2022年人教A版高中数学必修四2.3《平面向量的基本定理及坐标表示》教学设计 .pdf
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1、名师精编优秀教案2.3 平面向量的基本定理及坐标表示教学设计【教学目标】1了解平面向量基本定理;2理解平面里的任何一个向量都可以用两个不共线的向量来表示,初步掌握应用向量解决实际问题的重要思想方法;3能够在具体问题中适当地选取基底,使其他向量都能够用基底来表达.【导入新课】复习引入:1 实数与向量的积实数 与向量a的积是一个向量,记作:a.(1)|a|=|a|;(2)0时,a与a方向相同;0 时,a与a方向相反;=0 时,a=0.2运算定律结合律:(a)=()a;分配律:(+)a=a+a,(a+b)=a+b.3.向量共线定理向量b与非零向量a共线的充要条件是:有且只有一个非零实数,使b=a.新
2、授课阶段一、平面向量基本定理:如果1e,2e是同一平面内的两个不共线向量,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=11e+22e.探究:(1)我们把不共线向量、叫做表示这一平面内所有向量的一组基底;(2)基底 不惟一,关键是不共线;(3)由定理可将任一向量a在给出基底、的条件下进行分解;(4)基底 给定时,分解形式惟一.1,2是被a,1e,2e唯一确定的数量.二、平面向量的坐标表示如图,在直角坐标系内,我们分别取与x轴、y轴方向相同的两个单位向量i、j作为名师精编优秀教案基底.任作一个向量a,由平面向量基本定理知,有且只有一对实数x、y,使得yjxia 1 1我们把),(y
3、x叫做向量a的(直角)坐标,记作),(yxa 22其中x叫做a在x轴上的坐标,y叫做a在y轴上的坐标,22式叫做向量的坐标表示.与a相等的向量的坐标也为),(yx.特别地,)0,1(i,)1,0(j,)0,0(0.如图,在直角坐标平面内,以原点O为起点作aOA,则点A的位置由a唯一确定.设yjxiOA,则向量OA的坐标),(yx就是点A的坐标;反过来,点A的坐标),(yx也就是向量OA的坐标.因此,在平面直角坐标系内,每一个平面向量都是可以用一对实数唯一表示.三、平面向量的坐标运算(1)若),(11yxa,),(22yxb,则ba),(2121yyxx,ba),(2121yyxx.两个向量和与
4、差的坐标分别等于这两个向量相应坐标的和与差.设基底为i、j,则ba)()(2211jyixjyixjyyixx)()(2121,即ba),(2121yyxx,同理可得ba),(2121yyxx.(2)若),(11yxA,),(22yxB,则1212,yyxxAB.一个向量的坐标等于表示此向量的有向线段的终点坐标减去始点的坐标.AB=OBOA=(x2,y2)-(x1,y1)=(x2 x1,y2 y1).(3)若),(yxa和实数,则),(yxa.实数与向量的积的坐标等于用这个实数乘原来向量的相应坐标.设基底为i、j,则a)(yjxiyjxi,即),(yxa.文档编码:CX8I1Y6V1C3 HH
5、5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA1
6、0H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码
7、:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3
8、HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 Z
9、A10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档
10、编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C
11、3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4名师精编优秀教案例 1 已知 A(x1,y1),B(x2,y2),求AB的坐标.例 2 已知a=(2,1),b=(-3,4),求a+b,a-b,3a+4b的坐标.例 3 已知平面上三点的坐标分别为A(2,1),B(1,3),C(3,4),求点 D的坐标使这四点构成平行四边形四个顶点.解:当平行四边形为
12、ABCD 时,由DCAB,得 D1=(2,2).当平行四边形为ACDB时,得 D2=(4,6),当平行四边形为DACB 时,得 D3=(6,0).例 4 已知三个力1F(3,4),2F(2,5),3F(x,y)的合力1F+2F+3F=0,求3F的坐标.解:由题设1F+2F+3F=0,得:(3,4)+(2,5)+(x,y)=(0,0),即:320,450,xy5,1.xy3F(5,1).例 5 已知a=(2,1),b=(3,4),求ab,ab,3a4b的坐标.解:ab(2,1)+(-3,4)=(1,5),ab(2,1)-(-3,4)=(5,3),3a4b3(2,1)+4(-3,4)=(6,3)+
13、(-12,16)=(6,19).点评:利用平面向量的坐标运算法则直接求解.例 6 已知平行四边形ABCD 的三个顶点A、B、C的坐标分别为(-2,1)、(-1,3)(3,4),求顶点D的坐标.解:设点 D的坐标为(x,y),即 3-x=1,4-y=2.解得 x=2,y=2.所以顶点 D的坐标为(2,2).(1,3)(2,1)(1,2),(3,4)(,)(3,4),ABDCx yxyABDC且(1,2)(3,4).xy文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8
14、I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N
15、4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H
16、3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:C
17、X8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH
18、5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA1
19、0H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码
20、:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4名师精编优秀教案另解:由平行四边形法则可得例 7 经过点(2,3)M的直线分别交x轴、y轴于点,A B,且|3|ABAM,求点,A B的坐标.解:由题设知,,A B M三点共线,且|3|ABAM,设(,0),(0,)A xBy,点M在,A B之间,则有3ABAM,(,)3(2,3)x yx.解之 得:3,3xy,点,A B的坐标分别为(3,0),(0,3).点M不在,A B之间,则有3ABAM,同理,可求得点,A B的坐标分别为3(,0)2
21、,(0,9).综上,点,A B的坐标分别为(3,0),(0,3)或3(,0)2,(0,9).例 8.已知三点(2,3),(5,4),(7,10)ABC,若AMABAC,试求实数的取值范围,使M落在第四象限.解:设点(,)M x y,由题设得(2,3)(3,)(5,7)(35,7)xy,33,4xy,要使M落在第四象限,则330,40 xy,解之得14.例 8 已知向量(8,2),(3,3),(6,12),(6,4)abcp,问是否存在实数,x y z同时满足两个条件:(1);(2)1pxaybzcxyz?如果存在,求出,x y z的值;如果不存在,请说明理由.(2(1),13)(3(1),43
22、)(3,1),BDBABC(1,3)(3,1)(2,2).ODOBBD文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH
23、5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA1
24、0H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码
25、:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3 HH5N4X3L7K6 ZA10H3T1X2K4文档编码:CX8I1Y6V1C3
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