两向量的数性积(数量积

( )
cos ∠ a, b = 0 ,
( )
若 a = 0 或 b = 0 ,则 a ⊥ b , 若 cos ∠ a, b = 0 ,则 a ⊥ b .
1
( )
2.代数性质(运算律) ① ②
a ⋅b = b ⋅ a
(λ a ) b = λ ( a ⋅ b) = a (λb) (a + b) c = a ⋅ c + b ⋅ c ( ) ( )
PA = a , PB = b , PC = c ，
则
AB = b − a , BC = c − b , CA = a − c ∵ PA ⊥ BC , PB ⊥ CA ， PA ⋅ BC = 0 , PB ⋅ CA = 0 a c − b = 0 , 即 ac = ab
B F
A
a
P
E
(
)
b
D
c
C
b a − c = 0 ，即 ab = bc
( ) ( ) ( )
X a Y a Z a = = =
cos α = cos β = cos γ =
X X +Y2 + Z2
2
Y X +Y2 + Z2
2
Z X +Y2 + Z2
2
且 cos 2 α + cos 2 β + cos 2 γ = 1 证明: ∵ a ⋅ i = X ⋅1 + Y ⋅ 0 + Z ⋅ 0 = X , 又 a ⋅ i = a i cos ∠ a, i = a cos α
P 1P 2 =
2.向量的方向余弦
( x1 − x2 ) + ( y1 − y2 ) + ( z1 − z2 )
2 2
2
方向角:矢量与坐标轴(或坐标矢量)所成的角.
3
⎧ ∠ a, i = α ⎪ ⎪ a 的方向角 ⎨∠ a, j = β ⎪ ⎪∠ a, k = γ ⎩
定理:设非零矢量 a = X i + Y j + Z k ,则 a 的方向余弦为:
X 1 X 2 + Y1Y2 + Z1Z 2
2 X 12 + Y12 + Z12 X 22 + Y22 + Z 2
.
a ⊥ b ⇔ X 1 X 2 + Y1Y2 + Z1Z 2 = 0
例 2.已知三点 A(1,0,0),B(3,1,1),C(2,0,1),求:①.AB 边长,.②.三角形的内角,.③. BC 在 AB 上的射影. 解: ①. AB = (2,1,1) , AB = 4 + 1 + 1 = 6 ,
∴a = b − c
2 2 2
A
(
)
2
= b − 2b ⋅ c + c = b + c − 2 b c cos ∠ b, c22ຫໍສະໝຸດ 22( )
B
a
b
C
即
a = b + c − 2 b c cos ∠ b, c
2 2 2
2
( )
c
所以, a = b + c − 2bc cos A . (2).设 ΔABC 的 BC 和 CA 边上的高 AD、BE 交于一点 P，连结 CP，再设
( )
∴ a cos α = X
即
cos α = cos β = cos γ =
X a Y a Z a
=
X X 2 +Y2 + Z2 Y X 2 +Y2 + Z2 Z X +Y2 + Z2
2
,
同理:
=
,
=
.
且
cos 2 α + cos 2 β + cos 2 γ = 1
另有结论: X = a cos α , Y = a cos β , Z = a cos γ , X : Y : Z = cos α : cos β : cos γ .
(
)
所以
ac = bc ，即 c b − a = 0 ，∴ PC ⊥ AB ，因此 P 在第三条高线 CF 上，命题得证。
(
)
③．设直线 a, b, c 在平面π 内，且 n ⊥ a , n ⊥ b , a, b 相交，分别在直线 a, b, c, n 上取非零矢 量 a , b, c , n , 因为 a, b 相交, a, b, c 共面,则有 c = λ a + μ b 又因为 n ⊥ a , n ⊥ b , 则有 所以,
cos ∠ a, b =
( )
a ⋅b a b
=
X 1 X 2 + Y1Y2 + Z1Z 2
2 2 + Y22 + Z 2 X + Y12 + Z12 X 2 2 1
证明: ∵ a ⋅ b = a b cos ∠ a, b , a b ≠ 0 ,
( )
=
∴ cos ∠ a, b =
推论:
( )
a ⋅b a b
∑
3
i =1
b i2
7
证明: 设 因为
§1.7 两向量的数性积(数量积、内积、点积)
一.定义 两个矢量 a 与 b 的模和它们夹角的余弦的乘积叫做矢量 a 与 b 的数性积， 记做 a ⋅ b ,或 ab . 即 二.性质 1.几何性质: ①(几何意义)设 a, b 均为非零矢量,有 a ⋅ b = a 射影 a b = b 射影 b a . a ⋅ b = a b cos ∠ a, b .
所以点 P 的坐标为 ( −1, 0, 0 ) 或 ( 9, 0, 0 ) . 例 5. 设矢量 a + 3b 垂直于矢量 7 a − 5b ,矢量 a − 4b 垂直于矢量 7 a − 2b ,试求 a 与 b 间的夹 角. 解: 由题意得,
( a + 3b )( 7a − 5b ) = 7 a ( a − 4b )( 7a − 2b ) = 7 a
b b b
而
a ⋅ λb = λb ⋅ a = λ b ⋅ a = λ a ⋅ b ,
( ) ( )
( ) ( )
所以
(λ a ) b = λ ( a ⋅ b) = a (λb) . ( a + b ) c = c 射影 ( a + b ) = c (射影 a + 射影 b )
c c c
③
= c 射影 c a + c 射影 c b = a ⋅ c + b ⋅ c . 推论: λ a + μ b ⋅ c = λ a ⋅ c + μ b ⋅ c . 例 1.证明平行四边形对角线的平方和等于它各边的平方和. 证明:如图,在平行四边形 ABCD 中,设
2
当 a = b 时, a ⋅ a = a ,记做: a ⋅ a = a . ②(正交条件) a ⊥ b ⇔ a ⋅ b = 0 . 证明: ⇒ 设 a ⊥ b ,则 cos ∠ a, b = 0 ,所以 a ⋅ b = 0 .
2
( )
⇐ 设 a ⋅ b = 0 , 即 a ⋅ b = a b cos ∠ a, b = 0 , 则 a = 0 , 或 b = 0 , 或 有
①-②得,
2
2
+ 16a ⋅ b − 15 b = 0
2
2
①
2
− 30a ⋅ b + 8 b = 0
2
②
−23 b + 46a ⋅ b = 0 ,
2 2
即
2ab = b ,
①×15+②×8 得, 161 a − 161 b = 0
2
即
a =b,
所以,
cos ∠ a, b =
( )
a ⋅b a b
1
=
A
② BC = {−1, −1, 0} , BC = 2 , AC = {1, 0,1} , AC = 2
cos ∠ AB, AC =
(
)
AB ⋅ AC AB AC
=
3 3 = 2 6⋅ 2
B
C
∴∠A = ∠ AB, AC =
(
)
π
6 ∴∠B = ∠A =
∵ BC = AC = 2 ,
π
6
, ∠C = π −
即
m + n =2 a + b
(
2
)
}
三. 坐标表示 在空间直角坐标系 o; i, j , k 下, (1). i ⋅ i = 1 , i ⋅ j = 0 , i ⋅ k = 0 , j ⋅ j = 1 , j ⋅ k = 0 , k ⋅ k = 1 , (2).设 a = X 1 i + Y1 j + Z1 k , b = X 2 i + Y2 j + Z 2 k ,则 a ⋅ b = X 1 X 2 + Y1Y2 + Z1Z 2 . 证明:
π
6
−
π
6
=
2π . 3
π⎞ 6 ⎛ . ③.射影 AB BC = BC cos ∠ AB, BC = 2 cos ⎜ π − ⎟ = − 6⎠ 2 ⎝
(
)
例 3. 已知矢量 a 与 x 轴, y 轴所成的方向角分别为α = 60 , β = 120 ,且 a = 2 ,试求 a 的 坐标. 解: 设 a 与 z 轴的夹角为 γ , a 的坐标为 ( X , Y , Z ) ,则
(
)
OA = a , OB = b , OC = m , BA = n ,则 m = a +b , n = a−b
m = a+b