2022年新教材高中数学第2章平面向量及其应用2从位移的合成到向量的加减法2.1向量的加法素养作业北师大版必修第二册

1、PAGE PAGE 6第二章2.1A组素养自测一、选择题1(多选)下列等式中正确的是(ABD)Aa0aBabbaC|ab|a|b|Deq o(AC,sup6()eq o(DC,sup6()eq o(AB,sup6()eq o(BD,sup6()解析当a与b方向不同时，|ab|a|b|.2.如图，四边形ABCD是梯形，ADBC，对角线AC与BD相交于点O，则eq o(OA,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(DO,sup6()等于(B)Aeq o(CD,sup6()Beq o(DC,sup6()Ceq o(DA,sup6()Deq o(DO,sup6()

2、解析eq o(OA,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(DO,sup6()eq o(DO,sup6()eq o(OA,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(DA,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(DB,sup6()eq o(BC,sup6()eq o(DC,sup6().3下列说法正确的个数为(B)如果非零向量a与b的方向相同或相反，那么ab的方向必与a与b的方向相同；在ABC中，必有eq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6()0

3、；若eq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6()0，则A，B，C一定为一个三角形的三个顶点A0B1C2D3解析错，若ab0，则ab的方向是任意的；正确；错，当A，B，C三点共线时，也满足eq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6()0.4如图，正六边ABCDEF中，eq o(BA,sup6()eq o(CD,sup6()eq o(FE,sup6()(B)A0Beq o(BE,sup6()Ceq o(AD,sup6()Deq o(CF,sup6()解析连接CF，取CF中点O，连接OE，OA则eq o(BA,sup6()

4、eq o(CD,sup6()eq o(FE,sup6()(eq o(BA,sup6()eq o(AF,sup6()eq o(FE,sup6()eq o(BE,sup6().5在ABC中，|eq o(AB,sup6()|eq o(BC,sup6()|eq o(AB,sup6()eq o(BC,sup6()|，则ABC是(B)A直角三角形B等边三角形C钝角三角形D等腰直角三角形解析eq o(AB,sup6()eq o(BC,sup6()eq o(AC,sup6()，则|eq o(AB,sup6()|eq o(BC,sup6()|eq o(AC,sup6()|，则ABC是等边三角形二、填空题6化简下

5、列各式：(1)eq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6() 0 ；(2)eq o(OA,sup6()eq o(OC,sup6()eq o(BO,sup6()eq o(CO,sup6() eq o(BA,sup6() .解析(1)eq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6()eq o(AC,sup6()eq o(CA,sup6()0.(2)eq o(OA,sup6()eq o(OC,sup6()eq o(BO,sup6()eq o(CO,sup6()(eq o(CO,sup6()eq o(OA,sup6()(eq o(

6、BO,sup6()eq o(OC,sup6()eq o(CA,sup6()eq o(BC,sup6()eq o(BA,sup6().7已知在菱形ABCD中，DAB60，|eq o(AB,sup6()|1，则|eq o(BC,sup6()eq o(CD,sup6()| 1 .解析在ABD中，ADAB1，DAB60，则BD1，所以|eq o(BC,sup6()eq o(CD,sup6()|eq o(BD,sup6()|18.如图所示，若P为ABC的外心，且eq o(PA,sup6()eq o(PB,sup6()eq o(PC,sup6()，则ACB 120 .解析因为P为ABC的外心，所以PAPB

7、PC，因为eq o(PA,sup6()eq o(PB,sup6()eq o(PC,sup6()，由向量的线性运算可得四边形PACB是菱形，且PAC60，所以ACB120.三、解答题9化简下列各式：(1)eq o(AB,sup6()eq o(DF,sup6()eq o(CD,sup6()eq o(BC,sup6()eq o(FA,sup6()；(2)(eq o(AB,sup6()eq o(MB,sup6()eq o(BO,sup6()eq o(OM,sup6().解析(1)eq o(AB,sup6()eq o(DF,sup6()eq o(CD,sup6()eq o(BC,sup6()eq o(F

8、A,sup6()eq o(AB,sup6()eq o(BC,sup6()eq o(CD,sup6()eq o(DF,sup6()eq o(FA,sup6()0.(2)方法一：(eq o(AB,sup6()eq o(MB,sup6()eq o(BO,sup6()eq o(OM,sup6()eq o(AB,sup6()eq o(BO,sup6()eq o(OM,sup6()eq o(MB,sup6()eq o(AB,sup6().方法二：(eq o(AB,sup6()eq o(MB,sup6()eq o(BO,sup6()eq o(OM,sup6()eq o(AB,sup6()(eq o(MB,s

9、up6()eq o(BO,sup6()eq o(OM,sup6()eq o(AB,sup6()0eq o(AB,sup6().方法三：(eq o(AB,sup6()eq o(MB,sup6()eq o(BO,sup6()eq o(OM,sup6()(eq o(AB,sup6()eq o(BO,sup6()eq o(OM,sup6()eq o(MB,sup6()eq o(AM,sup6()eq o(MB,sup6()eq o(AB,sup6().10如图所示，求：(1)ad；(2)cb；(3)ecb；(4)cfb.解析(1)addaeq o(DO,sup6()eq o(OA,sup6()eq o

10、(DA,sup6().(2)cbeq o(CO,sup6()eq o(OB,sup6()eq o(CB,sup6().(3)ecbe(cb)eeq o(CB,sup6()eq o(DC,sup6()eq o(CB,sup6()eq o(DB,sup6().(4)cfbeq o(CO,sup6()eq o(BA,sup6()eq o(OB,sup6()eq o(CA,sup6().B组素养提升一、选择题1已知|eq o(AB,sup6()|10，|eq o(AC,sup6()|7，则|eq o(BC,sup6()|的取值范围是(A)A3,17B(3,17)C(3,10)D3,10解析利用三角形两

11、边之和大于第三边，两边之差小于第三边的性质及eq o(AB,sup6()与eq o(AC,sup6()共线时的情况求解即|eq o(AB,sup6()|eq o(AC,sup6()|eq o(BC,sup6()|eq o(AC,sup6()|eq o(AB,sup6()|，故3|eq o(BC,sup6()|17.2如图所示的方格纸中有定点O，P，Q，E，F，G，H，则eq o(OP,sup6()eq o(OQ,sup6()(C)Aeq o(OH,sup6()Beq o(OG,sup6()Ceq o(FO,sup6()Deq o(EO,sup6()解析eq o(OP,sup6()eq o(OQ

12、,sup6()eq o(FO,sup6().3设P是ABC所在平面内的一点，eq o(BC,sup6()eq o(BA,sup6()2eq o(BP,sup6()，则(C)Aeq o(PA,sup6()eq o(PB,sup6()0Beq o(PB,sup6()eq o(PC,sup6()0Ceq o(PC,sup6()eq o(PA,sup6()0Deq o(PA,sup6()eq o(PB,sup6()eq o(PC,sup6()0解析eq o(BC,sup6()eq o(BA,sup6()2eq o(BP,sup6()，由平行四边形法则，点P为线段AC的中点，eq o(PC,sup6()

13、eq o(PA,sup6()0.故选C4若M为ABC的重心，则下列各向量中与eq o(AB,sup6()共线的是(C)Aeq o(AB,sup6()eq o(BC,sup6()eq o(AC,sup6()Beq o(AM,sup6()eq o(MB,sup6()eq o(BC,sup6()Ceq o(AM,sup6()eq o(BM,sup6()eq o(CM,sup6()D3eq o(AM,sup6()eq o(AC,sup6()解析由三角形重心性质得eq o(AM,sup6()eq o(BM,sup6()eq o(CM,sup6()0.二、填空题5某人在静水中游泳，速度为4eq r(3)

14、km/h.如果他向垂直于河对岸的方向游向河对岸，水的流速为4 km/h，他实际 沿与水流方向成60的(答案不唯一) 方向前进，速度为 8_km/h .解析OB4eq r(3)，OA4，OC8，COA60.6在菱形ABCD中，ABC120，向量|eq o(AB,sup6()|2，则eq blc|rc|(avs4alco1(o(AB,sup6()f(1,2)o(BC,sup6()o(CD,sup6() eq r(3) .解析因为在菱形ABCD中，ABC120，所以BAD60，又ABAD2，所以ABD为等边三角形，因此BD2，连接AC与BD且交于O点，则ABO为Rt，且AB2，BO1，AOBO，所以

15、AOeq r(AB2BO2)eq r(3)，所以eq blc|rc|(avs4alco1(o(AB,sup6()f(1,2)o(BC,sup6()o(CD,sup6()eq blc|rc|(avs4alco1(o(AB,sup6()f(1,2)o(BD,sup6()|eq o(AB,sup6()eq o(BO,sup6()|eq o(AO,sup6()|eq r(3).三、解答题7如图所示，P，Q是ABC的边BC上两点，且BPQC求证：eq o(AB,sup6()eq o(AC,sup6()eq o(AP,sup6()eq o(AQ,sup6().解析eq o(AB,sup6()eq o(AP

16、,sup6()eq o(PB,sup6()，eq o(AC,sup6()eq o(AQ,sup6()eq o(QC,sup6()，eq o(AB,sup6()eq o(AC,sup6()eq o(AP,sup6()eq o(PB,sup6()eq o(AQ,sup6()eq o(QC,sup6().eq o(PB,sup6()与eq o(QC,sup6()大小相等，方向相反，eq o(PB,sup6()eq o(QC,sup6()0.故eq o(AB,sup6()eq o(AC,sup6()eq o(AP,sup6()eq o(AQ,sup6()0eq o(AP,sup6()eq o(AQ,sup6().8.如图所示，已知矩形ABCD中，|eq o(AD,sup6()|4eq r(3)，设eq o(AB,sup6()a，eq o(BC,sup6()b，eq o(BD,sup6()c，试求|abc|的大小解析如图所示，过D作AC的平行线，交BC的延长线于点E.DEAC，ADBE，四