人教A版必修第二册622向量的减法运算作业

课时作业(三)向量的减法运算一、选择题1．化简eq\o(AB,\s\up14(→))＋eq\o(BD,\s\up14(→))－eq\o(AC,\s\up14(→))－eq\o(CD,\s\up14(→))＝()A．eq\o(AD,\s\up14(→)) B．eq\o(DA,\s\up14(→))C．eq\o(BC,\s\up14(→)) D．0答案：D2．(多项选择)对于菱形ABCD，以下结论正确的选项是()A．eq\o(AB,\s\up14(→))＝eq\o(BC,\s\up14(→))B．|eq\o(AB,\s\up14(→))|＝|eq\o(BC,\s\up14(→))|C．|eq\o(AB,\s\up14(→))－eq\o(CD,\s\up14(→))|＝|eq\o(AD,\s\up14(→))＋eq\o(BC,\s\up14(→))|D．|eq\o(AD,\s\up14(→))＋eq\o(CD,\s\up14(→))|＝|eq\o(CD,\s\up14(→))－eq\o(CB,\s\up14(→))|答案：BCD解析：易知向量eq\o(AB,\s\up14(→))与eq\o(BC,\s\up14(→))的方向是不同的，但它们的模是相等的，所以B结论正确，A结论错误；由于|eq\o(AB,\s\up14(→))－eq\o(CD,\s\up14(→))|＝|eq\o(AB,\s\up14(→))＋eq\o(DC,\s\up14(→))|＝2|eq\o(AB,\s\up14(→))|，|eq\o(AD,\s\up14(→))＋eq\o(BC,\s\up14(→))|＝2|eq\o(BC,\s\up14(→))|，且|eq\o(AB,\s\up14(→))|＝|eq\o(BC,\s\up14(→))|，所以|eq\o(AB,\s\up14(→))－eq\o(CD,\s\up14(→))|＝|eq\o(AD,\s\up14(→))＋eq\o(BC,\s\up14(→))|，即C结论正确；由于|eq\o(AD,\s\up14(→))＋eq\o(CD,\s\up14(→))|＝|eq\o(BC,\s\up14(→))＋eq\o(CD,\s\up14(→))|＝|eq\o(BD,\s\up14(→))|，|eq\o(CD,\s\up14(→))－eq\o(CB,\s\up14(→))|＝|eq\o(CD,\s\up14(→))＋eq\o(BC,\s\up14(→))|＝|eq\o(BD,\s\up14(→))|，所以D结论正确．3．(多项选择)以下结果为零向量的是()A．eq\o(AB,\s\up14(→))－(eq\o(BC,\s\up14(→))＋eq\o(CA,\s\up14(→))) B．eq\o(AB,\s\up14(→))－eq\o(AC,\s\up14(→))＋eq\o(BD,\s\up14(→))－eq\o(CD,\s\up14(→))C．eq\o(OA,\s\up14(→))－eq\o(OD,\s\up14(→))＋eq\o(AD,\s\up14(→)) D．eq\o(NO,\s\up14(→))＋eq\o(OP,\s\up14(→))＋eq\o(MN,\s\up14(→))－eq\o(MP,\s\up14(→))答案：BCD解析：A项，eq\o(AB,\s\up14(→))－(eq\o(BC,\s\up14(→))＋eq\o(CA,\s\up14(→)))＝eq\o(AB,\s\up14(→))－eq\o(BA,\s\up14(→))＝2eq\o(AB,\s\up14(→))；B项，eq\o(AB,\s\up14(→))－eq\o(AC,\s\up14(→))＋eq\o(BD,\s\up14(→))－eq\o(CD,\s\up14(→))＝eq\o(CB,\s\up14(→))＋eq\o(BC,\s\up14(→))＝0；C项，eq\o(OA,\s\up14(→))－eq\o(OD,\s\up14(→))＋eq\o(AD,\s\up14(→))＝eq\o(DA,\s\up14(→))＋eq\o(AD,\s\up14(→))＝0；D项，eq\o(NO,\s\up14(→))＋eq\o(OP,\s\up14(→))＋eq\o(MN,\s\up14(→))－eq\o(MP,\s\up14(→))＝eq\o(NP,\s\up14(→))＋eq\o(PN,\s\up14(→))D．4．在▱ABCD中，假设|eq\o(AB,\s\up14(→))＋eq\o(AD,\s\up14(→))|＝|eq\o(AB,\s\up14(→))－eq\o(AD,\s\up14(→))|，那么有()A．eq\o(AD,\s\up14(→))＝0B．eq\o(AB,\s\up14(→))＝0或eq\o(AD,\s\up14(→))＝0C．四边形ABCD是矩形D．四边形ABCD是菱形答案：C5．假设|eq\o(AB,\s\up14(→))|＝8，|eq\o(AC,\s\up14(→))|＝5，那么|eq\o(BC,\s\up14(→))|的取值范围是()A．[3,8] B．(3,8)C．[3,13] D．(3,13)答案：C二、填空题6．eq\o(OA,\s\up14(→))＝a，eq\o(OB,\s\up14(→))＝b，假设|eq\o(OA,\s\up14(→))|＝12，|eq\o(OB,\s\up14(→))|＝5，且∠AOB＝90°，那么|a－b|＝________.答案：13解析：∵|eq\o(OA,\s\up14(→))|＝12，|eq\o(OB,\s\up14(→))|＝5，∠AOB＝90°，∴|eq\o(OA,\s\up14(→))|2＋|eq\o(OB,\s\up14(→))|2＝|eq\o(AB,\s\up14(→))|2，∴|eq\o(AB,\s\up14(→))|＝13.∵eq\o(OA,\s\up14(→))＝a，eq\o(OB,\s\up14(→))＝b，∴a－b＝eq\o(OA,\s\up14(→))－eq\o(OB,\s\up14(→))＝eq\o(BA,\s\up14(→))，∴|a－b|＝|eq\o(BA,\s\up14(→))|＝13.7．如图，在正六边形ABCDEF中，与eq\o(OA,\s\up14(→))－eq\o(OC,\s\up14(→))＋eq\o(CD,\s\up14(→))相等的向量有________．(填序号)①eq\o(CF,\s\up14(→))；②eq\o(AD,\s\up14(→))；③eq\o(BE,\s\up14(→))；④eq\o(DE,\s\up14(→))－eq\o(FE,\s\up14(→))＋eq\o(CD,\s\up14(→))；⑤eq\o(CE,\s\up14(→))＋eq\o(BC,\s\up14(→))；⑥eq\o(CA,\s\up14(→))－eq\o(CD,\s\up14(→))；⑦eq\o(AB,\s\up14(→))＋eq\o(AE,\s\up14(→)).答案：①④解析：由于四边形ACDF是平行四边形，所以eq\o(OA,\s\up14(→))－eq\o(OC,\s\up14(→))＋eq\o(CD,\s\up14(→))＝eq\o(CA,\s\up14(→))＋eq\o(CD,\s\up14(→))＝eq\o(CF,\s\up14(→))，eq\o(DE,\s\up14(→))－eq\o(FE,\s\up14(→))＋eq\o(CD,\s\up14(→))＝eq\o(CD,\s\up14(→))＋eq\o(DE,\s\up14(→))＋eq\o(EF,\s\up14(→))＝eq\o(CF,\s\up14(→))，eq\o(CE,\s\up14(→))＋eq\o(BC,\s\up14(→))＝eq\o(BC,\s\up14(→))＋eq\o(CE,\s\up14(→))＝eq\o(BE,\s\up14(→))，eq\o(CA,\s\up14(→))－eq\o(CD,\s\up14(→))＝eq\o(DA,\s\up14(→)).由于四边形ABDE是平行四边形，所以eq\o(AB,\s\up14(→))＋eq\o(AE,\s\up14(→))＝eq\o(AD,\s\up14(→)).综上，与eq\o(OA,\s\up14(→))－eq\o(OC,\s\up14(→))＋eq\o(CD,\s\up14(→))相等的向量是①④.三、解答题8.如图，解答以下各题：(1)用a，d，e表示eq\o(DB,\s\up14(→))；(2)用b，c表示eq\o(DB,\s\up14(→))；(3)用a，b，e表示eq\o(EC,\s\up14(→))；(4)用d，c表示eq\o(EC,\s\up14(→)).解：由题图知，eq\o(AB,\s\up14(→))＝a，eq\o(BC,\s\up14(→))＝b，eq\o(CD,\s\up14(→))＝c，eq\o(DE,\s\up14(→))＝d，eq\o(EA,\s\up14(→))＝e，故(1)eq\o(DB,\s\up14(→))＝eq\o(DE,\s\up14(→))＋eq\o(EA,\s\up14(→))＋eq\o(AB,\s\up14(→))＝d＋e＋a＝a＋d＋e.(2)eq\o(DB,\s\up14(→))＝eq\o(CB,\s\up14(→))－eq\o(CD,\s\up14(→))＝－eq\o(BC,\s\up14(→))－eq\o(CD,\s\up14(→))＝－b－c.(3)eq\o(EC,\s\up14(→))＝eq\o(EA,\s\up14(→))＋eq\o(AB,\s\up14(→))＋eq\o(BC,\s\up14(→))＝a＋b＋e.(4)eq\o(EC,\s\up14(→))＝－eq\o(CE,\s\up14(→))＝－(eq\o(CD,\s\up14(→))＋eq\o(DE,\s\up14(→)))＝－c－d.9．△ABC为等腰直角三角形，∠ACB＝90°，M为斜边AB的中点，eq\o(CM,\s\up14(→))＝a，eq\o(CA,\s\up14(→))＝b.求证：(1)|a－b|＝|a|；(2)|a＋(a－b)|＝|b|.证明：如图，在等腰Rt△ABC中，由于M是斜边AB的中点，那么|eq\o(CM,\s\up14(→))|＝|eq\o(AM,\s\up14(→))|，|eq\o(CA,\s\up14(→))|＝|eq\o(CB,\s\up14(→))|.(1)在△ACM中，eq\o(AM,\s\up14(→))＝eq\o(CM,\s\up14(→))－eq\o(CA,\s\up14(→))＝a－b.由|eq\o(AM,\s\up14(→))|＝|eq\o(CM,\s\up14(→))|，得|a－b|＝|a|.(2)由于eq\o(MB,\s\up14(→))＝eq\o(AM,\s\up14(→))＝a－b，所以eq\o(CB,\s\up14(→))＝eq\o(MB,\s\up14(→))－eq\o(MC,\s\up14(→))＝a－b＋a＝a＋(a－b)．由|eq\o(CB,\s\up14(→))|＝|eq\o(CA,\s\up14(→))|，得|a＋(a－b)|＝|b|.10．P，Q是△ABC的边BC上的两点，且eq\o(BP,\s\up14(→))＝eq\o(QC,\s\up14(→))，求证：eq\o(AB,\s\up14(→))＋eq\o(AC,\s\up14(→))＝Aeq\o(P,\s\up14(→))＋eq\o(AQ,\s\up14(→)).证明：如图．eq\o(AB,\s\up14(→))＝eq\o(AP,\s\up14(→))＋eq\o(PB,\s\up14(→))，①eq\o(AC,\s\up14(→))＝eq\o(AQ,\s\up14(→))＋eq\o(QC,\s\up14(→)).②由①＋②，得eq\o(AB,\s\up14(→))＋eq\o(AC,\s\up14(→))＝eq\o(AP,\s\up14(→))＋eq\o(AQ,\s\up14(→))＋eq\o(PB,\s\up14(→))＋eq\o(QC,\s\up14(→)).∵eq\o(BP,\s\up14(→))＝eq\o(QC,\s\up14(→))，∴eq\o(PB,\s\up14(→))＋eq\o(QC,\s\up14(→))＝0，∴eq\o(AB,\s\up14(→))＋eq\o(AC,\s\up14(→))＝eq\o(AP,\s\up14(→))＋eq\o(AQ,\s\up14(→)).11．三个大小相同的力a，b，c作用在同一物体P上，使物体P沿a方向做匀速运动，设eq\o(PA,\s\up14(→))＝a，eq\o(PB,\s\up14(→))＝b，eq\o(PC,\s\up14(→))＝c，推断△ABC的外形．解：由题意，得|a|＝|b|＝|c|，由于合力作用后做匀速运动，故合力为0，即a＋b＋c＝0.所以a＋c＝－b.如图，作平行四边形APCD为菱形．eq\o(PD,\s\up14(→))＝a＋c＝－b，所以∠APC＝120°，同理，∠APB＝∠BPC＝120°，又由于|a|＝|b|＝|c|，所以△ABC为等边三角形．12．(2020·浙江杭州二中高二期中考试)某人在静水中游泳时，速度的大小为2eq\r(3)km/h，当他在水流速度的大小为2km/h的河中游泳时，(1)假如他垂直游向河对岸，那么他实际沿什么方向前进(精确到1°)？实际前进速度的大小为多少？(2)他必需朝哪个方向游，才能沿与水流垂直的方向前进(精确到1°)？实际前进速度的大小为多少？eq\b\lc\(\rc\)(\a\vs4\al\co1(c