()meq o(AM,sup16()，eq o(AC,sup16()neq o(AN,sup16()，则eq o(

12、AO,sup16()eq f(1,2)meq o(AM,sup16()eq f(1,2)neq o(AN,sup16().由于M，O，N三点共线，则eq f(1,2)meq f(1,2)n1，从而mn2.三、解答题9设e1，e2是两个不共线的向量，如果eq o(AB,sup16()2e1e2，eq o(BC,sup16()3e1e2，eq o(CD,sup16()7e16e2.(1)求证：A，B，D三点共线；(2)试确定的值，使2e1e2和e1e2共线；(3)若e1e2与e1e2不共线，试求的取值范围解(1)证明：因为eq o(BD,sup16()eq o(BC,sup16()eq o(CD,

13、sup16()3e1e27e16e210e15e25(2e1e2)5eq o(AB,sup16()，所以eq o(AB,sup16()与eq o(BD,sup16()共线因为eq o(AB,sup16()与eq o(BD,sup16()有公共点B，所以A，B，D三点共线(2)因为2e1e2与e1e2共线，所以存在实数，使2e1e2(e1e2)因为e1，e2不共线，所以eq blcrc (avs4alco1(2，,1.)所以eq f(r(2),2).(3)假设e1e2与e1e2共线，则存在实数，使e1e2(e1e2)因为e1，e2不共线，所以eq blcrc (avs4alco1(1，,，)所以

14、1.所以当1时，e1e2与e1e2不共线B级：“四能”提升训练1如图所示，向量eq o(OA,sup16()，eq o(OB,sup16()，eq o(OC,sup16()的终点A，B，C在一条直线上，且eq o(AC,sup16()3eq o(CB,sup16().设eq o(OA,sup16()p，eq o(OB,sup16()q，eq o(OC,sup16()r，则以下等式中成立的是()Areq f(1,2)peq f(3,2)q Brp2qCreq f(3,2)peq f(1,2)q Drq2p答案A解析eq o(OC,sup16()eq o(OB,sup16()eq o(BC,sup

15、16()，eq o(AC,sup16()3eq o(CB,sup16()3eq o(BC,sup16()，eq o(BC,sup16()eq f(1,3)eq o(AC,sup16().eq o(OC,sup16()eq o(OB,sup16()eq f(1,3)eq o(AC,sup16()eq o(OB,sup16()eq f(1,3)(eq o(OC,sup16()eq o(OA,sup16()rqeq f(1,3)(rp)req f(1,2)peq f(3,2)q.2设D，E分别是ABC的边AB，BC上的点，ADeq f(1,2)AB，BEeq f(2,3)BC.若eq o(DE,sup16()1eq o(AB,sup16()2eq o(AC,sup16()(1，2为实数)，则12的值为_答案eq f(1,2)解析由已知eq