2022年新教材高中數(shù)學第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2.1向量的加法素養(yǎng)作業(yè)北師大版必修第二冊

1、PAGE PAGE 6第二章2.1A組素養(yǎng)自測一、選擇題1(多選)下列等式中正確的是(ABD)Aa0aBabbaC|ab|a|b|Deq o(AC,sup6()eq o(DC,sup6()eq o(AB,sup6()eq o(BD,sup6()解析當a與b方向不同時，|ab|a|b|.2.如圖，四邊形ABCD是梯形，ADBC，對角線AC與BD相交于點O，則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下列說法正確的個數(shù)為(B)如果非零向量a與b的方向相同或相反，那么ab的方向必與a與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，則A，B，C一定為一個三角形的三個頂點A0B1C2D3解析錯，若ab0，則ab的方向是任意的；正確；錯，當A，B，C三點共線時，也滿足eq o(AB,sup6()eq o(BC,sup6()eq o(CA,sup6()0.4如圖，正六邊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()解析連接CF，取CF中點O，連接OE，OA則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()|，則ABC是(B)A直角三角形B等邊三角形C鈍角三角形D等腰直角三角形解析eq o(AB,sup6()eq o(BC,sup6()eq o(AC,sup6()，則|eq o(AB,sup6()|eq o(BC,sup6()|eq o(AC,sup6()|，則ABC是等邊三角形二、填空題6化簡下

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，則|eq o(BC,sup6()eq o(CD,sup6()| 1 .解析在ABD中，ADAB1，DAB60，則BD1，所以|eq o(BC,sup6()eq o(CD,sup6()|eq o(BD,sup6()|18.如圖所示，若P為ABC的外心，且eq o(PA,sup6()eq o(PB,sup6()eq o(PC,sup6()，則ACB 120 .解析因為P為ABC的外心，所以PAPB

7、PC，因為eq o(PA,sup6()eq o(PB,sup6()eq o(PC,sup6()，由向量的線性運算可得四邊形PACB是菱形，且PAC60，所以ACB120.三、解答題9化簡下列各式：(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如圖所示，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組素養(yǎng)提升一、選擇題1已知|eq o(AB,sup6()|10，|eq o(AC,sup6()|7，則|eq o(BC,sup6()|的取值范圍是(A)A3,17B(3,17)C(3,10)D3,10解析利用三角形兩

11、邊之和大于第三邊，兩邊之差小于第三邊的性質(zhì)及eq o(AB,sup6()與eq o(AC,sup6()共線時的情況求解即|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如圖所示的方格紙中有定點O，P，Q，E，F(xiàn)，G，H，則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設(shè)P是ABC所在平面內(nèi)的一點，eq o(BC,sup6()eq o(BA,sup6()2eq o(BP,sup6()，則(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()，由平行四邊形法則，點P為線段AC的中點，eq o(PC,sup6()

13、eq o(PA,sup6()0.故選C4若M為ABC的重心，則下列各向量中與eq o(AB,sup6()共線的是(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質(zhì)得eq o(AM,sup6()eq o(BM,sup6()eq o(CM,sup6()0.二、填空題5某人在靜水中游泳，速度為4eq r(3)

14、km/h.如果他向垂直于河對岸的方向游向河對岸，水的流速為4 km/h，他實際 沿與水流方向成60的(答案不唯一) 方向前進，速度為 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，則eq blc|rc|(avs4alco1(o(AB,sup6()f(1,2)o(BC,sup6()o(CD,sup6() eq r(3) .解析因為在菱形ABCD中，ABC120，所以BAD60，又ABAD2，所以ABD為等邊三角形，因此BD2，連接AC與BD且交于O點，則ABO為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、解答題7如圖所示，P，Q是ABC的邊BC上兩點，且BPQC求證：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()與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.如圖所示，已知矩形ABCD中，|eq o(AD,sup6()|4eq r(3)，設(shè)eq o(AB,sup6()a，eq o(BC,sup6()b，eq o(BD,sup6()c，試求|abc|的大小解析如圖所示，過D作AC的平行線，交BC的延長線于點E.DEAC，ADBE，四