2018年高考数学平面向量：平面向量的概念及线性运算题源探究

1、PAGE PAGE 6平面向量的概念及线性运算【考点梳理】1向量的有关概念(1)向量：既有大小又有方向的量叫做向量，向量的大小叫做向量的长度(或模)(2)零向量：长度为0的向量，其方向是任意的(3)单位向量：长度等于1个单位的向量(4)平行向量：方向相同或相反的非零向量平行向量又叫共线向量规定：0与任一向量平行(5)相等向量：长度相等且方向相同的向量(6)相反向量：长度相等且方向相反的向量2向量的线性运算向量运算定义法则(或几何意义)运算律加法求两个向量和的运算三角形法则平行四边形法则(1)交换律：abba；(2)结合律：(ab)ca(bc)减法求a与b的相反向量b的和的运算叫做a与b的差三角

2、形法则aba(b)数乘求实数与向量a的积的运算(1)|a|a|；(2)当0时，a的方向与a的方向相同；当0，故qeq o(,sup0(/)p.p是q的充分不必要条件，故选A. 5(必修4 P90练习T4(2)改编)向量e1与e2不共线，若ae1e2与b2e1e2共线，则的值为() A1 B2 C1D2答案 B解析 e1与e2不共线，且ae1e2与b2e1e2共线，存在R，使e1e2(2e1e2)2e1e2，得eq blc(avs4alco1(12,1)，2. 6(必修4 P91A组T4改编)下列四个结论：eq o(AB,sup16()eq o(BC,sup16()eq o(CA,sup16()

3、0；eq avs4al()eq o(AB,sup16()eq o(MB,sup16()eq o(BO,sup16()eq o(OM,sup16()0；eq avs4al()eq o(AB,sup16()eq o(AC,sup16()eq o(BD,sup16()eq o(CD,sup16()0；eq avs4al()eq o(NQ,sup16()eq o(QP,sup16()eq o(MN,sup16()eq o(MP,sup16()0，其中一定正确的结论个数是() A1 B2 C3 D4答案 C解析 eq o(AB,sup16()eq o(BC,sup16()eq o(CA,sup16()e

4、q o(AC,sup16()eq o(CA,sup16()0，正确；eq o(AB,sup16()eq o(MB,sup16()eq o(BO,sup16()eq o(OM,sup16()eq o(AB,sup16()eq o(MO,sup16()eq o(OM,sup16()eq o(AB,sup16()，错；eq o(AB,sup16()eq o(AC,sup16()eq o(BD,sup16()eq o(CD,sup16()eq o(CB,sup16()eq o(BD,sup16()eq o(DC,sup16()eq o(CB,sup16()eq o(BC,sup16()0，正确；eq

5、o(NQ,sup16()eq o(QP,sup16()eq o(MN,sup16()eq o(MP,sup16()eq o(NP,sup16()eq o(PN,sup16()0，正确故正确 7(必修4 P92B组T5改编)O为四边形ABCD所在平面内一点，若eq o(OA,sup16()eq o(OC,sup16()eq o(OB,sup16()eq o(OD,sup16()，则四边形ABCD一定为() A正方形 B矩形 C菱形 D平行四边形答案 D解析 由eq o(OA,sup16()eq o(OC,sup16()eq o(OB,sup16()eq o(OD,sup16()，得eq o(OA

6、,sup16()eq o(OB,sup16()eq o(OD,sup16()eq o(OC,sup16()，即eq o(BA,sup16()eq o(CD,sup16()，BACD，且|BA|CD|.四边形ABCD一定为平行四边形，故选D. 8(必修4 P118A组T2(5)改编)M是ABCD的对角线的交点，O为平面内任一点，则eq o(OA,sup16()eq o(OB,sup16()eq o(OC,sup16()eq o(OD,sup16()等于() A.eq o(OM,sup16() B2eq o(OM,sup16() C3eq o(OM,sup16() D4eq o(OM,sup16(

7、)答案 D解析 如图，eq o(OA,sup16()eq o(OB,sup16()eq o(OC,sup16()eq o(OD,sup16()eq o(OM,sup16()eq o(MA,sup16()eq o(OM,sup16()eq o(MB,sup16()eq o(OM,sup16()eq o(MC,sup16()eq o(OM,sup16()eq o(MD,sup16()4eq o(OM,sup16()(eq o(MA,sup16()eq o(MC,sup16()(eq o(MB,sup16()eq o(MD,sup16()4eq o(OM,sup16()，故选D. 9(必修4 P89

8、例6改编)向量a，b不共线，eq o(OA,sup16()ab，eq o(OB,sup16()a2b，eq o(OC,sup16()2a3b，若A，B，C三点共线，则的值为_答案 解析 eq o(AB,sup14()eq o(OB,sup14()eq o(OA,sup14()a2b(ab)(1)ab，eq o(AC,sup14()eq o(OC,sup14()eq o(OA,sup14()2a3b(ab)a2b，A，B，C三点共线，即eq o(AB,sup14()与eq o(AC,sup14()共线，存在实数，使eq o(AB,sup14()eq o(AC,sup14()，即(1)ab(a2b

9、)a2b，eq blcrc (avs4alco1(1,12)eq f(3,2)，eq f(1,2). 10(必修4 P86例4改编)如图，ABCD中，eq o(AC,sup14()a，eq o(BD,sup14()b，则eq o(AB,sup14()_，eq o(AD,sup14()_. 答案 eq f(1,2)aeq f(1,2)b eq f(1,2)aeq f(1,2)b解析 法一：设AC与BD交于M(图略)，eq o(AB,sup14()eq o(MB,sup14()eq o(MA,sup14()eq f(1,2)eq o(BD,sup14()eq blc(rc)(avs4alco1(f

10、(1,2)o(AC,sup14()eq f(1,2)beq blc(rc)(avs4alco1(f(1,2)a)eq f(1,2)aeq f(1,2)b.eq o(AD,sup14()eq o(MD,sup14()eq o(MA,sup14()eq f(1,2)eq o(BD,sup14()eq blc(rc)(avs4alco1(f(1,2)o(AC,sup14()eq f(1,2)beq blc(rc)(avs4alco1(f(1,2)a)eq f(1,2)aeq f(1,2)b.法二：设eq o(AB,sup14()x，eq o(AD,sup14()y，则eq blcrc (avs4al

11、co1(xya，,yxb，)解得xeq f(1,2)aeq f(1,2)b，yeq f(1,2)aeq f(1,2)b.11(必修4 P120B组T4改编)如图，EF是等腰梯形ABCD的中位线，M，N是E，F上的两个三等分点，若eq o(AB,sup14()a，eq o(BC,sup14()b，eq o(AB,sup14()2eq o(DC,sup14().(1)用a，b表示eq o(AM,sup6()；(2)证明A，M，C三点共线解析 (1)eq o(AD,sup14()eq o(AB,sup14()eq o(BC,sup14()eq o(CD,sup14()abeq blc(rc)(avs

12、4alco1(f(1,2)a)eq f(1,2)ab，又E为AD中点，eq o(AE,sup14()eq f(1,2)eq o(AD,sup14()eq f(1,4)aeq f(1,2)b，EF是梯形的中位线，且eq o(AB,sup14()2eq o(DC,sup14()，eq o(EF,sup14()eq f(1,2)(eq o(AB,sup14()eq o(DC,sup14()eq f(1,2)eq blc(rc)(avs4alco1(af(1,2)a)eq f(3,4)a，又M，N是E，F的三等分点，eq o(EM,sup14()eq f(1,3)eq o(EF,sup14()eq f(1,4)a，eq o(AM,sup14()eq o(AE,sup14()eq o(EM,sup14()eq f(1,4)aeq f(1,2)beq f(1,4)aeq f(1,2)aeq f(1