2022年新教材高中數(shù)學(xué)第2章平面向量及其應(yīng)用2從位移的合成到向量的加減法2.2向量的減法素養(yǎng)作業(yè)北師大版必修第二冊(cè)

1、PAGE PAGE 6第二章2.2A組素養(yǎng)自測(cè)一、選擇題1如圖，在平行四邊形ABCD中，下列結(jié)論錯(cuò)誤的是(C)Aeq o(AB,sup6()eq o(DC,sup6()Beq o(AD,sup6()eq o(AB,sup6()eq o(AC,sup6()Ceq o(AB,sup6()eq o(AD,sup6()eq o(BD,sup6()Deq o(AD,sup6()eq o(CB,sup6()0解析A項(xiàng)顯然正確，由平行四邊形法則知B正確；C項(xiàng)中eq o(AB,sup6()eq o(AD,sup6()eq o(DB,sup6()，故C錯(cuò)誤；D項(xiàng)中eq o(AD,sup6()eq o(CB,su

2、p6()eq o(AD,sup6()eq o(DA,sup6()0，故選C2(多選)化簡(jiǎn)以下各式，結(jié)果為零向量的是(ABCD)Aeq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6()Beq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()eq o(CD,sup6()Ceq o(OA,sup6()eq o(OD,sup6()eq o(AD,sup6()Deq o(NQ,sup6()eq o(QP,sup6()eq o(MN,sup6()eq o(MP,sup6()解析Aeq o(AB,sup6()eq o(BC,sup6()eq o(C

3、A,sup6()eq o(AC,sup6()eq o(CA,sup6()eq o(AC,sup6()eq o(AC,sup6()0；Beq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()eq o(CD,sup6()(eq o(AB,sup6()eq o(BD,sup6()(eq o(AC,sup6()eq o(CD,sup6()eq o(AD,sup6()eq o(AD,sup6()0；Ceq o(OA,sup6()eq o(OD,sup6()eq o(AD,sup6()(eq o(OA,sup6()eq o(AD,sup6()eq o(OD,sup6()eq

4、o(OD,sup6()eq o(OD,sup6()0；Deq o(NQ,sup6()eq o(QP,sup6()eq o(MN,sup6()eq o(MP,sup6()eq o(NP,sup6()eq o(PM,sup6()eq o(MN,sup6()eq o(NM,sup6()eq o(NM,sup6()0.3如圖，D，E，F(xiàn)是ABC的邊AB，BC，CA的中點(diǎn)，則eq o(AF,sup6()eq o(DB,sup6()(D)Aeq o(FD,sup6()Beq o(FC,sup6()Ceq o(FE,sup6()Deq o(BE,sup6()解析由圖可知，eq o(AF,sup6()eq o

5、(DB,sup6()eq o(AF,sup6()eq o(AD,sup6()eq o(DF,sup6()eq o(BE,sup6().4已知D，E，F(xiàn)分別是ABC的邊AB，BC，CA的中點(diǎn)，則(A)Aeq o(AD,sup6()eq o(BE,sup6()eq o(CF,sup6()0Beq o(BD,sup6()eq o(CF,sup6()eq o(DF,sup6()0Ceq o(AD,sup6()eq o(CE,sup6()eq o(CF,sup6()0Deq o(BD,sup6()eq o(BE,sup6()eq o(FC,sup6()05O是四邊形ABCD所在平面上任一點(diǎn)，eq o(A

6、B,sup6()eq o(CD,sup6()，且|eq o(OA,sup6()eq o(OB,sup6()|eq o(OC,sup6()eq o(OD,sup6()|，則四邊形ABCD一定為(D)A菱形B任意四邊形C矩形D平行四邊形解析由|eq o(OA,sup6()eq o(OB,sup6()|eq o(OC,sup6()eq o(OD,sup6()|知|eq o(BA,sup6()|eq o(DC,sup6()|，且eq o(AB,sup6()eq o(CD,sup6()故四邊形ABCD是平行四邊形6已知eq o(OA,sup6()a，eq o(OB,sup6()b，eq o(OC,sup

7、6()c，eq o(OD,sup6()d，且四邊形ABCD為平行四邊形，則(B)Aabcd0Babcd0Cabcd0Dabcd0解析如圖，abeq o(OA,sup6()eq o(OB,sup6()eq o(BA,sup6()，cdeq o(OC,sup6()eq o(OD,sup6()eq o(DC,sup6()，又四邊形ABCD為平行四邊形，則eq o(BA,sup6()eq o(CD,sup6()，即eq o(BA,sup6()eq o(CD,sup6()0，所以eq o(BA,sup6()eq o(DC,sup6()0，即abcd0.故選B二、填空題7若非零向量a與b互為相反向量，給出

8、下列結(jié)論：ab；ab；|a|b|；ba.其中所有正確命題的序號(hào)為 .解析非零向量a、b互為相反向量時(shí)，模一定相等，因此不正確8若向量a、b方向相反，且|a|b|1，則|ab| 2 .9如圖，在正六邊形ABCDEF中，與eq o(OA,sup6()eq o(OC,sup6()eq o(CD,sup6()相等的向量有 .eq o(CF,sup6()；eq o(AD,sup6()；eq o(DA,sup6()；eq o(BE,sup6()；eq o(CE,sup6()eq o(BC,sup6()；eq o(CA,sup6()eq o(CD,sup6()；eq o(AB,sup6()eq o(AE,s

9、up6().解析eq o(OA,sup6()eq o(OC,sup6()eq o(CD,sup6()eq o(CA,sup6()eq o(CD,sup6()eq o(CF,sup6()；eq o(CE,sup6()eq o(BC,sup6()eq o(BC,sup6()eq o(CE,sup6()eq o(BE,sup6()eq o(CF,sup6()；eq o(CA,sup6()eq o(CD,sup6()eq o(DA,sup6()eq o(CF,sup6()；eq o(AB,sup6()eq o(AE,sup6()eq o(AD,sup6()eq o(CF,sup6().三、解答題10.

10、如圖，在五邊形ABCDE中，若四邊形ACDE是平行四邊形，且eq o(AB,sup6()a，eq o(AC,sup6()b，eq o(AE,sup6()c，試用a，b，c表示向量eq o(BD,sup6()，eq o(BC,sup6()，eq o(BE,sup6()，eq o(CD,sup6()及eq o(CE,sup6().解析四邊形ACDE是平行四邊形，eq o(CD,sup6()eq o(AE,sup6()c，eq o(BC,sup6()eq o(AC,sup6()eq o(AB,sup6()ba，eq o(BE,sup6()eq o(AE,sup6()eq o(AB,sup6()ca，

11、eq o(CE,sup6()eq o(AE,sup6()eq o(AC,sup6()cb，eq o(BD,sup6()eq o(BC,sup6()eq o(CD,sup6()bac.B組素養(yǎng)提升一、選擇題1在平面上有A、B、C三點(diǎn)，設(shè)meq o(AB,sup6()eq o(BC,sup6()，neq o(AB,sup6()eq o(BC,sup6()，若m與n的長(zhǎng)度恰好相等，則有(C)AA，B，C三點(diǎn)必在一條直線上BABC必為等腰三角形且B為頂角CABC必為直角三角形且B為直角DABC必為等腰直角三角形解析以eq o(BA,sup6()，eq o(BC,sup6()為鄰邊作平行四邊形，則meq

12、 o(AB,sup6()eq o(BC,sup6()eq o(AC,sup6()，neq o(AB,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(AD,sup6()eq o(DB,sup6()，由m，n的長(zhǎng)度相等可知，兩對(duì)角線相等，因此平行四邊形一定是矩形，故選C2下列各式結(jié)果是eq o(AB,sup6()的是(B)Aeq o(AM,sup6()eq o(MN,sup6()eq o(MB,sup6()Beq o(AC,sup6()eq o(BF,sup6()eq o(CF,sup6()Ceq o(AB,sup6()eq o(DC,sup6()eq o(CB,su

13、p6()Deq o(AB,sup6()eq o(FC,sup6()eq o(BC,sup6()解析eq o(AC,sup6()eq o(BF,sup6()eq o(CF,sup6()eq o(AC,sup6()eq o(CF,sup6()eq o(BF,sup6()eq o(AF,sup6()eq o(BF,sup6()eq o(AF,sup6()eq o(FB,sup6()eq o(AB,sup6().3若|eq o(AB,sup6()|8，|eq o(AC,sup6()|5，則|eq o(BC,sup6()|的取值范圍是(C)A3,8B(3,8)C3,13D(3,13)解析由于eq o(B

14、C,sup6()eq o(AC,sup6()eq o(AB,sup6()，則有|eq o(AB,sup6()|eq o(AC,sup6()|eq o(BC,sup6()|eq o(AB,sup6()|eq o(AC,sup6()|，即3|eq o(BC,sup6()|13.4平面上有一個(gè)ABC和一點(diǎn)O，設(shè)eq o(OA,sup6()a，eq o(OB,sup6()b，eq o(OC,sup6()c.又eq o(OA,sup6()，eq o(BC,sup6()的中點(diǎn)分別為D，E，則向量eq o(DE,sup6()等于(B)Aeq f(1,2)(abc)Beq f(1,2)(abc)Ceq f(1

15、,2)(abc)Deq f(1,2)(abc)解析eq o(DE,sup6()eq o(DO,sup6()eq o(OE,sup6()eq f(1,2)aeq f(1,2)(bc)eq f(1,2)(abc)二、填空題5已知O為四邊形ABCD所在平面外的一點(diǎn)，且向量eq o(OA,sup6()，eq o(OB,sup6()，eq o(OC,sup6()，eq o(OD,sup6()滿足eq o(OA,sup6()eq o(OC,sup6()eq o(OB,sup6()eq o(OD,sup6()，則四邊形ABCD的形狀為 平行四邊形 .解析eq o(OA,sup6()eq o(OC,sup6(

16、)eq o(OB,sup6()eq o(OD,sup6()，eq o(OA,sup6()eq o(OD,sup6()eq o(OB,sup6()eq o(OC,sup6()，eq o(DA,sup6()eq o(CB,sup6().|eq o(DA,sup6()|eq o(CB,sup6()|，且DACB，四邊形ABCD是平行四邊形6已知|a|7，|b|2，且ab，則|ab| 5或9 .解析當(dāng)a與b方向相同時(shí)，|ab|a|b|725；當(dāng)a與b方向相反時(shí)，|ab|a|b|729.三、解答題7已知點(diǎn)B是ACDE內(nèi)一點(diǎn)，且eq o(AB,sup6()a，eq o(AC,sup6()b，eq o(AE

17、,sup6()c，試用a、b、c表示向量eq o(CD,sup6()、eq o(BC,sup6()、eq o(BE,sup6()、eq o(CE,sup6()及eq o(BD,sup6().解析四邊形ACDE為平行四邊形eq o(CD,sup6()eq o(AE,sup6()c；eq o(BC,sup6()eq o(AC,sup6()eq o(AB,sup6()ba；eq o(BE,sup6()eq o(AE,sup6()eq o(AB,sup6()ca；eq o(CE,sup6()eq o(AE,sup6()eq o(AC,sup6()cb；eq o(BD,sup6()eq o(BC,sup

18、6()eq o(CD,sup6()bac.8如圖，已知eq o(OA,sup6()a，eq o(OB,sup6()b，eq o(OC,sup6()c，eq o(OD,sup6()d，eq o(OF,sup6()f，試用a，b，c，d，f表示以下向量：(1)eq o(AC,sup6()；(2)eq o(AD,sup6()；(3)eq o(AD,sup6()eq o(AB,sup6()；(4)eq o(AB,sup6()eq o(CF,sup6()；(5)eq o(BF,sup6()eq o(BD,sup6().解析(1)eq o(AC,sup6()eq o(OC,sup6()eq o(OA,sup6()ca.(2)eq o(AD,sup6()eq o(AO,s