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wb9130com:高中数学新人教A版必修4 2.4.2《平面向量数量积的坐标表示、模、夹角》PPT课件_图文

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2.4.2《平面向量数量积
的坐标表示、模、夹角》

1

教学目标
? 1.掌握平面向量数量积运算规律; ? 2.能利用数量积的5个重要性质及数量积运 算规律解决有关问题; ? 3.掌握两个向量共线、垂直的几何判断,会 证明两向量垂直,以及能解决一些简单问 题. ? 教学重点: ? 平面向量数量积及运算规律. ? 教学难点: ? 平面向量数量积的应用 2

一、复习引入
(1) a ? b ? a ? b cos ? ( 2) a ? a ? a 或 a ?
2

a ? a; a ?b a ?b .

a ? b ? a ? b ? 0; cos ? ?

我们学过两向量的和与差可以转 化为它们相应的坐标来运算,那么怎 样用 a和b的坐标表示a ? b呢?

3

二、新课学习
1、平面向量数量积的坐标表示 如图,i是x轴上的单位向量, j 是y 轴上的单位向量, 由于 a ? b ? a ? b cos? 所以
1 1 j ? j ? . . i ?i ?
i ? j ? j ?i ?0 .

y A(x ,y ) 1 1
j

B(x2,y2)
b

a

o i
4

x

下面研究怎样用

a和b的坐标表示a ? b.
设两个非零向量
a

=(x1,y1),

b=(x2,y2),则

a ? x1 i ? y1 j
2

b ? x2 i ? y2 j ,
2

a ? b ? ( x1 i ? y1 j ) ? ( x2 i ? y2 j ) ? x1 x2 i ? x1 y2 i ? j ? x2 y1 i ? j ? y1 y2 j ? x1 x2 ? y1 y2
5

故两个向量的数量积等于它们对应 坐标的乘积的和。即 y A(x ,y )
1 1

a ? b ? x1 x2 ? y1 y2 .

B(x2,y2)
b
j

a
i

o

x

根据平面向量数量积的坐标表示,向 量的数量积的运算可转化为向量的坐标运 算。

6

2、向量的模和两点间的距离公式
(1) a ? a ? a 或 a ?
2

a ? a;

(1)向量的模 设a ? ( x, y ), 则 a ? x ? y , 或 a ?
2 2 2

x ?y ;
2 2

(2)两点间的距离公式 设A(x1 , y1 )、B ( x2 , y2 ), 则 AB ? (x1 ? x2 ) ? (y1 ? y2 )
2 2
7

3、两向量垂直和平行的坐标表示 (1)垂直 a ? b ? a ? b ? 0

设a ? (x1 , y1 ), b ? ( x2 , y2 ), 则 a ? b ? x1 x2 ? y1 y2 ? 0
(2)平行

设a ? (x1 , y1 ), b ? ( x2 , y2 ), 则 a// b ? x1 y2 ? x2 y1 ? 0
8

4、两向量夹角公式的坐标运算
设a与b 的夹角为?(0 ? ? ? 180 ),
? ?

则 cos ? ?

a ?b ab

设a ? (x1 , y1 ), b ?( x2 , y2 ), 且a与b夹角为?, (0 ? ? ? 180 )则 cos ? ?
? ? 2 1 2 1 2 2

x1 x2 ? y1 y2 x ?y ? x ?y
2 1 2 1 2 2 2 2 2 2
9

.

其中 x ? y ? 0, x ? y ? 0.

三、基本技能的形成与巩固
例1 (1)已知a ? (?1,2 ? 3 ), b ? (1,1), 求a ? b, a ? b, a与b的夹角? .

a ? b ? 1 ? 3, a ?b

a ? b ? 2 4 ? 2 3 ? 2(1 ? 3),

1 ? ? ? cos? ? ? ,? 0 ? ? ? 180 ,?? ? 60 . a?b 2
10

(2)已知a ? (2,3), b ? (?2,4), 则(a ? b) ( ? a ? b) ? .
法一: a ? b ? (0,7), a ? b ? (4,?1) ? (a ? b) ( ? a ? b) ? 0 ? 4 ? 7 ? (?1) ? ?7. 法二:(a ? b) ( ? a ? b) ? a ?b
2 2 2 2

? a ? b ? 13 ? 20 ? ?7
练习:课本P1191、2、3.
11

例2 已知A(1,2),B(2,3),C(-2,5),
试判断?ABC的形状,并给出证明.

y

C(-2,5) 证明 :?AB ? (2 ?1,3 ? 2) ? (1,1)
AC ? (?2 ?1,5 ? 2) ? (?3,3)
?AB? AC ? 1? (?3) ? 1? 3 ? 0

B(2,3)
A(1,2) 0
12

? AB ? AC

x

?三角形 ABC是直角三角形 .

练习2:以原点和A(5,2)为两 个顶点作等腰直角三角形OAB, ?B=90?,求点B的坐标.
y 3 7 B 答案:B的坐标为( , ) 2 2 7 3 或( , ? ) 2 2 O

A
x
13

四、逆向及综合运用
例3 (1)已知 a =(4,3),向量 b 是 垂直于 a 的单位向量,求 b .
3? (3)已知a ? (3,0), b ? (k ,5),且a与b的夹角为 , 4 求k的值.
3 4 3 4 答案:( 1 ) b ? ( ,? )或b ? (? , ). 5 5 5 5 14 (2)( 2, 2 2)或( ? 2, ? 2 2);(3)k ? ?5.

(2)已知a ? 10, b ? (1,2),且a // b,求a的坐标.

提高练习
1、已知OA ? (?3,1), OB ? (0,5),且 AC // OB, BC ? AB ,则点C的坐标为
29 C (?3, ) 3

2、已知A(1,2)、B(4、0)、C(8,6)、 D(5,8),则四边形ABCD的形状是矩形 .

b = (-3,2), a = (1,2), 若k a +2 b 与 2 a - 4 b 平行,则k = - 1 .
3、已知
15

小结

1、理解各公式的正向及逆向运用; 2、数量积的运算转化为向量的坐 标运算; 3、掌握平行、垂直、夹角及距离 公式,形成转化技能。
作业

课本P119A组5(1),9,10,11.

16

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