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9988777:2.4.2平面向量数量积的坐标表示,模,夹角公开课课件[精编文档]_图文

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一复习引入

??
1.非零向量 a 与 b的数量积的定义是什么?

几何意义是什么?

??

??

a b ?| a || b | cos?,其中?是a 与b的夹角

B

b

o?

B1 ?

| b | cos? a

A

2.平面向量的数量积满足的运算律?
(1)a·b=b·a;
(2)(λa)·b=λ(a·b)=a·(λb);
(3)(a+b)·c=a·c+b·c;
3.设向量a与b都是非零向量,则
(1)a ? b ? a b ? 0; (2)a a ? a 2 或 a ? a a ;
(3) a b ? a b .

3.平面向量的表示方法有几何法和坐标 法,向量的坐标表示,对向量的加、减、 数乘运算带来了很大的方便.若已知向量 a与b的坐标,则其数量积是唯一确定的, 因此,如何用坐标表示向量的数量积就 成为我们需要研究的课题.

探究(一):平面向量数量积的坐标表示

i ? i ?1

y

j ? j ?1 i ? j ?0

b

a

j

oi

x

已知两个非零向量a=(x1,y1),b=(x2,y2), 怎样用a与b的坐标表示a·b?

则:a ? x1i ? y1 j, b ? x2i ? y2 j

a ?b ? (x1i ? y1 j) ? (x2i ? y2 j)

2

2

? x1x2i ? x1y2i ? j ? x2 y1i ? j ? y1y2 j

因为i 2

? 1, i ?

j

?

0,

2
j

?1

所以a ?b ? x1x2 ? y1y2

两个向量的数量积等于它们对应坐标的 乘积的和.

?

?

练习1:已知向量 a ? (3, ?1),b =(1,-2)

求:(1)

?
a

?
b

?? ??
(2) (a? b) (a? b)

??
a b?5

?? ??
(a? b) (a? b) ? 5

变式:已知向量

?
a

?

?
(2, ?3), b

?

(x, 2x),

?
且a

?
b

?

4,

则x=

3

-1 3

探究(二):向量的模和夹角的坐标表示
a ? a ? a 2 或 a ? a ? a;

(1)向量的模



?
a

?

(x,

y), 则

a

2

?

x2

?

y2,或

a

?

x2 ? y2

(2)设 A(x1, y1)、B(x2, y2 ),

则 AB ? (x2 ? x1)2 ?(y2 ? y1)2

(3)平行
设 a ?(x11, y11), b ? (x22 , y22 ), 则 a / /b ? x11 y22 ? x22 y11 ? 0, (b ? 0) (4)垂直 a ? b ? a ? b ? 0
设 a ? (x1, y1),b ? (x2, y2),(a ? 0,b ? 0)则
a ? b ? x1x2 ? y1y2 ? 0

(5)设 a, b 是两个非零向量,其夹角为
θ,若 a ? (x1, y1),b ? (x2, y2那) 么
cosθ如何用坐 标表示?

cos? ? a b ?

x1x2 ? y1 y2

ab

x12 ? y12 x22 ? y22

例题讲解
例1:设a=(5,-7),b=(-6,-4),求a·b及a、b间的 夹角θ(精确到1°)
解 a·b = 5×(-6)+(-7) ×(-4) = -30+28 = -2
a ? 52 ? 72 ? 74, b ? ?? 6?2 ? ?? 4?2 ? 52
cos? ? ? 2 ? ?0.03
74 ? 52
? ? 1.6rad ? 92?

例2:已知向量 a ? (1, 2),b ? (x,1).
(1) 当 a ? 2b与2a ? b平行 时,求x?
(2) 当 a ? 2b与2a ? b垂直 时,求x?
解:a ? 2b ? (1? 2x, 4), 2a ? b ? (2 ? x,3)
(1)若a ? 2b与2a ? b平行,则
3(1? 2x) ? 4(2 ? x) ? 0, 得x ? 1 2
(2)若a ? 2b与2a ? b垂直,则
(1? 2x)(2 ? x) ? 4 ?3 ? 0, 得x= - 2或 7 2

例3 (1) 已知a ? (?1, 3),b ? (1,1),
a与b的夹角? ,求cos?

cos? ? a ?b ? 6 ? 2 .

ab

4

变式:已知向量 a=(λ ,-2), b=(-3,5),若向量a 与b的夹角为钝 角,求λ 的取值范围.
(- 10 , 6) U (6 , + ? ) 35 5

例4 已知A(1,2),B(2,3),C(-2,5),

试判断?ABC的形状,并给出证明. 向量的数量

积是否为零,

y

证明:?AB ? (2是?判1,断3 ?相2应) ? (1,1)

C(-2,5)

的两条线段
AC ? (?2或垂?1直直,5线的?是重2)否要? (?3,3)

B(2,3) ?AB? AC ?1方?法(?之3)一?1?3 ? 0

?AB ? AC
A(1,2) ??ABC是直角三角形. x

0

思考:还有

其他证明方

法吗?

变式:已知?ABC为直角三角形,
?C ? 90?,AB=(1,3),AC=(2,k),求k值?
解:
若?C ? 90 ,则CA ? CB,?CA CB=0, -2? (?1)+(-k)(3-k)=0,
?k=1或2.

练习

已知i=(1,0),j=(0,1),与2i+j垂直的向量是[ B ]

A. 2i-j C. 2i+j

B . i-2j D . i+2j

已知a=(λ,2),b=(-3,5),且a和b的夹角是钝角,则λ的
范围是[ ]A

A. ? ? 10
3
C. ? ? 10
3

B. ? ? 10
3
D. ? ? 10
3

练习
已知a ? ?2m ?1,2 ? m?,若 a ? 10,则m的取值

范围为[ B ]

A. ??1,1?
C. ?? 2, 2?

B. ??1,1?
D. ?- ?,-1?? ?1,???

练习
? ? ? ? 已知a ? 1, 3 ,b ? 3 ?1, 3 ?1 ,则a与b的夹角是多少?

分析:为求a与b夹角,需先求a·b及|a||b|,再结合夹

角θ的范围确定其值. 0≤θ≤π

? ? ? ? 解 a ? 1, 3 ,b ? 3 ?1, 3 ?1

? ? a ? b ? 3 ?1? 3 3 ?1 ? 4

a ? 2, b ? 2 2
记a与b的夹角为θ
cos? ? a ? b ? 2
ab 2

知三角形函 数值求角时, 应注重角的 范围的确定

又0≤θ≤π

?? ? ?
4

练习
已知a=(3,4),b=(4,3),求x,y的值使 (xa+yb)⊥a,且|xa+yb|=1.

???x

?

? ??

y

? ?

24 35 ?5
7

和????x

? ??

y

? ?

?
5 7

24 35

小结

?? ? ? ? ? a ?b ? (x1i ? y1 j ) ? (x2i ? y2 j )

? x1x2 ? y1y2

? a?

? x12 ? y12 , b ?

x22 ? y22 .

A、B两点间的距离公式:已知 A(x1, y1), B(x2, y2 ),

AB ? (x2 ? x1)2 ? ( y2 ? y1)2 ,

小结
cos? ?

x1x2 ? y1 y2 x12 ? y12 ? x22 ? y22

a // b ?x1y2 ? x2 y1 ? 0
a ? b ? x1x2 ? y1y2 ? 0
2.向量的坐标运算沟通了向量与解析几 何的内在联系,解析几何中与角度、距 离、平行、垂直有关的问题,可以考虑 用向量方法来解决.

作业
课本第121页习题2.4A组题6,7,8


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