新人教版高中數(shù)學(xué)必修第二冊(cè)全套課時(shí)作業(yè)：課時(shí)跟蹤檢測（二） 向量的加法運(yùn)算

1、課時(shí)跟蹤檢測（二） 向量的加法運(yùn)算A級(jí)學(xué)考合格性考試達(dá)標(biāo)練1在四邊形ABCD中，eq o(AB,sup7()eq o(AD,sup7()eq o(AC,sup7()，則四邊形ABCD是()A梯形B矩形C正方形 D平行四邊形解析：選D由平行四邊形法則可得，四邊形ABCD是以AB，AD為鄰邊的平行四邊形故選D.2已知a，b，c是非零向量，則(ac)b，b(ac)，b(ca)，c(ab)，c(ba)中，與向量abc相等的向量的個(gè)數(shù)為()A5 B4C3 D2解析：選A向量加法滿足交換律，所以五個(gè)向量均等于abc.故選A.3向量(eq o(AB,sup7()eq o(MB,sup7()(eq o(BO,

2、sup7()eq o(BC,sup7()eq o(OM,sup7()()A.eq o(BC,sup7() B.eq o(AB,sup7()C.eq o(AC,sup7() D.eq o(AM,sup7()解析：選C(eq o(AB,sup7()eq o(MB,sup7()(eq o(BO,sup7()eq o(BC,sup7()eq o(OM,sup7()(eq o(AB,sup7()eq o(BO,sup7()(eq o(MB,sup7()eq o(BC,sup7()eq o(OM,sup7()eq o(AO,sup7()eq o(MC,sup7()eq o(OM,sup7()(eq o(A

3、O,sup7()eq o(OM,sup7()eq o(MC,sup7()eq o(AM,sup7()eq o(MC,sup7()eq o(AC,sup7().故選C.4.如圖，正六邊形ABCDEF中，eq o(BA,sup7()eq o(CD,sup7()eq o(FE,sup7()()A0 B.eq o(BE,sup7()C.eq o(AD,sup7() D.eq o(CF,sup7()解析：選B連接BE，取BE中點(diǎn)O，連接OF，BF.eq o(CD,sup7()eq o(AF,sup7()，則eq o(BA,sup7()eq o(CD,sup7()eq o(FE,sup7()(eq o(B

4、A,sup7()eq o(AF,sup7()eq o(FE,sup7()eq o(BE,sup7().故選B.5已知向量ab，且|a|b|0，則向量ab的方向()A與向量a的方向相同B與向量a的方向相反C與向量b的方向相同D不確定解析：選A若a和b方向相同，則它們的和的方向應(yīng)該與a(或b)的方向相同；若它們的方向相反，而a的模大于b的模，則它們的和的方向與a的方向相同故選A.6.如圖所示，四邊形ABCD是梯形，ADBC，則eq o(OA,sup7()eq o(BC,sup7()eq o(AB,sup7()_.解析：eq o(OA,sup7()eq o(BC,sup7()eq o(AB,sup7

5、()eq o(OA,sup7()eq o(AB,sup7()eq o(BC,sup7()eq o(OC,sup7().答案：eq o(OC,sup7()7在菱形ABCD中，DAB60，|eq o(AB,sup7()|1，則|eq o(BC,sup7()eq o(DC,sup7()|_.解析：如圖，|eq o(BC,sup7()eq o(DC,sup7()|eq o(AC,sup7()|，在RtAOB中，AB1，OAB30，AC2AO2ABcos 30eq r(3).答案：eq r(3)8若|a|b|2，則|ab|的取值范圍為_，當(dāng)|ab|取得最大值時(shí)，向量ab的方向_解析：由|a|b|ab|a

6、|b|知0|ab|4，當(dāng)|ab|取得最大值時(shí)，向量a，b的方向相同答案：0,4相同9.如圖所示，求：(1)ad；(2)cb；(3)ecb；(4)cfb.解：(1)addaeq o(DO,sup7()eq o(OA,sup7()eq o(DA,sup7()；(2)cbeq o(CO,sup7()eq o(OB,sup7()eq o(CB,sup7()；(3)ecbe(cb)eeq o(CB,sup7()eq o(DC,sup7()eq o(CB,sup7()eq o(DB,sup7()；(4)cfbeq o(CO,sup7()eq o(OB,sup7()eq o(BA,sup7()eq o(CA

7、,sup7().10.如圖，點(diǎn)D，E，F(xiàn)分別為ABC的三邊AB，BC，CA的中點(diǎn)求證：(1)eq o(AB,sup7()eq o(BE,sup7()eq o(AC,sup7()eq o(CE,sup7()；(2)eq o(EA,sup7()eq o(FB,sup7()eq o(DC,sup7()0.證明：(1)由向量加法的三角形法則，eq o(AB,sup7()eq o(BE,sup7()eq o(AE,sup7()，eq o(AC,sup7()eq o(CE,sup7()eq o(AE,sup7()，eq o(AB,sup7()eq o(BE,sup7()eq o(AC,sup7()eq o

8、(CE,sup7().(2)由向量加法的平行四邊形法則，eq o(EA,sup7()eq o(EF,sup7()eq o(ED,sup7()，eq o(FB,sup7()eq o(FE,sup7()eq o(FD,sup7()，eq o(DC,sup7()eq o(DF,sup7()eq o(DE,sup7()，eq o(EA,sup7()eq o(FB,sup7()eq o(DC,sup7()eq o(EF,sup7()eq o(ED,sup7()eq o(FE,sup7()eq o(FD,sup7()eq o(DF,sup7()eq o(DE,sup7()(eq o(EF,sup7()eq

9、 o(FE,sup7()(eq o(ED,sup7()eq o(DE,sup7()(eq o(FD,sup7()eq o(DF,sup7()0000.B級(jí)面向全國卷高考高分練1如圖所示的方格紙中有定點(diǎn)O，P，Q，E，F(xiàn)，G，H，則eq o(OP,sup7()eq o(OQ,sup7()()A.eq o(OH,sup7() B.eq o(OG,sup7()C.eq o(FO,sup7() D.eq o(EO,sup7()解析：選Ceq o(OP,sup7()eq o(OQ,sup7()eq o(FO,sup7().故選C.2已知ABC的三個(gè)頂點(diǎn)A，B，C及平面內(nèi)一點(diǎn)P滿足eq o(PA,sup7

10、()eq o(PB,sup7()eq o(PC,sup7()，則下列結(jié)論中正確的是()AP在ABC的內(nèi)部BP在ABC的邊AB上CP在AB邊所在的直線上DP在ABC的外部解析：選Deq o(PA,sup7()eq o(PB,sup7()eq o(PC,sup7()，根據(jù)平行四邊形法則，如圖，則點(diǎn)P在ABC外故選D.3若在ABC中，eq o(AB,sup7()a，eq o(BC,sup7()b，且|a|b|1，|ab|eq r(2)，則ABC的形狀是()A正三角形 B銳角三角形C斜三角形 D等腰直角三角形解析：選D由于eq o(AB,sup7()|a|1，|eq o(BC,sup7()|b|1，|

11、eq o(AC,sup7()|ab|eq r(2)，所以ABC為等腰直角三角形故選D.4已知|eq o(AB,sup7()|10，|eq o(AC,sup7()|7，則|eq o(BC,sup7()|的取值范圍是()A3,17 B(3,17) C(3,10) D3,10解析：選A利用三角形兩邊之和大于第三邊，兩邊之差小于第三邊的性質(zhì)及eq o(AB,sup7()與eq o(AC,sup7()共線時(shí)的情況求解即|eq o(AB,sup7()|eq o(AC,sup7()|eq o(BC,sup7()|eq o(AC,sup7()|eq o(AB,sup7()|，故3|eq o(BC,sup7()

12、|17.故選A.5在菱形ABCD中，DAB60，向量|eq o(AB,sup7()|1，則|eq o(BC,sup7()eq o(CD,sup7()|_.解析： 在ABD中，ADAB1，DAB60，ABC是等邊三角形，則BD1，則|eq o(BC,sup7()eq o(CD,sup7()|eq o(BD,sup7()|1.答案：16.如圖，已知電線AO與天花板的夾角為60，電線AO所受拉力|F1|24 N. 繩BO與墻壁垂直，所受拉力|F2|12 N，則F1與F2的合力大小為_，方向?yàn)開. 解析：以eq o(OA,sup7()，eq o(OB,sup7()為鄰邊作平行四邊形BOAC，則F1F2

13、F，即eq o(OA,sup7()eq o(OB,sup7()eq o(OC,sup7()，則OAC60，|eq o(OA,sup7()|24，|eq o(AC,sup7()|eq o(OB,sup7()|12，ACO90，|eq o(OC,sup7()|12eq r(3).F1與F2的合力大小為12eq r(3) N，方向?yàn)樨Q直向上答案：12eq r(3) N豎直向上7.如圖所示，AOBBOC120，|eq o(OA,sup7()|eq o(OB,sup7()|eq o(OC,sup7()|，求eq o(OA,sup7()eq o(OB,sup7()eq o(OC,sup7().解：如圖所示

14、，以O(shè)A，OB為鄰邊作平行四邊形OADB，由向量加法的平行四邊形法則知eq o(OA,sup7()eq o(OB,sup7()eq o(OD,sup7().由|eq o(OA,sup7()|eq o(OB,sup7()|，AOB120，知BOD60，|eq o(OB,sup7()|eq o(OD,sup7()|.又COB120，且|eq o(OB,sup7()|eq o(OC,sup7()|.eq o(OD,sup7()eq o(OC,sup7()0，故eq o(OA,sup7()eq o(OB,sup7()eq o(OC,sup7()0.C級(jí)拓展探索性題目應(yīng)用練如圖，已知向量a，b，c，d.(1)求作abcd.(2)設(shè)|a|2，e為單位向量，試探索|ae|的最大值解：(1)在平面內(nèi)任取一點(diǎn)O，作eq o(OA,sup7()a，eq o(A