6.2.2 向量的减法运算-【金版教程】2024-2025学年高中数学必修第二册作业与测评课件PPT（人教A版2019）

第六章　平面向量及其应用 6.2　平面向量的运算 6.2.2　向量的减法运算 知识对点练 40分钟综合练 目录 知识对点练 1 2 3 4 5 6 7 8 知识对点练 4 ① 1 2 3 4 5 6 7 8 知识对点练 5 1 2 3 4 5 6 7 8 知识对点练 6 1 2 3 4 5 6 7 8 知识对点练 7 1 2 3 4 5 6 7 8 知识对点练 8 1 2 3 4 5 6 7 8 知识对点练 9 1 2 3 4 5 6 7 8 知识对点练 10 1 2 3 4 5 6 7 8 知识对点练 11 1 2 3 4 5 6 7 8 知识对点练 12 40分钟综合练 一、单项选择题 1．若非零向量a，b互为相反向量，则下列说法错误的是(　　) A．a∥b B．a≠b C．|a|≠|b| D．b＝－a 解析：a，b互为相反向量，则a，b长度相等、方向相反，从而a∥b，|a|＝|b|，b＝－a，a≠b都是正确的．故选C． 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 14 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 15 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 17 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 18 二、多项选择题 6．(2024·安徽淮北阶段练习)对于任意三个向量a，b，c，下列命题中错误的是(　　) A．|a－b|≤|a|－|b| B．|a＋b|≤|a|＋|b| C．若a，b满足|a|<|b|，且a与b反向，则a>b D．若a∥b，b∥c，则a∥c 解析：对于A，|a|－|b|≤|a－b|，故A错误；对于B，|a＋b|≤|a|＋|b|，故B正确；对于C，因为向量不能比较大小，故C错误；对于D，取b＝0，则对于任意的向量a，c，都有a∥b，b∥c，故a∥c不一定成立，故D错误．故选ACD． 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 19 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 20 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 21 ①②③ 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 22 13 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 23 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 24 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 25 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 26 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 27 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 28 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 29 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 30 14．已知向量a，b，c的模分别为3，4，5，则|a－b＋c|的最大值为_____，最小值为_____． 12 0 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 31 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 32 1 2 3 4 5 6 7 8 9 10 40分钟综合练 11 12 13 14 15 33 　　　　　　　　　　　　　 R 知识点一　向量减法的几何意义 1．若平行四边形ABCD的对角线AC和BD相交于点O，且eq \o(OA,\s\up14(→))＝a，eq \o(OB,\s\up14(→))＝b，用a，b表示向量eq \o(BC,\s\up14(→))为(　　) A．a＋b B．－a－b C．－a＋b D．a－b 解析：由平行四边形对角线互相平分的性质知eq \o(OA,\s\up14(→))＝－eq \o(OC,\s\up14(→))，即eq \o(OC,\s\up14(→))＝－a，eq \o(BC,\s\up14(→))＝eq \o(OC,\s\up14(→))－eq \o(OB,\s\up14(→))＝－a－b．故选B． 2．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____(只填序号)． ①eq \o(CF,\s\up14(→))；②eq \o(AD,\s\up14(→))；③eq \o(DA,\s\up14(→))；④eq \o(BE,\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(→))． 解析：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(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(CF,\s\up14(→))；eq \o(CA,\s\up14(→))－eq \o(CD,\s\up14(→))＝eq \o(DA,\s\up14(→))≠eq \o(CF,\s\up14(→))；eq \o(AB,\s\up14(→))＋eq \o(AE,\s\up14(→))＝eq \o(AD,\s\up14(→))≠eq \o(CF,\s\up14(→))． 3．如图，O为△ABC内一点，eq \o(OA,\s\up14(→))＝a，eq \o(OB,\s\up14(→))＝b，eq \o(OC,\s\up14(→))＝c．求作： (1)b＋c－a；(2)a－b－c． 解：(1)如图1，以eq \o(OB,\s\up14(→))，eq \o(OC,\s\up14(→))为邻边作▱OBDC，连接OD，AD，则eq \o(OD,\s\up14(→))＝eq \o(OB,\s\up14(→))＋eq \o(OC,\s\up14(→))＝b＋c，所以b＋c－a＝eq \o(OD,\s\up14(→))－eq \o(OA,\s\up14(→))＝eq \o(AD,\s\up14(→))． (2)由a－b－c＝a－(b＋c)，如图2，作▱OBEC，连接OE，则eq \o(OE,\s\up14(→))＝eq \o(OB,\s\up14(→))＋eq \o(OC,\s\up14(→))＝b＋c，连接AE，则eq \o(EA,\s\up14(→))＝eq \o(OA,\s\up14(→))－eq \o(OE,\s\up14(→))＝a－(b＋c)＝a－b－c． 知识点二　向量的减法运算 4．(2024·河南省许平汝漯联盟高一5月大联考)eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))－eq \o(AD,\s\up14(→))＝(　　) A．eq \o(CD,\s\up14(→)) B．eq \o(DC,\s\up14(→)) C．eq \o(AC,\s\up14(→)) D．eq \o(CA,\s\up14(→)) 解析：eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))－eq \o(AD,\s\up14(→))＝eq \o(AC,\s\up14(→))－eq \o(AD,\s\up14(→))＝eq \o(DC,\s\up14(→))．故选B． 5．[多选]下列各式中化简结果为0的是(　　) A．eq \o(AD,\s\up14(→))－eq \o(OD,\s\up14(→))－eq \o(AO,\s\up14(→)) B．eq \o(AB,\s\up14(→))－eq \o(CD,\s\up14(→))＋eq \o(BD,\s\up14(→))－eq \o(AC,\s\up14(→)) C．eq \o(NQ,\s\up14(→))－eq \o(MP,\s\up14(→))＋eq \o(QP,\s\up14(→))＋eq \o(MN,\s\up14(→)) D．eq \o(AB,\s\up14(→))＋eq \o(DA,\s\up14(→))＋eq \o(BD,\s\up14(→))－eq \o(BC,\s\up14(→))－eq \o(CA,\s\up14(→)) 解析：对于A，eq \o(AD,\s\up14(→))－eq \o(OD,\s\up14(→))－eq \o(AO,\s\up14(→))＝eq \o(AD,\s\up14(→))＋eq \o(DO,\s\up14(→))＋eq \o(OA,\s\up14(→))＝eq \o(AO,\s\up14(→))＋eq \o(OA,\s\up14(→))＝0；对于B，eq \o(AB,\s\up14(→))－eq \o(CD,\s\up14(→))＋eq \o(BD,\s\up14(→))－eq \o(AC,\s\up14(→))＝eq \o(AB,\s\up14(→))＋eq \o(BD,\s\up14(→))－(eq \o(AC,\s\up14(→))＋eq \o(CD,\s\up14(→)))＝eq \o(AD,\s\up14(→))－eq \o(AD,\s\up14(→))＝0；对于C，eq \o(NQ,\s\up14(→))－eq \o(MP,\s\up14(→))＋eq \o(QP,\s\up14(→))＋eq \o(MN,\s\up14(→))＝eq \o(NQ,\s\up14(→))＋eq \o(QP,\s\up14(→))＋eq \o(MN,\s\up14(→))－eq \o(MP,\s\up14(→))＝eq \o(NP,\s\up14(→))＋eq \o(PN,\s\up14(→))＝0；对于D，eq \o(AB,\s\up14(→))＋eq \o(DA,\s\up14(→))＋eq \o(BD,\s\up14(→))－eq \o(BC,\s\up14(→))－eq \o(CA,\s\up14(→))＝eq \o(AB,\s\up14(→))＋eq \o(DA,\s\up14(→))＋eq \o(BD,\s\up14(→))＋eq \o(CB,\s\up14(→))＋eq \o(AC,\s\up14(→))＝(eq \o(AB,\s\up14(→))＋eq \o(BD,\s\up14(→)))＋(eq \o(AC,\s\up14(→))＋eq \o(CB,\s\up14(→)))＋eq \o(DA,\s\up14(→))＝eq \o(AD,\s\up14(→))＋eq \o(AB,\s\up14(→))＋eq \o(DA,\s\up14(→))＝eq \o(AD,\s\up14(→))＋eq \o(DA,\s\up14(→))＋eq \o(AB,\s\up14(→))＝0＋eq \o(AB,\s\up14(→))＝eq \o(AB,\s\up14(→))．故选ABC． 8eq \r(7) 知识点三　向量减法的应用 6．如图所示，已知在矩形ABCD中，|eq \o(AD,\s\up14(→))|＝4eq \r(3)，|eq \o(AB,\s\up14(→))|＝8．设eq \o(AB,\s\up14(→))＝a，eq \o(BC,\s\up14(→))＝b，eq \o(BD,\s\up14(→))＝c，则|a－b－c|＝______． 解析：如图，延长AD至D′，使DD′＝AD，延长AB至B′，使BB′＝AB，连接B′D′，BD′．b＋c＝eq \o(BD′,\s\up14(→))，a－b－c＝a－(b＋c)＝a－eq \o(BD′,\s\up14(→))＝eq \o(BB′,\s\up14(→))－eq \o(BD′,\s\up14(→))＝eq \o(D′B′,\s\up14(→))，则|a－b－c|＝|eq \o(D′B′,\s\up14(→))|＝eq \r(（2×4\r(3)）2＋（2×8）2)＝8eq \r(7)． 7．已知a，b是两个非零向量，且|a|＝|b|＝|a－b|，求eq \f(|a＋b|,|a－b|)． 解：如图，设eq \o(OA,\s\up14(→))＝a，eq \o(OB,\s\up14(→))＝b，则eq \o(BA,\s\up14(→))＝eq \o(OA,\s\up14(→))－eq \o(OB,\s\up14(→))＝a－b， eq \o(OC,\s\up14(→))＝eq \o(OA,\s\up14(→))＋eq \o(OB,\s\up14(→))＝a＋b． ∵|a|＝|b|＝|a－b|， ∴OA＝OB＝BA． ∴△OAB为正三角形，且四边形AOBC是菱形．设△OAB的边长为1，则|a－b|＝|eq \o(BA,\s\up14(→))|＝1，|a＋b|＝2×eq \f(\r(3),2)＝eq \r(3)．∴eq \f(|a＋b|,|a－b|)＝eq \f(\r(3),1)＝eq \r(3)． 8．如图所示，O是▱ABCD的对角线AC，BD的交点，若eq \o(AB,\s\up14(→))＝a，eq \o(DA,\s\up14(→))＝b，eq \o(OC,\s\up14(→))＝c． 证明：b＋c－a＝eq \o(OA,\s\up14(→))． 证明：证法一：因为b＋c＝eq \o(DA,\s\up14(→))＋eq \o(OC,\s\up14(→))＝eq \o(OC,\s\up14(→))＋eq \o(CB,\s\up14(→))＝eq \o(OB,\s\up14(→))，eq \o(OA,\s\up14(→))＋a＝eq \o(OA,\s\up14(→))＋eq \o(AB,\s\up14(→))＝eq \o(OB,\s\up14(→))，所以b＋c＝eq \o(OA,\s\up14(→))＋a，即b＋c－a＝eq \o(OA,\s\up14(→))． 证法二：eq \o(OA,\s\up14(→))＝eq \o(OC,\s\up14(→))＋eq \o(CA,\s\up14(→))＝eq \o(OC,\s\up14(→))＋eq \o(CB,\s\up14(→))＋eq \o(CD,\s\up14(→))＝c＋eq \o(DA,\s\up14(→))＋eq \o(BA,\s\up14(→))＝b＋c－eq \o(AB,\s\up14(→))＝b＋c－a． 证法三：因为c－a＝eq \o(OC,\s\up14(→))－eq \o(AB,\s\up14(→))＝eq \o(OC,\s\up14(→))－eq \o(DC,\s\up14(→))＝eq \o(OD,\s\up14(→))＝eq \o(OA,\s\up14(→))＋eq \o(AD,\s\up14(→))＝eq \o(OA,\s\up14(→))－eq \o(DA,\s\up14(→))＝eq \o(OA,\s\up14(→))－b，所以b＋c－a＝eq \o(OA,\s\up14(→))． 2．(2024·陕西延安期末)下列不能化简为eq \o(PQ,\s\up14(→))的是(　　) A．eq \o(QC,\s\up14(→))－eq \o(QP,\s\up14(→))＋eq \o(CQ,\s\up14(→)) B．eq \o(AB,\s\up14(→))＋(eq \o(PA,\s\up14(→))＋eq \o(BQ,\s\up14(→))) C．(eq \o(AB,\s\up14(→))＋eq \o(PC,\s\up14(→)))＋(eq \o(BA,\s\up14(→))－eq \o(QC,\s\up14(→))) D．eq \o(PA,\s\up14(→))＋eq \o(AB,\s\up14(→))－eq \o(BQ,\s\up14(→)) 解析：对于A，eq \o(QC,\s\up14(→))－eq \o(QP,\s\up14(→))＋eq \o(CQ,\s\up14(→))＝eq \o(PC,\s\up14(→))＋eq \o(CQ,\s\up14(→))＝eq \o(PQ,\s\up14(→))，故A不符合题意；对于B，eq \o(AB,\s\up14(→))＋(eq \o(PA,\s\up14(→))＋eq \o(BQ,\s\up14(→)))＝eq \o(PA,\s\up14(→))＋eq \o(AB,\s\up14(→))＋eq \o(BQ,\s\up14(→))＝eq \o(PB,\s\up14(→))＋eq \o(BQ,\s\up14(→))＝eq \o(PQ,\s\up14(→))，故B不符合题意；对于C，(eq \o(AB,\s\up14(→))＋eq \o(PC,\s\up14(→)))＋(eq \o(BA,\s\up14(→))－eq \o(QC,\s\up14(→)))＝eq \o(AB,\s\up14(→))＋eq \o(BA,\s\up14(→))＋eq \o(PC,\s\up14(→))＋eq \o(CQ,\s\up14(→))＝eq \o(PQ,\s\up14(→))，故C不符合题意；对于D，eq \o(PA,\s\up14(→))＋eq \o(AB,\s\up14(→))－eq \o(BQ,\s\up14(→))＝eq \o(PB,\s\up14(→))－eq \o(BQ,\s\up14(→))≠eq \o(PQ,\s\up14(→))，故D符合题意．故选D． 3．在四边形ABCD中，设eq \o(AB,\s\up14(→))＝a，eq \o(AD,\s\up14(→))＝b，eq \o(BC,\s\up14(→))＝c，则eq \o(DC,\s\up14(→))＝(　　) A．a－b＋c B．b－(a＋c) C．a＋b＋c D．b－a＋c 解析：eq \o(DC,\s\up14(→))＝eq \o(DB,\s\up14(→))＋eq \o(BC,\s\up14(→))＝eq \o(AB,\s\up14(→))－eq \o(AD,\s\up14(→))＋eq \o(BC,\s\up14(→))＝a－b＋c．故选A． 4．若|eq \o(AB,\s\up14(→))|＝5，|eq \o(AC,\s\up14(→))|＝8，则|eq \o(BC,\s\up14(→))|的取值范围是(　　) A．[3，8] B．(3，8) C．[3，13] D．(3，13) 解析：∵|eq \o(BC,\s\up14(→))|＝|eq \o(AC,\s\up14(→))－eq \o(AB,\s\up14(→))|且||eq \o(AC,\s\up14(→))|－|eq \o(AB,\s\up14(→))||≤|eq \o(AC,\s\up14(→))－eq \o(AB,\s\up14(→))|≤|eq \o(AC,\s\up14(→))|＋|eq \o(AB,\s\up14(→))|，∴3≤|eq \o(AC,\s\up14(→))－eq \o(AB,\s\up14(→))|≤13，∴3≤|eq \o(BC,\s\up14(→))|≤13．故选C． 5．在平面上有A，B，C三点，设m＝eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))，n＝eq \o(AB,\s\up14(→))－eq \o(BC,\s\up14(→))，若m与n的长度恰好相等，则有(　　) A．A，B，C三点必在一条直线上 B．△ABC必为等腰三角形且∠B为顶角 C．△ABC必为直角三角形且∠B为直角 D．△ABC必为等腰直角三角形 解析：以BA，BC为邻边作平行四边形，第四个顶点为D，则m＝eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))＝eq \o(AC,\s\up14(→))，n＝eq \o(AB,\s\up14(→))－eq \o(BC,\s\up14(→))＝eq \o(AB,\s\up14(→))－eq \o(AD,\s\up14(→))＝eq \o(DB,\s\up14(→))，由m，n的长度相等可知，两对角线相等，因此平行四边形一定是矩形．故选C． 7．如图，已知D，E，F分别是△ABC的边AB，BC，CA的中点，则(　　) A．eq \o(AD,\s\up14(→))＋eq \o(BE,\s\up14(→))＋eq \o(CF,\s\up14(→))＝0 B．eq \o(BD,\s\up14(→))－eq \o(CF,\s\up14(→))＋eq \o(DF,\s\up14(→))＝0 C．eq \o(AD,\s\up14(→))＋eq \o(CE,\s\up14(→))－eq \o(CF,\s\up14(→))＝0 D．eq \o(BE,\s\up14(→))－eq \o(BD,\s\up14(→))－eq \o(FC,\s\up14(→))＝0 解析：因为D，E，F分别是△ABC的边AB，BC，CA的中点，所以eq \o(AD,\s\up14(→))＝eq \o(DB,\s\up14(→))，eq \o(CF,\s\up14(→))＝eq \o(ED,\s\up14(→))，eq \o(FC,\s\up14(→))＝eq \o(DE,\s\up14(→))，eq \o(FE,\s\up14(→))＝eq \o(DB,\s\up14(→))，所以eq \o(AD,\s\up14(→))＋eq \o(BE,\s\up14(→))＋eq \o(CF,\s\up14(→))＝eq \o(DB,\s\up14(→))＋eq \o(BE,\s\up14(→))＋eq \o(ED,\s\up14(→))＝0，故A正确；eq \o(BD,\s\up14(→))－eq \o(CF,\s\up14(→))＋eq \o(DF,\s\up14(→))＝eq \o(BD,\s\up14(→))＋eq \o(DF,\s\up14(→))－eq \o(CF,\s\up14(→))＝eq \o(BF,\s\up14(→))＋eq \o(FC,\s\up14(→))＝eq \o(BC,\s\up14(→))≠0，故B不正确；eq \o(AD,\s\up14(→))＋eq \o(CE,\s\up14(→))－eq \o(CF,\s\up14(→))＝eq \o(AD,\s\up14(→))＋eq \o(FE,\s\up14(→))＝eq \o(AD,\s\up14(→))＋eq \o(DB,\s\up14(→))＝eq \o(AB,\s\up14(→))≠0，故C不正确；eq \o(BE,\s\up14(→))－eq \o(BD,\s\up14(→))－eq \o(FC,\s\up14(→))＝eq \o(DE,\s\up14(→))－eq \o(DE,\s\up14(→))＝0，故D正确．故选AD． 三、填空题 8．(2024·河南夏邑第一高级中学高一期中)下列四个等式：①a＋b＝b＋a；②－(－a)＝a；③eq \o(BA,\s\up14(→))－eq \o(BC,\s\up14(→))－eq \o(CA,\s\up14(→))＝0；④a＋(－a)＝0，其中正确的是________．(填序号) 解析：由向量的运算律及相反向量的性质可知①②正确；eq \o(BA,\s\up14(→))－eq \o(BC,\s\up14(→))－eq \o(CA,\s\up14(→))＝eq \o(CA,\s\up14(→))－eq \o(CA,\s\up14(→))＝0，故③正确；对于④，向量的加法运算，结果应为向量，故④错误． 9．已知eq \o(OA,\s\up14(→))＝a，eq \o(OB,\s\up14(→))＝b，若|eq \o(OA,\s\up14(→))|＝12，|eq \o(OB,\s\up14(→))|＝5，且∠AOB＝90°，则|a－b|的值为_____． 解析：a，b，a－b构成了一个直角三角形，则|a－b|＝eq \r(|a|2＋|b|2)＝eq \r(122＋52)＝13． 10．已知O为四边形ABCD所在平面内一点，且向量eq \o(OA,\s\up14(→))，eq \o(OB,\s\up14(→))，eq \o(OC,\s\up14(→))，eq \o(OD,\s\up14(→))满足等式eq \o(OA,\s\up14(→))＋eq \o(OC,\s\up14(→))＝eq \o(OB,\s\up14(→))＋eq \o(OD,\s\up14(→))．若E为AC的中点，则eq \f(S△EAB,S△BCD)＝_____． 解析：∵向量eq \o(OA,\s\up14(→))，eq \o(OB,\s\up14(→))，eq \o(OC,\s\up14(→))，eq \o(OD,\s\up14(→))满足等式eq \o(OA,\s\up14(→))＋eq \o(OC,\s\up14(→))＝eq \o(OB,\s\up14(→))＋eq \o(OD,\s\up14(→))，∴eq \o(OA,\s\up14(→))－eq \o(OB,\s\up14(→))＝eq \o(OD,\s\up14(→))－eq \o(OC,\s\up14(→))，即eq \o(BA,\s\up14(→))＝eq \o(CD,\s\up14(→))，则四边形ABCD为平行四边形．∵E为AC的中点，∴E为对角线AC与BD的交点，∴S△EAB＝S△ECB＝S△ADE＝S△DCE，则eq \f(S△EAB,S△BCD)＝eq \f(1,2)． eq \f(1,2) 四、解答题 11．如图，在▱ABCD中，eq \o(AB,\s\up14(→))＝a，eq \o(AD,\s\up14(→))＝b． (1)当a，b满足什么条件时，a＋b与a－b所在的直线互相垂直？ (2)a＋b与a－b有可能为相等向量吗？为什么？ 解：(1)eq \o(AC,\s\up14(→))＝eq \o(AB,\s\up14(→))＋eq \o(AD,\s\up14(→))＝a＋b，eq \o(DB,\s\up14(→))＝eq \o(AB,\s\up14(→))－eq \o(AD,\s\up14(→))＝a－b． 若a＋b与a－b所在的直线互相垂直，则AC⊥BD． 因为当|a|＝|b|时，四边形ABCD为菱形，此时AC⊥BD，故当a，b满足|a|＝|b|时，a＋b与a－b所在的直线互相垂直． (2)不可能．因为▱ABCD的两条对角线不可能平行，所以a＋b与a－b不可能为共线向量，更不可能为相等向量． 12．如图，已知正方形ABCD的边长等于1，eq \o(AB,\s\up14(→))＝a，eq \o(BC,\s\up14(→))＝b，eq \o(AC,\s\up14(→))＝c，试作下列向量并分别求模． (1)a＋b＋c；(2)a－b＋c． 解：(1)由已知得a＋b＝eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))＝eq \o(AC,\s\up14(→))， 又eq \o(AC,\s\up14(→))＝c， ∴如图，延长AC到E， 使|eq \o(CE,\s\up14(→))|＝|eq \o(AC,\s\up14(→))|， 则a＋b＋c＝eq \o(AE,\s\up14(→))， 且|eq \o(AE,\s\up14(→))|＝2eq \r(2)， ∴|a＋b＋c|＝2eq \r(2)． (2)如图，作eq \o(BF,\s\up14(→))＝eq \o(AC,\s\up14(→))，连接CF， ∴a－b＋c＝eq \o(AB,\s\up14(→))－eq \o(AD,\s\up14(→))＋eq \o(AC,\s\up14(→))＝eq \o(DB,\s\up14(→))＋eq \o(AC,\s\up14(→))＝eq \o(DB,\s\up14(→))＋eq \o(BF,\s\up14(→))＝eq \o(DF,\s\up14(→))，且|eq \o(DF,\s\up14(→))|＝2， ∴|a－b＋c|＝2． 13．[多选](2024·内蒙古包头期末)已知A，B，C，D四点不共线，下列等式能判断四边形ABCD为平行四边形的是(　　) A．eq \o(AB,\s\up14(→))＝eq \o(DC,\s\up14(→)) B．eq \o(OB,\s\up14(→))－eq \o(OA,\s\up14(→))＝eq \o(OC,\s\up14(→))－eq \o(OD,\s\up14(→))(O为平面内任意一点) C．eq \o(AB,\s\up14(→))＋eq \o(AD,\s\up14(→))＝eq \o(AC,\s\up14(→)) D．eq \o(OA,\s\up14(→))＋eq \o(OD,\s\up14(→))＝eq \o(OB,\s\up14(→))＋eq \o(OC,\s\up14(→))(O为平面内任意一点) 解析：A，B，C，D四点不共线，对于A，因为eq \o(AB,\s\up14(→))＝eq \o(DC,\s\up14(→))，所以AB∥DC且AB＝DC，所以四边形ABCD为平行四边形，故A符合题意；对于B，因为eq \o(OB,\s\up14(→))－eq \o(OA,\s\up14(→))＝eq \o(OC,\s\up14(→))－eq \o(OD,\s\up14(→))，所以eq \o(AB,\s\up14(→))＝eq \o(DC,\s\up14(→))，所以AB∥DC且AB＝DC，所以四边形ABCD为平行四边形，故B符合题意；对于C，因为eq \o(AB,\s\up14(→))＋eq \o(AD,\s\up14(→))＝eq \o(AC,\s\up14(→))，即eq \o(AB,\s\up14(→))＋eq \o(AD,\s\up14(→))＝eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))，所以eq \o(AD,\s\up14(→))＝eq \o(BC,\s\up14(→))，所以AD∥BC且AD＝BC，所以四边形ABCD为平行四边形，故C符合题意；对于D，因为eq \o(OA,\s\up14(→))＋eq \o(OD,\s\up14(→))＝eq \o(OB,\s\up14(→))＋eq \o(OC,\s\up14(→))，所以eq \o(OA,\s\up14(→))－eq \o(OB,\s\up14(→))＝eq \o(OC,\s\up14(→))－eq \o(OD,\s\up14(→))，所以eq \o(BA,\s\up14(→))＝eq \o(DC,\s\up14(→))，所以四边形ABDC为平行四边形，故D不符合题意．故选ABC． 解析：向量a，b，c的模分别为3，4，5，则向量可共线，又|c|2＝|a|2＋|b|2，则以|c|，|a|，|b|为边长可构成直角三角形，则当a，－b，c同向时，|a－b＋c|最大，所以|a－b＋c|max＝|a|＋|－b|＋|c|＝3＋4＋5＝12；当a，－b，c和为0时，|a－b＋c|最小，由于以|c|，|a|，|b|为边长可构成直角三角形，设a＝eq \o(AB,\s\up14(→))，b＝eq \o(CB,\s\up14(→))，c＝eq \o(CA,\s\up14(→))，所以此时a－b＋c＝eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))＋eq \o(CA,\s\up14(→))＝0，故|a－b＋c|min＝0． 15．如图所示，在平行四边形ABCD中，E，F分别 为边AB和BC的中点，G为AC与BD的交点． (1)若|eq \o(AB,\s\up14(→))|＝|eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))＋eq \o(CD,\s\up14(→))|，则四边形ABCD是什么特 殊的平行四边形？说明理由； (2)化简eq \o(DA,\s\up14(→))－eq \o(BE,\s\up14(→))－eq \o(CG,\s\up14(→))，并在图中作出表示该化简结果的向量． 解：(1)由条件知|eq \o(AB,\s\up14(→))|＝|eq \o(AB,\s\up14(→))＋eq \o(BC,\s\up14(→))＋eq \o(CD,\s\up14(→))|＝|eq \o(AD,\s\up14(→))|，即AB＝AD， 又四边形ABCD是平行四边形，故四边形ABCD是菱形． (2)由平行四边形及三角形中位线的性质可知eq \o(CG,\s\up14(→))＝eq \o(FE,\s\up14(→))． 因为E为AB的中点，所以eq \o(BE,\s\up14(→))＝eq \o(EA,\s\up14(→))， 所以eq \o(DA,\s\up14(→))－eq \o(BE,\s\up14(→))－eq \o(CG,\s\up14(→))＝eq \o(DA,\s\up14(→))－eq \o(EA,\s\up14(→))－eq \o(FE,\s\up14(→))＝eq \o(DA,\s\up14(→))＋eq \o(AE,\s\up14(→))＋eq \o(EF,\s\up14(→))＝eq \o(DF,\s\up14(→))． 作出向量eq \o(DF,\s\up14(→))如图所示． $$