2.3.1 平面向量基本定理3

1、第6课时平面向量基本定理教学过程一、 问题情境 1. 情境:火箭在升空的某一时刻,速度可以分解成竖直向上和水平向前的两个分速度(如下图所示).在力的分解的平行四边形法则中,我们看到一个力可以分解为两个不共线方向的力的和.(图1) 2. 问题:平面内任一向量是否可以用两个不共线的向量来表示?二、 数学建构设e1, e2是平面内两个不共线的向量.活动1请同学们作出向量 QUOTE * MERGEFORMAT =2.5e1+1.5e2.4活动2a是平面内的任一向量,能否通过作图用e1, e2表示呢?5如图2,在平面内任取一点O,作 QUOTE * MERGEFORMAT =e1, QUOTE * M

2、ERGEFORMAT =e2, QUOTE * MERGEFORMAT =a.过点C作平行于OB的直线,交直线OA于M;过点C作平行于OA的直线,交直线OB于N,则有且只有一对实数1, 2,使得 QUOTE * MERGEFORMAT =1e1, QUOTE * MERGEFORMAT =2e2.因为 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT ,所以a=1e1+2e2.(图2)问题1是不是平面内每一个向量都可以分解成两个不共线的向量?这样的分解是否唯一?问题2对于平面上两个不共线的向量e1, e2,是不是平面

3、上所有的向量都可以用它们来表示?6平面向量基本定理如果e1, e2是同一平面内两个不共线的向量,那么对于这一平面内的任一向量a,有且只有一对实数1, 2,使a=1e1+2e2.我们把不共线的向量e1, e2,叫做表示这一平面内所有向量的一组基底.一个平面向量用一组基底e1, e2表示成a=1e1+2e2的形式,我们称它为向量的分解.当e1, e2互相垂直时,就称为向量的正交分解.定理理解(1)基底e1, e2必须不共线;(2)1,2是被e1, e2, a唯一确定的实数对.思考平面向量基本定理与前面所学的向量共线定理,在内容和表述形式上有什么区别和联系?(平面向量基本定理是向量共线定理的推广)三

4、、 数学运用【例1】(教材第75页例1)如图,ABCD的对角线AC和BD交于点M, QUOTE * MERGEFORMAT =a, QUOTE * MERGEFORMAT =b,试用基底a, b表示 QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT 和 QUOTE * MERGEFORMAT .7(见学生用书P45)(例1)处理建议引导学生利用关系式 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT 和 QUOTE * MERGEFORMA

5、T = QUOTE * MERGEFORMAT - QUOTE * MERGEFORMAT ,以及平行四边形的对角线互相平分来求解.规范板书解 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =a+b, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT - QUOTE * MERGEFORMAT =a-b. 平行四边形的对角线互相平分, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT a+ QUOTE

6、* MERGEFORMAT b, QUOTE * MERGEFORMAT =- QUOTE * MERGEFORMAT =- QUOTE * MERGEFORMAT a- QUOTE * MERGEFORMAT b, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT a- QUOTE * MERGEFORMAT b, QUOTE * MERGEFORMAT =- QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT b- QUOTE * MERGEFORMAT a.题后反思根据平面向量基

7、本定理,选择适当的基底,将有关向量用基底线性表示,这样可通过向量运算来解决有关问题.变式在ABCD中, E和F分别是边CD和BC的中点, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT ,其中, R ,则+= QUOTE * MERGEFORMAT .8处理建议引导学生选择适当的基底解决问题;并讨论不同基底的解法,激发学生的创新意识.规范板书解设 QUOTE * MERGEFORMAT =b, QUOTE * MERGEFORMAT =a,则 QUOTE * MERGEFORMAT = QUOTE * MERGEF

8、ORMAT b-a, QUOTE * MERGEFORMAT =b- QUOTE * MERGEFORMAT a, QUOTE * MERGEFORMAT =b-a. QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT 解得= QUOTE * MERGEFORMAT , += QUOTE * MERGEFORMAT .【例2】(教材第75页例3)设e1, e2是平面内的一组基底,如果 QUOTE * MERGEFORMAT =3e1-2e2, QUOTE * MERGEFOR

9、MAT =4e1+e2, QUOTE * MERGEFORMAT =8e1-9e2,求证:A, B, D三点共线.9(见学生用书P46)处理建议引导学生分析:欲证A, B, D共线,只需证明共起点的两个向量共线,让学生讨论“共起点”的必要性,提高学生思维的活跃性.规范板书证明 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =(3e1-2e2)+(4e1+e2)+(8e1-9e2)=15e1-10e2=5(3e1-2e2)=5 QUOTE * MERGEFORMAT ,

10、QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 共线.又 QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 有公共的起点A, A, B, D三点共线.题后反思当出现三点共线时,转化为共起点的两个向量共线问题求解;选择适当的基底是寻求两向量共线的基础,即将共起点的两向量化归到一组基底线性表示,共线问题就很容易解决了.变式设两个非零向量e1和e2不共线.(1) 如果 QUOTE * MERGEFORMAT =e1-e2, QUOTE * MERGEFORMAT =3e1+2e2, QUOTE * MERGEFORMAT =-8e

11、1-2e2,求证:A, C, D三点共线;(2) 如果 QUOTE * MERGEFORMAT =e1+e2, QUOTE * MERGEFORMAT =2e1-3e2, QUOTE * MERGEFORMAT =2e1-ke2,且A, C, D三点共线,求k的值.规范板书解(1) QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =4e1+e2=- QUOTE * MERGEFORMAT (-8e1-2e2)=- QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT 与 QUOTE *

12、 MERGEFORMAT 共线.又 QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 有公共点C, A, C, D三点共线.(2) QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =(e1+e2)+(2e1-3e2)=3e1-2e2. A, C, D三点共线, QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 共线,从而存在实数,使得 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT ,即3e1-2e2=(2e1-

13、ke2), QUOTE * MERGEFORMAT 解得= QUOTE * MERGEFORMAT ,k= QUOTE * MERGEFORMAT .【例3】如图,在ABC中,M是BC的中点,点N在AC上,且AN=2NC, AM与BN相交于点P,求APPM的值.10(见学生用书P46)(例3)处理建议分析题设中的条件,A, P, M三点共线,B, P, N三点共线,可得向量共线,从而可由向量共线定理和平面向量基本定理建立方程组求解.规范板书解法一设 QUOTE * MERGEFORMAT =e1, QUOTE * MERGEFORMAT =e2,则 QUOTE * MERGEFORMAT =

14、QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =-3e2-e1, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =2e1+e2. A, P, M和B, P, N分别共线, 存在实数, , 使得 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT =-3e2-e1, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT =2e1+e2, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMA

15、T - QUOTE * MERGEFORMAT =(+2)e1+(3+)e2.又 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =2e1+3e2, QUOTE * MERGEFORMAT 解得 QUOTE * MERGEFORMAT QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT , APPM=41.解法二设 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT

16、 . QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT ( QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT )= QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT . QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT QUOTE * MERGEFORMAT . B, P, N三点共线, 存在实数t,使得 QUOTE * MERGEFORMAT =t QUOTE * MERGEFORMAT ,即 QUOTE * ME

17、RGEFORMAT - QUOTE * MERGEFORMAT =t( QUOTE * MERGEFORMAT - QUOTE * MERGEFORMAT ), QUOTE * MERGEFORMAT =(1+t) QUOTE * MERGEFORMAT -t QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT 解得= QUOTE * MERGEFORMAT , APPM=41.题后反思解题的关键是由点共线转化为向量共线,体现了转化的数学思想.(例4)*【例4】如图,在OAB中, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT

18、 , QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT ,AD与BC交于点M,设 QUOTE * MERGEFORMAT =a, QUOTE * MERGEFORMAT =b.(1) 用a, b表示 QUOTE * MERGEFORMAT ;(2) 在线段AC上取一点E,在线段BD上取一点F,使EF过M点,设 QUOTE * MERGEFORMAT =p QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT =q QUOTE * MERGEFORMAT ,求证: QUOTE * MERGEFORMAT + QUOTE * MERG

19、EFORMAT =1.规范板书解(1) 设 QUOTE * MERGEFORMAT =ma+nb,则 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =(m-1)a+nb, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =-a+ QUOTE * MERGEFORMAT b. 点A, M, D共线, QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 共线, QUOTE * MERGEFORMAT = Q

20、UOTE * MERGEFORMAT , m+2n=1. QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT - QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT a+nb, QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT - QUOTE * MERGEFORMAT =- QUOTE * MERGEFORMAT a+b. 点C, M, B共线, QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 共线, QUOTE * MERGEFORMAT = QUOTE *

21、 MERGEFORMAT , 4m+n=1.联立可得m= QUOTE * MERGEFORMAT , n= QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT a+ QUOTE * MERGEFORMAT b.(2) QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT a+ QUOTE * MERGEFORMAT b, QUOTE * MERGEFORMAT =-pa+qb, QUOTE * MERGEFORMAT 与 QUOTE * MERGEFORMAT 共线, QUOTE * ME

22、RGEFORMAT = QUOTE * MERGEFORMAT , QUOTE * MERGEFORMAT q-pq=- QUOTE * MERGEFORMAT p,即 QUOTE * MERGEFORMAT + QUOTE * MERGEFORMAT =1.四、 课堂练习 1. 在ABCD中, QUOTE * MERGEFORMAT =a, QUOTE * MERGEFORMAT =b, QUOTE * MERGEFORMAT =3 QUOTE * MERGEFORMAT , M为BC的中点,则 QUOTE * MERGEFORMAT =- QUOTE * MERGEFORMAT a+ QUOTE * MERGEFORMAT b(用a, b表示).提示由 QUOTE * MERGEFORMAT =3 QUOTE * MERGEFORMAT ,得4 QUOTE * MERGEFORMAT =3 QUOTE * MERGEFORMAT =3(a+b).由M是BC的中点得 QUOTE * MERGEFORMAT =a+ QUOTE * MERGEFORMAT b,所以 QUOTE * MERGEFORMAT = QUOTE * MERGEFORMAT (a+b)-(a+ QUOTE * MERGEFORMAT b)=-