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www.629554.com:2017-2018学年高中数学苏教版必修四教学案:第2章 2.2 向量的线性运算 Word版含答案

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第 1 课时 向量的加法 在大型生产车间里,一重物被天车从 A 处搬运到 B 处,如图所示.它的 实际位移 AB ,可以看作水平运动的分位移 AC 与竖直运动的分位移 AD 的合位移. 问题 1:根据物理中位移的合成与分解,你认为 AB , AD , AC 之间 有什么关系? 提示: AB = AC + AD . 问题 2: AD 与 CB 之间有什么关系? 提示: AD = CB . 问题 3:向量 AB , AC , CB 之间有什么关系? 提示: AB = AC + CB . ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? 1.向量加法的定义 求两个向量和的运算叫做向量的加法. 2.向量加法的运算法则 (1)三角形法则: 已知向量 a 和 b, 在平面内任取一点 O, 作 OA =a,AB =b, 则向量 OB 叫做 a 与 b 的和,记作 a+b,即 a+b= OA + AB = OB . (2)平行四边形法则: 已知两个不共线的非零向量 a, b, 作 OA =a,OC =b, 以 OA , ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? OC 为邻边作?OABC,则以 O 为起点的对角线上的向量 OB =a+b, 如 图.这个法则叫做两个向量求和的平行四边形法则. 问题 1:如图, AC = AB + BC =a+b;同理 AC = AD + DC = ??? ? ??? ? ??? ? ??? ? ??? ? ???? b+a.由此你能得出什么结论? 提示:a+b=b+a. 问题 2:如图, AD = AB + BC + CD =a+b+c; AD = AB + ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ? ??? ? ??? ? ??? ? ??? BD =a+(b+c); AD = AC + CD =(a+b)+c.由此你又能得出什 么结论? 提示:a+b+c=a+(b+c)=(a+b)+c. 向量加法的交换律和结合律 (1)交换律:a+b=b+a; (2)结合律:(a+b)+c=a+(b+c); (3)a+0=0+a=a; (4)a+(-a)=(-a)+a=0. 1.向量加法的三角形法则是从位移求和引出的,使用三角形法则特别要注意“首尾相接”, 和向量是从第一个向量的起点指向第二个向量的终点,当两个向量平行(或共线)时,三角形法则 同样适用. 2.向量加法的平行四边形法则是从力的合成引出的,使用该法则关键是将向量 a,b 的起点 移到同一点 A,并以 a,b 为邻边作平行四边形 ABCD,则向量 AC 即为 a+b. ??? ? [例 1] 化简下列各式: (1) PB + OP + BO ; (2)( AB + MB )+ BO + OM ; ??? ? ??? ? ??? ? ??? ? ???? ??? ? ???? ? (3) AB + BC + CD + DE . [思路点拨] 多个向量相加,可以利用向量加法的三角形法则求解,也可直接运算. [精解详析] (1) PB + OP + BO =( OP + PB )+ BO = OB + BO =0; (2)( AB + MB )+ BO + OM =( AB + BO )+( OM + MB )= AO + OB = AB ; (3) AB + BC + CD + DE = AC + CD + DE = AD + DE = AE . [一点通] 在进行向量加法运算时,利用运算律转化为“顺次首尾相接的形式相加”,即 ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ???? ??? ? ???? ? ??? ? ??? ? ???? ? ??? ? ???? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ??? ? ? ??? ? ??? ??? ? ??? ? AB + BC + CD = AD 的形式,计算简捷且不易出错. 1.在平行四边形 ABCD 中, BC + CD + DA =________. 解析: BC + CD + DA =( BC + CD )+ DA = BD + DA = BA . 答案: BA 2.下列各式中结果为 0 的是________. ① AB + BC + CA ;② AB + MA + BO + OM ; ③ OA + OC + BO + CO ;④ AB + CA + BD + DC . 解析:①原式= AC + CA =0; ②原式=( AB + BO )+( OM + MA )= AO + OA =0; ③原式=( BO + OA )+( CO + OC )= BA +0= BA . ④原式=( AB + BD )+( DC + CA )= AD + DA =0. 故①②④符合. 答案:①②④ 3.化简或计算: (1) CD + BC + AB ; (2) AB + DF + CD + BC + FA . 解:(1) CD + BC + AB =( AB + BC )+ CD = AC + CD = AD . (2) AB + DF + CD + BC + FA =( AB + BC )+( CD + DF )+ FA

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