wwwpi881com:(北师大版)数学必修四:2.6《平面向量数量积的坐标表示》ppt课件(3)_图文

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平面向量数量积 的坐标表示

复习
1、向量加法 三角形法则 a + b = ( x 1+ x 2, y 1+ y 2) 2、向量减法 三角形法则
a+b b

a
b a ma a a-b

a – b = ( x 1 – x 2, y 1 – y 2 )
3、实数与向量的积

ma = (mx1, my1)

4、向量的数量积 a ? b = | a | | b | cos θ 5、共线的充要条件 a ∥ b (a≠0) , 即a、b共线 ? 存在实数m,使b = ma ? x1y2 = x2y1 6、垂直的充要条件 a?b ? a?b=0

b

θ
a

平面向量数量积的坐标表示 中在直角坐标系中,已知两个非零向量a =(x1,y1),

b = (x2,y2), 如何用 a 与 b 的坐标表示 a · b 呢?
设x轴上的单位向量 i,y轴上的单位向量 j 则 i i = | i |2 = 1,j j = | j |2 = 1

? i? j = j ? i = 0

?

B(x2,y2)

y

∵a = x1 i + y1 j ,b = x2 i + y2 j ∴a ·b =( x1 i + y1 j)· (x2 i + y2 j) =x1x2+y1y2.

b
j O i

A(x1,y1)

a
x

=x1x2 i ·i + x1y2i · j + y1x2 j · i + y1y2 j · j

结论:两个向量的数量积等于它们的对应坐标乘积的和 a ·b = x1x2+y1y2
2 2 | a | ? x ? y | ab |= ?x ? y 11 、设 aa == (x , ,则 或 例 、设 ( 5y) , 7), (6, 4),求a · b

2

2

2

2、设a = AB,若A(x1,y1),B(x2,y2),则

| a |?| AB |? ( x2 ? x1 ) ? ( y2 ? y1 )
2

2

即是平面内两点间的距离公式 3、设a = (x1,y1),b = (x2,y2),则

a ⊥b ? x1x2 ? y1 y2 ? 0

例2、已知A(1、2),B(2,3),C(2,5),

求证Δ ABC是直角三角形
证明: ∵AB = (2 - 1,3 - 2)= (1,1)
y
C

AC = (2 - 1 ,5 - 2 )= (3 ,3 )

∴AB ? AC = 1╳(3 )+ 1 ╳ 3 = 0
∴AB⊥AC ∴Δ ABC是直角三角形
O

B

A x

例3、已知正方形OABC的边长为1,点D、E分别为

AB、 BC的中点,求∠DOE的值.
解:以 OA 和 OC 所在直线为坐标轴建立 直角坐标系,如图所示. 则由已知条件,可得
1 OD ? (1 , ) , OE ? ( 1 , 1) 2 2

y

C

E

B

D

O

A

x





1 1 1? ? ? 1 4 OD ? OE ? cos ?DOE ? ? 2 2 5 5 5 | OD || OE | ? 2 2 ? 4 0 ? ?DOE ? 所以 ?DOE ? arccos 2 5

例4 已知四点坐标:A(-1,3)、B(1,1)、C(4,4)、 D(3,5). (1)求证:四边形ABCD是直角梯形; (2)求∠DAB的大小. (1) 证明: AB = (1 – (-1), 1 – 3) = (2, -2), DC = (4 – 3, 4 – 5) = (1, -1), BC = (4 – 1,4 – 1) = (3, 3).
∵ AB = 2DC, ∴ AB//DC. A B

y

D C

x

∵ AB· BC = 2×3 +(-2) ×3 = 0, ∴ AB⊥BC. 又∵ AB≠DC, ∴ ABCD是直角梯形.

(2)解:AD = (3 – (-1), 5 – 3) = (4, 2)

y

D C

| AB |? (1 - (-1)) 2 ? (1 – (3) 2 ? 2 2
| AD | ? (3 - (-1)) 2 ? (5 - (3) 2 ? 2 5

A B

x

AD· AB = 4×2 + 2× (-2) = 4,
4 10 cos ?DAB ? ? ? 10 | AD | ? | AB | 2 5 ? 2 2 AD ? AB

10 ? ?DAB ? arccos 10

例5 M是?OAB中AB边上的中点,且|OA| = |OB|, 利用向量证明: OM ⊥ AB .
O

证明: 设OA = a, OB = b, 则AB = b – a,
1 OM= 2 (a + b). ∵ |OA| = |OB|, ∴ |a| = |b|. 1 ∴ OM· AB = 2 (a + b)(b – a) 1 2 2 = 2 (b – a ) = 0, ∴ OM⊥AB.

a
A
M

b
B





1、数量积的坐标表示 2、垂直的充要条件 3、平面内的两点间距离


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