2016届高考数学一轮复习题组层级快练30含解析

1、PAGE PAGE 6题组层级快练(三十)1对于非零向量a，b，“ab0”是“ab”的()A充分不必要条件B必要不充分条件C充分必要条件 D既不充分也不必要条件答案A解析若ab0，则ab，所以ab；若ab，则ab，ab0不一定成立，故前者是后者的充分不必要条件2设a是任一向量，e是单位向量，且ae，则下列表示形式中正确的是()Aeeq f(a,|a|) Ba|a|eCa|a|e Da|a|e答案D解析对于A，当a0时，eq f(a,|a|)没有意义，错误；对于B，C，D当a0时，选项B，C，D都对；当a0时，由ae可知，a与e同向或反向，选D.3(2015北京东城期中)已知ABCD为平行四边形

2、，若向量eq o(AB,sup10()a，eq o(AC,sup10()b，则向量eq o(BD,sup10()为()Aab BabCb2a Dab答案C4.如图所示，在正六边形ABCDEF中，eq o(BA,sup10()eq o(CD,sup10()eq o(EF,sup10()()A0 B.eq o(BE,sup10()C.eq o(AD,sup10() D.eq o(CF,sup10()答案D解析由于eq o(BA,sup10()eq o(DE,sup10()，故eq o(BA,sup10()eq o(CD,sup10()eq o(EF,sup10()eq o(CD,sup10()eq

3、 o(DE,sup10()eq o(EF,sup10()eq o(CF,sup10().5(2015广东惠州二中模拟)已知点O，A，B不在同一条直线上，点P为该平面上一点，且eq o(OP,sup10()eq f(3o(OA,sup10()o(OB,sup10(),2)，则()A点P在线段AB上B点P在线段AB的反向延长线上C点P在线段AB的延长线上D点P不在直线AB上答案B解析eq o(OP,sup10()eq f(3o(OA,sup10()o(OB,sup10(),2)eq f(3,2)eq o(OA,sup10()eq f(1,2)eq o(OB,sup10()eq o(OA,sup10

4、()eq f(1,2)(eq o(OA,sup10()eq o(OB,sup10()eq o(OA,sup10()eq f(1,2)eq o(BA,sup10()，即eq o(OP,sup10()eq o(OA,sup10()eq o(AP,sup10()eq f(1,2)eq o(BA,sup10()，所以点P在线段AB的反向延长线上，故选B.6在ABC中，点D在边AB上，CD平分ACB.若eq o(CB,sup10()a，eq o(CA,sup10()b，|a|1，|b|2，则eq o(CD,sup10()()A.eq f(1,3)aeq f(2,3)b B.eq f(2,3)aeq f(

5、1,3)bC.eq f(3,5)aeq f(4,5)b D.eq f(4,5)aeq f(3,5)b答案B解析由内角平分线定理，得eq f(|CA|,|CB|)eq f(|AD|,|DB|)2.eq o(CD,sup10()eq o(CA,sup10()eq o(AD,sup10()eq o(CA,sup10()eq f(2,3)eq o(AB,sup10()eq o(CA,sup10()eq f(2,3)(eq o(CB,sup10()eq o(CA,sup10()eq f(2,3)eq o(CB,sup10()eq f(1,3)eq o(CA,sup10()eq f(2,3)aeq f(1

6、,3)b.故B正确7已知向量i与j不共线，且eq o(AB,sup10()imj，eq o(AD,sup10()nij，若A，B，D三点共线，则实数m，n应该满足的条件是()Amn1 Bmn1Cmn1 Dmn1答案C解析由A，B，D共线可设eq o(AB,sup10()eq o(AD,sup10()，于是有imj(nij)nij.又i，j不共线，因此eq blcrc (avs4alco1(n1，,m，)即有mn1.8O是平面上一定点，A，B，C是该平面上不共线的三个点，一动点P满足：eq o(OP,sup10()eq o(OA,sup10()(eq o(AB,sup10()eq o(AC,su

7、p10()，(0，)，则直线AP一定通过ABC的()A外心 B内心C重心 D垂心答案C解析取BC中点M.eq o(OP,sup10()eq o(OA,sup10()(eq o(AB,sup10()eq o(AC,sup10()，eq o(OP,sup10()eq o(OA,sup10()(eq o(AB,sup10()eq o(AC,sup10()，eq o(AP,sup10()2eq o(AD,sup10().A，P，D三点共线，AP一定通过ABC的重心，C正确9在四边形ABCD中，eq o(AB,sup10()a2b，eq o(BC,sup10()4ab，eq o(CD,sup10()5a

8、3b，则四边形ABCD的形状是()A矩形 B平行四边形C梯形 D以上都不对答案C解析由已知eq o(AD,sup10()eq o(AB,sup10()eq o(BC,sup10()eq o(CD,sup10()8a2b2(4ab)2eq o(BC,sup10().eq o(AD,sup10()eq o(BC,sup10().又eq o(AB,sup10()与eq o(CD,sup10()不平行，四边形ABCD是梯形10已知四边形ABCD是菱形，点P在对角线AC上(不包括端点A，C)的充要条件是eq o(AP,sup10()(eq o(AB,sup10()eq o(AD,sup10()，则的取值

9、范围是()A(0,1) B(1,0)C(0，eq f(r(2),2) D(eq f(r(2),2)，0)答案A解析如图所示，点P在对角线AC上(不包括端点A，C)，eq o(AP,sup10()eq o(AC,sup10()(eq o(AB,sup10()eq o(AD,sup10()由eq o(AP,sup10()与eq o(AC,sup10()同向知，0.又|eq o(AP,sup10()|eq o(AC,sup10()|，eq f(|o(AP,sup10()|,|o(AC,sup10()|)1，(0,1)反之亦然11设A1，A2，A3，A4是平面直角坐标系中两两不同的四点，若eq o(A

10、1A3,sup10()eq o(A1A2,sup10()(R)，eq o(A1A4,sup10()eq o(A1A2,sup10()(R)，且eq f(1,)eq f(1,)2，则称A3，A4调和分割A1，A2.已知平面上的点C，D调和分割点A，B，则下列说法正确的是()AC可能是线段AB的中点BD可能是线段AB的中点CC，D可能同时在线段AB上DC，D不可能同时在线段AB的延长线上答案D解析若A成立，则eq f(1,2)，而eq f(1,)0，不可能；同理B也不可能；若C成立，则01，且02，与已知矛盾；若C，D同时在线段AB的延长线上时，1，且1，eq f(1,)eq f(1,)2，与已知

11、矛盾，故C，D不可能同时在线段AB的延长线上，故D正确12.如图所示，下列结论不正确的是_eq o(PQ,sup10()eq f(3,2)aeq f(3,2)b；eq o(PT,sup10()eq f(3,2)aeq f(3,2)b；eq o(PS,sup10()eq f(3,2)aeq f(1,2)b；eq o(PR,sup10()eq f(3,2)ab.答案解析由abeq f(2,3)eq o(PQ,sup10()，知eq o(PQ,sup10()eq f(3,2)aeq f(3,2)b，正确；由eq o(PT,sup10()eq f(3,2)aeq f(3,2)b，从而错误；eq o(P

12、S,sup10()eq o(PT,sup10()b，故eq o(PS,sup10()eq f(3,2)aeq f(1,2)b，正确；eq o(PR,sup10()eq o(PT,sup10()2beq f(3,2)aeq f(1,2)b，错误故正确的为.13.如图所示，已知B30，AOB90，点C在AB上，OCAB，用eq o(OA,sup10()和eq o(OB,sup10()来表示向量eq o(OC,sup10()，则eq o(OC,sup10()等于_答案eq f(3,4)eq o(OA,sup10()eq f(1,4)eq o(OB,sup10()解析eq o(OC,sup10()eq

13、 o(OA,sup10()eq o(AC,sup10()eq o(OA,sup10()eq f(1,4)eq o(AB,sup10()eq o(OA,sup10()eq f(1,4)(eq o(OB,sup10()eq o(OA,sup10()eq f(3,4)eq o(OA,sup10()eq f(1,4)eq o(OB,sup10().14设a和b是两个不共线的向量，若eq o(AB,sup10()2akb，eq o(CB,sup10()ab，eq o(CD,sup10()2ab，且A，B，D三点共线，则实数k的值等于_答案4解析A，B，D三点共线，eq o(AB,sup10()eq o(

14、BD,sup10().eq o(AB,sup10()2akb，eq o(BD,sup10()eq o(BC,sup10()eq o(CD,sup10()a2b，k4.故填4.15已知O为ABC内一点，且eq o(OA,sup10()eq o(OC,sup10()2eq o(OB,sup10()0，则AOC与ABC的面积之比是_答案12解析如图所示，取AC中点D.eq o(OA,sup10()eq o(OC,sup10()2eq o(OD,sup10().eq o(OD,sup10()eq o(BO,sup10().O为BD中点，面积比为高之比16已知向量a2e13e2，b2e13e2，其中e1

15、，e2不共线，向量c2e19e2.问是否存在这样的实数，使向量dab与c共线？答案当2时共线解析d(2e13e2)(2e13e2)(22)e1(33)e2.要使d与c共线，则应有实数k，使dkc.即(22)e1(33)e22ke19ke2.即eq blcrc (avs4alco1(222k，,339k，)得2.故存在这样的实数，只要2，就能使d与c共线17如图所示，已知点G是ABO的重心(1)求eq o(GA,sup10()eq o(GB,sup10()eq o(GO,sup10()；(2)若PQ过ABO的重心G，且eq o(OA,sup10()a，eq o(OB,sup10()b，eq o(

16、OP,sup10()ma，eq o(OQ,sup10()nb，求证：eq f(1,m)eq f(1,n)3.答案(1)eq o(GA,sup10()eq o(GB,sup10()eq o(GO,sup10()0(2)略解析(1)如图所示，延长OG交AB于M点，则M是AB的中点eq o(GA,sup10()eq o(GB,sup10()2eq o(GM,sup10().G是ABO的重心，eq o(GO,sup10()2eq o(GM,sup10().eq o(GA,sup10()eq o(GB,sup10()eq o(GO,sup10()0.(2)M是AB边的中点，eq o(OM,sup10()eq f(1,2)(eq o(OA,sup10()eq o(OB,sup10()eq f(1,2)(ab)又G是ABO的重心，eq o(OG,sup10()eq f(2,3)eq o(OM,sup10()eq f(1,3)(ab)eq o(PG,sup10()eq o(OG,sup10()eq o(OP,sup10()eq f(1,3)(