b 向量 p - web.cjcu.edu.twweb.cjcu.edu.tw/~ykchen/physics/vector.pdf · 1 向量...

18
1 向量 向量(vector,有些專科教科書稱為矢量)是一個具有方向性的數值,它具 有兩個基本特性:大小與方向;一般所使用的數值只能表是大小的特性,稱為純 量(scalar)。在數學中,向量是幾何與代數間的橋梁。 向量的表示法 向量有數種表示方式,每種方式都有不同的用途。 向量符號與名稱 在印刷文件中,我們以粗體英文字母,如 A,代表向量的名稱;但在手寫 時,則以上方帶有箭頭的英文字母,如 A 代表向量。 注意: 在手寫時, A r 代表名稱為 A 的向量;A 則代表名稱為 A 的純量,或者 是向量 A r 的大小。在印刷文件中,A 代表名稱為 A 的向量;A (斜體)則代 表名稱為 A 的純量,或者是向量 A 的大小。 圖示法 在圖形中,向量通常使用如下具有箭頭的線段表示: A 線段的長度代表向量的大小,箭頭所指的方向代表向量的方向。這個向量的起點 在沒有箭頭的一端;終點則在有箭頭的一端。 端點法 在某些狀況下,我們也會將向量定義如下: P = AB 以手寫方式表示則是 AB P = r 這代表向量 P 由點 A 到點 B 的向量,也就是以點 A 為起點,以點 B 為終點: 2 P A B 分量法 分量法是向量最標準的表示方法。如下圖所示,在 x-y 座標中,一向量 A 終點開始沿 x 方向前進 4 個單位;沿 y 方向前進 2 個單位後到達終點: 4 2 A x y 在印刷文件終,依據標準的寫法,是: A = 4i + 2j 若是手寫,則是: j i A r r r 2 4 + = 其中, i j (手寫則分別是 i r j r )分別是沿 x y 方向的單位向量(unit vector,也就是大小為 1 的向量) 注意: 不能將 j i A r r r 2 4 + = 寫成 A = 4i + 2j 首先,A 不是向量,其次 i j 也不是單位向量。特別是 i 有可能被誤為 1 或者是電流。

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1

vector

scalar

A A

Ar

A A A A

r A A A

A A

A

P = AB

ABP =r

P A B A B

2

P

A

B

x-y A x 4 y 2

4

2 A

x

y

A = 4i + 2j

jiArrr

24 +=

i j ir

jr x y unit

vector 1

jiArrr

24 +=

A = 4i + 2j

A i j i 1

3

A = 4i + 2j

A = (4, 2)

A = (4i, 2j)

4

24

24

2

A = 4i + 2j i 4 x component

projection x j 2 y y

-3i + 3j4i + 2j

-4i - 2j

i - 4j

4

A = Axi + Ayj

Ax x Ax Ay y Ay x y x y 1 i + 4j 1i + 4j

x y 0

4i -2i

j -4j

0 4i + 0j

i j x y 1 x-y-z k k

r z

A = i -3j + 4k x 1 y -3 y 3 z 4

x

y

z

O

4

1-3

1-3

A = i -3j + 4k

5

x-y x-y-z z = 0 x-y A = 4i + 2j A = 4i +2j + 0k

A = Axi + Ayj + Azk A

x y z

x z

y

i j k x y z

A = Axi + Ayj + Azk AxAy Az scalar component

magnitude x-y

A = Axi + Ayj

22yx AAA +== A

Ax Ay A

6

Ax

Ay

A

x

y

| |

A = Axi + Ayj + Azk

222zyx AAAA ++== A

( ) 22222 zyxzxy AAAAAA ++=+== A

x

y

z

O

Az

AxAy

A = 2i 3j

( ) 133232 22 =+== jiA

7

B = i +4j -6k

( ) 5364164 222 =++=+= kjiA

+x

+x 0

x

y

A

Ax

Ay

A

0

x

y

AA

=tan

x

y

AA1tan=

-90 o 90 o /2 /2

A = 2i + 3j = tan-1(3/2) = 56.3 o

B = 2i 3j = tan-1(-3/2) = -56.3 o

8

> 90 o < -90 o > /2 < -/2 Ax < 0

o180tan 1 = x

y

AA

180 o

C = -2i + 3j tan-1(3/-2) 180 o = 123.7 o -236.3 o

360 o 123.7 o

D = -2i - 3j tan-1(-3/-2) 180 o = -123.7 o 236.3 o

360 o -123.7 o

A = 2i + 3j

x

y

= 56.3

= -56.3

B = 2i - 3j

C = -2i + 3j

D = -2i - 3j

56.3

= 180 - 56.3= 123.7

56.3= -180 + 56.3= -123.7

= 56.3 o = -123.7 o tan = 1.5 = 3/2 = tan-1 (Ay/Ax) = tan-1(3/2) tan = 3/2 Ax 180 o

A = Axi + Ayj

9

tan-1 x Ax

x

y

A

Ax

Ay

A

Ax = Acos

Ay = Asin

A A

A = Axi + Ayj = Acos i + Asin j

5 50 o

A = 5cos50 o i + 5sin50 o j = 3.21i + 3.83j

cos50 o sin50 o

10

x

y A

A = 335

x

y A

A = 335

= 90 - 35 = 65

A y 35 o x +x 90 o - 35 o = 65 o A = 3

A = 3cos65 o i + 3sin65 o j = 1.27i + 2.72j

+x

50

= 180 - 50 = 130

x

y

= 90 + 45 = 135

x

y

45

20

= -90 - 20 = -110

15

= -180 + 15 = -165

11

r b a

a

b

rb

=sin

ra

=cos

ab

=tan

b = rsin Ay = Asin

a = rcos Ax = Acos

b = atan

cos cos

sin2 + cos2 = (b/r)2 + (a/r)2 = (a2 + b2)/r2 = r2/r2 = 1

a2 + b2 = r2

tan = b/a = (b/r)/(a/r) = sin /cos

sin(90 o - ) = cos 90 o -

cos(90 o - ) = sin 90 o -

tan sin cos tan

12

sin0 = 0 b = 0

cos0 = 1 b = 0 a = r

sin90 o = 1 a = 0 b = r

cos90 o = 0 a = 0

0 o 90 o sin cos

x = rcos r

y = rsin r

= 150

r

x

y30

= 150 o II x y cos = x/r sin = y/r |cos150 o | = |x|/r = cos 30 o = sin60 o cos150 o = cos30 o ;| sin150 o |=|y|/r = sin30 o sin150 o = sin30 o = cos60 o

sin210 o = -sin30 o = -cos60 o cos210 o = -cos30 o = -sin60 o

sin(-) = -sin

cos(-) = cos

x cos = x/r y sin = y/r

13

x y

-y

r sin cos -1 1 r tan = 90 o -90 o tan tan = sin /cos cos 0

CBA =

B = AC

cos = a/r a = rcos

= / => =

= / => =

= / => =

= / => =

= / => =

14

position r displacement r velocityv accelerationa forceF momentump impulseI torque

r r = |r| r |r| r

P (x, y, z) O P

r = OP = xi + yj +zk

P position vector P xi + yj + zk (x, y, z)

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

A + B = (Axi + Ayj + Azk) + (Bxi + Byj + Bzk)

= (Ax + Bx)i + (Ay + By)j +(Az + Bz)k

A = 2i -3kB = -i + 2j A + B = (2i + 0j 3k) + (-i + 2j + 0k) = (2 -1)i + (0 + 2)j + (-3 + 0)k = i + 2j 3k

x-y

15

A = 2i + 3j

B = -i +2j

AB

A + B

16

AB

B

B A A + B

A

B B

A

A B A + B B A

AB A B + BC B C = AC A C

17

A = 2.00i + 3.00j m B = i + 2.00j m C = A + B = (2.00 1.00)i + (3.00 + 2.00)j m = 1.00i + 5.00j m

F1 = 2.00i + 3.00j N F2 = -i + 2.00j N net forceresultant force F = F1 + F2 = 1.00i + 5.00j N

F1

F2

F1+F2

A = 2.00 mi + 3.00 mj A = 2.00i + 3.00j m

(2.00 m, 3.00 m) (2.00, 3.00) m

A + B = B + A

(A + B) + C = A + (B + C)

18

A

B

A

B

C

A

B

C

A + B = B + A A + (B + C) = (A + B) + C

A = 2i + 3j 2i 3j

2i 3j A x y vector component 2 3 scalar component

2i

3j

A A A = Axi + Ayj A = -Axi AyjA = 2i 3j -A = -2i + 3j

19

A

-A

Ax

Ay

-Ax

-Ay

i i i i x 1

F12 2 1 F21 1 2 F12 = -F21

A = B A = Axi + Ayj + AzkB = Bxi + Byj + Bzk A = B Ax = BxAy = By Az = Bz

A

BAx

Ay

Bx

By

20

0

0 0 0i + 0j + 0k0

A B A + (-B) A B A = 2i + 3jB = i - 5jA B = A + (-B) = (2i + 3j) + (-i + 5j) = (2 1)i + (3 (-5))j = i + 8j

A B = (Ax Bx)i + (Ay By)j + (Az Bz)k

A B

A

B

-B

A B = A + (-B)

A - B

B -B B A A B = A + (-B) A B B A A B

A B A B B A

A B A B

P1 P2 (x1, y1, z1) (x2, y2, z2) r1 = OP1 = x1i + y1j + z1k r2 = OP2 =

21

x2i + y2j + z2kP2 P1 P1 P2 P1P2 = OP2 OP1 = r2 r1 = (x2 x1)i + (y2 y1)j + (z2 z1)k OP2 = OP1 + P1P2 r21 = r2 r1 P2 P1 P1 P2

(1) A (2, -1) B (3, 2) A B AB = (3 2)i + (2 (-1))j = i + 3j BA = -i 3j AB

(2) OP1 (x1 0)i + (y1 - 0)j + (z1 0)k P1 0

ri rf i f initial final r = rf ri

A vA B vB B A vBA = vB vA

(1) F1F2 F3

equilibrium 0 F = F1 + F2 + F3 = 0 F1 F2 F3 = -F1 F2

(2) r1r2 r3 r1 + r2 + r3 = 0 r1 r2 r3 = - r1 - r2

a A aA A |a| A a < 0

22

A 2A

-2A

A/2

-A/2

aA Aa a A

3i 4j 3i +x 1 -4j +y y 4

F = ma F m

a p = mv v

1 ij k A A A A 1

eA = A/A = A/|A|

A k

keA = kA/A

A = 2i 6j + 2k

A = |A| = ( ) 44262 222 =++

A

eA = A/A = (2i 6j + 2k)/ 44 = 0.302i 0.905j + 0.302k

23

A 5.00

5eA = (5.00)( 0.302i 0.905j + 0.302k) = 1.51i - 4.52j + 1.51k

+x +y O 10.0 km 5.00 km A 50 o 8.00 km B C 20 o 6.00 km

CB CB =

x

y

O

A

B

50 C

10.0 km

5.00

km

20

?

?

(1) A (10.0, 5.00) km (2) OA = 10.0i + 5.00j km (3) AB +x = 90 o + 50 o = 140 o

(4) AB = (8.00 km)(cos140 o i + sin140 o j) = -6.13i + 5.14j km (5) OB = OA + AB = (10.0i + 5.00j) + (-6.13i + 5.14j)

= 3.87i + 10.14j km (6) B (3.87, 10.14) km (7) OC 90 o - 20 o = 70 o (8) OC = (6.00 km)(cos70 o i + sin70 o j) = 2.05i + 5.64j km (9) (2.05, 5.64) km (10) CB = OB OC = (3.87i + 10.14j) (2.05i + 5.64j) = 1.82i + 4.50j km

24

(11) CB = |CB| = 22 50.482.1 + = 4.85 km (12) CB = tan-1(4.50/1.82) = 68 o 68 o CB

x 180 o

+= cos2222 abbac

ba

= sinsin

ab c

A(-2, 3) m B(2, 1) m F = 5.00 N F =

x

y

A

B

A B B A A B BA = OA OB = (-2i + 3j) (2i + j) = -4i + 2j m

AB A B AB = ( ) 2024 22 =+ m AB F e = AB/AB = (-4i + 2j)/ 20 = -0.894i + 0.447j F = Fe = (5.00)(-0.894i + 0.447j) = -4.47i + 2.24j N

BA = tan-1(2/-4) + 180 o = 153.4 o BA F AB F = F(cos153.4 o i + sin153.4 o j) = (5.00)( cos153.4 o i + sin153.4 o j) = -4.47i + 2.24j N

25

x

y

A

B

153.4

A (1, 3, -2) mB (2, 0, 5) m F 5.00 NF = -0.651i + 1.95j 4.56k N

= cosABBA

A = |A|B = |B| dot product

scalar product

BA AB

A Bcos B Acos

B

AABcos

B

B

AA

B

26

BA//= BA

AB//= BA

= cos// AA A B = cos// BB B A

A B = 0 ocos = cos0 o = 1 A

B = 90 ocos = cos90 o = 0 0

ij k 1 1 |i| = |j| = |k| = 1

ij k

i j = j i = j k = k j = k i = i k = 0

i i = |i||i|cos0 o = (1)(1)(1) = 1

j j = k k = 1

A = Axi + Ayj + AzkB = Bxi + Byj + Bzk

A B = (Axi + Ayj + Azk) ( Bxi + Byj + Bzk)

= AxBx (i i) + AxBy (i j) + AxBz (i k) + AyBx (j i) + AyBy (j j)

+ AyBz (j k) + AzBx (k i) + AzBy (k j) + AzBz (k k)

= AxBx (i i) + AyBy (j j) + AzBz (k k) 0

= AxBx + AyBy+ AzBz

A = -i + 3kB = 2i + j + 2k A B = (-i + 0j + 3k) (2i + j + 2k) = (-1)(2)

+ (0)(1) + (3)(2) = -2 + 0 + 6 = 4

A B = B A

A B = ABcos A B = ABcos(-) = ABcos F

27

r work W = F r energy F W = 0 F = 0r = 0W 0

A(1, 0, 2)B(0, 2, 3) C(1, 4, -1) A A AB = OB OA = (0-1)i + (2-0)j + (3-2)k = -i + 2j + kAC = OC OA = (1-1)i + (4 0)j + (-1-2)k = 4j 3k

AB AC = (-i + 2j + k) (0i + 4j 3k) = (-1)(0)+(2)(4) +(1)(-3) = 5

AB = |AB| = ( ) 6121 222 =++

AC = |AC| = ( ) 25340 222 =++ = 5

AB AC = |AB||AC|cos

AB AC A

cos = ( )( ) 61

565

==ACABACAB

-1 1

A = = o9.656

1cos 1 =

0 0 -90 o < < 90 o cos < 0 cos 0 0

BA = 0 A B 0 A B BA > 0A B BA < 0A B

A B A B

28

|A B| = ABsin

A = |A|B = |B| A B A B

A B A B A B A B

B A = -(A B)

A B = 0 o 180 o sin o = 0 0 |A B| = ABsin = 0 0 0

29

B A = A B

B A = -(A B)

cross product

vector product

()() = ()

() () = ()

() () = ()

ij k sin0 o = 0

i i = j j = k k = 0 0

i j x y = 90 osin90 o = 1

| i j | = |i||j|sin90 o = (1)(1)(1) = 1

x-y-z i j z 1

i j = k

30

i j = k j k = i k i = j

j i = -k k j =- i i k = -j

A = -i + 3j + 2kB = 2j 3k

A B = (-i + 3j + 2k) (2j 3k) = (-1)(2)(i j) + (-1)(-3)( i k)

+ (3)(2)(j j) + (3)(-3)(j k) + (2)(2)( k j) + (2)(-3)( k k)

= (-1)(2)(i j) + (-1)(-3)( i k) + (3)(-3)(j k) + (2)(2)( k j)

= -2k + 3(-j) 9i +4(-i) = -13i -3j -2k

A B 0 0 A B

(A B) A = (-13i -3j -2k) (-i + 3j + 2k) = (-13)(-1) + (-3)(3)+(-2)(2)

= 13 9 4 = 0

(A B) B = (-13i -3j -2k) (0i + 2j 3k) = (-13)(0) + (-3)(2)+(-2)(-3)

= 0 6 + 6 = 0

31

ijk

ij k

ij k j k j k i +i

i

j+

k

i x j = k

i

j+

k

j x k = i

i

j+

k

k x i = j

i

j

k

j x i = -k

i

j

k

k x j = -i

i

j

k

i x k = -j

ijkijk

i j k i j k i x j = k

i j k i j k j x k = k

i j k i j k k x i = j

i j k i j k k x j = -i

i j k i j k j x i = -k

i j k i j k i x k = -j

i j k i j k

A = Axi + Ayj + AzkB = Axi + Ayj +

Azk

( ) ( ) ( )kjikji

BA xyyxzxxzyzzyzyx

zyx BABABABABABABBBAAA ++==

32

kjikji

BA 2313320

231 =

=

A B 6

A B x-y

x-y z A = 2i jB = -i + 3j A B = 6k k = 5k z z 5k k z x-y

x

y

z

A

B

A x B

x

y

z

B

A

A x B

A B |A B | = ABsin A B ABsin = A Bsin B Asin

33

A

B

A

= ABsin

A

B

A

= Absin/2

= |A B| A B |A B|/2

OABC OAB C (0, 0), (3, 0), (4, 3) (1, 4))

x

y

O(0, 0) A(3, 0)

B(4, 3)

C(1, 4)

OABC OAC ABC OABC (1) OA = 3i (2) OC = i + 4j(3) OA OC = (3i) ( i + 4j) = (3)(4)(i j) = 12k (4) |OA OC | = 12(5) OAC = |OA OC |/2 = 6 (6) AB = OB OA = (4i + 3j) 3i = i + 3j (7) AC = OC OA = (i + 4j) 3i =-2i + 4j (8) AC AB = (-2i + 4j) (i + 3j) = (-2)(3)(k) + (1)(4)(-k) = -10k(9) |AB AC| = 10 (10) ABC = |AB AC |/2 = 5 (11) OABC = OAC + ABC =

6 + 5 = 11 A B A B

34

A 3.00 m D

O

xy

z

A(0, 0, 3) m

B(8, 0, 8) mC(0, 4, 6) m

D

(1) AB = OB OA = (8i + 8k) (3k) = 8i + 5k m (2) AC = OC OA = (4j + 6k) (3k) = 4j + 3k m (3) AD AB AC

AB AC AB AC = (8i + 5k) (4j + 3k) = (8)(4)k + (8)(3)(-j) + (5)(4)(-i) = -20i 24j + 32k m2 AB AC AC AB = -(AB AC) = 20i + 24j - 32k m2 -20i 24j + 32k z 32 AD z z

(4) |AB AC| = ( ) ( ) 7.442000322420 222 ==++ m (5) AD AB AC

eAD = 2000

322420 kjiACAC +

= = -0.447i - 0.537j + 0.716k

(6) AD = (3.00 m)eAD = (3.00 m)( -0.447i - 0.537j + 0.716k) = -1.34i 1.61j + 2.15k m

(7) OD = OA + AD = 3k + (-1.34i 1.61j + 2.15k) = -1.34i 1.61j +5.15k m (8) D (-1.34, -1.61, 5.15) m (9) ABC =

= r Fr F

r F x-y z

35

v = r r

F = qv B q v B

0 0 0

A = (Ax, Ay, Az) A = Axi + Ayj + Azk