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2019版同步优化探究文数(北师大版)练习:第四章 第一节 平面向量的概念及其线性运算 Word版含解析

课时作业 A 组——基础对点练 → 1 A. a-b 2 1 C.a- b 2 → → → → 1 B. a+b 2 1 D.a+ b 2 → → ) 1.(2017· 杭州模拟)在△ABC 中,已知 M 是 BC 中点,设CB=a,CA=b,则AM=( → 1 1 解析:AM=AC+CM=-CA+ CB=-b+ a,故选 A. 2 2 答案:A → A.A,B,C C.B,C,D → → → → → → ) B.A,B,D D.A,C,D → → → 2.已知AB=a+2b,BC=-5a+6b,CD=7a-2b,则下列一定共线的三点是( 解析:因为AD=AB+BC+CD=3a+6b=3(a+2b)=3AB,又AB,AD有公共点 A.所以 A,B, D 三点共线. 答案:B 3.已知向量 a,b,c 中任意两个都不共线,但 a+b 与 c 共线,且 b+c 与 a 共线,则向量 a+b+c=( A.a C.c ) B.b D.0 解析:依题意,设 a+b=mc,b+c=na,则有(a+b)-(b+c)=mc-na,即 a-c=mc-na. 又 a 与 c 不共线,于是有 m=-1,n=-1,a+b=-c,a+b+c=0. 答案:D → → A.BC → C.AD 解析: → 1 B. AD 2 → 1 D. BC 2 → ) 4.设 D,E,F 分别为△ABC 的三边 BC,CA,AB 的中点,则EB+FC=( → → → → 1 1 如图,EB+FC=EC+CB+FB+BC=EC+FB= (AC+AB)= · 2AD=AD. 2 2 答案:C → → → 2 1 A. OA- OB 3 3 → → → → → → → → → C.2 OA-OB → → ) 5.已知 O,A,B,C 为同一平面内的四个点,若 2 AC+CB=0,则向量OC等于( → → 1 2 B.- OA+ OB 3 3 → → → → → → → → D.-OA+2 OB → → → → → → → → 解析: 因为AC=OC-OA, CB=OB-OC, 所以 2 AC+CB=2(OC-OA)+(OB-OC)=OC- → → → → → 2 OA+OB=0,所以OC=2 OA-OB. 答案:C 6.已知点 G 是△ABC 的重心,过点 G 作一条直线与 AB,AC 两边分别交于 M,N 两点, xy 且AM=x AB,AN=y AC,则 的值为( x+y A.3 C.2 → → → → ) 1 B. 3 1 D. 2 → → → → → 解析:由已知得 M,G,N 三点共线,所以AG=λ AM+(1-λ)AN=λx AB+(1-λ)y AC.∵点 G 是 △ ABC ?λx=3, 2 1 1 的 重 心 , ∴ AG = × ( AB + AC ) = ( AB + AC ) , ∴ ? 3 2 3 1 ??1-λ?y=3, → → → → → x+y 1 1 1 1 xy 1 得 + =1,即 + =3,通分得 =3,∴ = . 3x 3y x y xy x+y 3 1 即 ?λ=3x, ? 1 ?1-λ=3y, 答案:B 1 → → 1 7.在△ABC 中,已知 D 是 AB 边上的一点,若AD=2DB,CD= CA+λCB,则 λ 等于( 3 2 A. 3 1 C.- 3 → → → → → → 1 B. 3 2 D.- 3 → → → ) 解析:∵AD=2DB,即CD-CA=2(CB-CD), → → 1 2 2 ∴CD= CA+ CB,∴λ= . 3 3 3 答案:A a b 8.设 a,b 都是非零向量,下列四个条件中,使 = 成立的充分条件是( |a| |b| A.a=-b C.a=2b B.a∥b D.a∥b 且|a|=|b| ) → a b |a|b 解析: = ?a= ?a 与 b 共线且同向?a=λb 且 λ>0.B,D 选项中 a 和 b 可能反向.A |a| |b| |b| 选项中 λ<0,不符合 λ>0. 答案:C → → → 1 4 A.AD=- AB+ AC 3 3 → → 4 1 C.AD= AB+ AC 3 3 → → → → → → → ) → → 1 4 B.AD= AB- AC 3 3 → → 4 1 D.AD= AB- AC 3 3 → → 9.设 D 为△ABC 所在平面内一点,BC=3CD,则( → → → → → → 1 1 1 1 4 解析:由题意得AD=AC+CD=AC+ BC=AC+ AC- AB=- AB+ AC,故选 A. 3 3 3 3 3 答案:A → y=________. → 2 解析:∵AM=2MC,∴AM= AC. 3 → → 1 ∵BN=NC,∴AN= (AB+AC), 2 → → → → → → → → → → → → 10.在△ABC 中,点 M,N 满足AM=2MC,BN=NC.若MN=xAB+yAC,则 x=________; → → → 1 2 ∴MN=AN-AM= (AB+AC)- AC 2 3 → → 1 1 = AB- AC. 2 6 1 1 又MN=xAB+yAC,∴x= ,y=- . 2 6 1 1 答案: - 2 6 → → → → → → → → → → → → → → → → → → 11. 已知 O 为四边形 ABCD 所在平面内一点,且向量OA, OB,OC, OD满足等式OA+OC= OB+OD,则四边形 ABCD 的形状为________. 解析:由OA+OC=OB+OD得OA-OB=OD-OC,所以BA=CD,所以四边形 ABCD 为平 行四边形. 答案:平行四边形 → e2 表示) → → → → → 1 1 1 解析:在矩形 ABCD 中,因为 O 是对角线的交点,所以OC= AC= (AB+AD)= (DC+BC) 2 2 2 1 = (5e1+3e2).

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