1. Vector Algebra

1

전자기학 (Electromagnetics):

전장과 자장을 연구하는 물리학 분야의 기초 학문.

응용 분야는 전기와 자석을 사용하는 장치.

좁은 의미의 전자기학:

Maxwell Equation을 유도, 이해, 응용하는 학문.

1.1 Introduction

2

1.2 A preview of the book

]m/C[densityeargchVolume:

]m/A[densityCurrent:J

]m/F[10854.8

)(tyPermittivi:]m/F[36/10

)(tyPermeabili:]m/H[104

]m/volt[fieldElectric:E

]m/C[densityfluxElectric:ED

]Gauss000,10m/WbTesla[fluxMagnetic:B

]m/A[fieldMagnetic:/BH

EqAmperet

DJH

EqFaradayt

BE

EqGauss0B

EqGaussD

EquationMaxwell

3v

2

12

90

70

2

2

v

유전율

투자율

cgs 단위 ↔ MKS 단위(cm, g, sec) (m, kg, sec)

3

• A Scalar is a quantity that has only magnitude

ex) 1, -23.56, 30, 2 ∙ ∙ ∙

• A Vector is a quantity that has both magnitude and direction

ex) (1,2.1,3) (1,3) ∙ ∙ ∙

• A Field : 들판, 공간 ∙ ∙ ∙

Scalar 장: ex: 건물 내의 온도.

Vector 장: ex: 대기중의 빗방울 속도.

1.3 Scalars and Vectors

4

1.4 Unit Vector

)9.1(AAA

aAaAaAa

)8.1(AAAA:

)7.1(aAaAaA

)7.1()A,A,A(

)6.1(aAA

)5.1(A||

a:AofVectorUnit

2z

2y

2x

zzyyxxA

2z

2y

2x

zzyyxx

zyx

A

A

절대값

5

Fig. 1.1 (a) Unit vectors ax, ay, and az,

(b) components of A along ax, ay, and az.

6

)5

3,

5

4(

5

)3,4(

||a

:AofVectorUnit

5

34

AAA:

)3,4(A

A

22

2y

2x

절대값

x

y

2

1

3

1 2 3 4

aA

A

7

)3

1,

3

1,

3

1(

3

)1,1,1(

||a

:AofVectorUnit

3

111

AAAA:

)1,1,1(A

A

222

2z

2y

2x

절대값

x

y

z

1

1

1

aA

A

8

1.5 Vector Addition and Subtraction

Vector Addition

)11.1(a)BA(a)BA(a)BA(C

)B,B,B(B

)A,A,A(A

)10.1(BAC

zzzyyyxxx

zyx

zyx

Fig. 1.2 Vector Addition

CA

B

(a)

CA

B

(b)

9

Vector Subtraction

)12.1(a)BA(

a)BA(

a)BA(

)B(A

BAD

zzz

yyy

xxx

Fig. 1.3 Vector Subtraction

A

B

(a)

D

−B

A

B

(b)

D

10

)3,12(

)8,3()5,15(BA

nSubtractioVector

)13,18(

)8,3()5,15(BA

)8,3(B

)5,15(A

AdditionVector

x

y

10

5

15

5 10 15 2520

B = (3,8)A + B = (18,13)

A = (15,5)

x

y

10

5

5 15

A = (15,5)

B = (3,8)

A − B = (12,−3)

−B = (−3,−8)

11

Law Addition Multiplication

Commutative A + B = B + A kA = Ak

Associative A + B + C = (A + B) + C k(jA) = (kj)A

Distributive k A + B = kA + kB

12

1.6 Position and Distance Vectors

The position vector 𝐫𝐩 (or radius vector) of point P is defined as the directed

distance from the origin O to P.

rp = OP = xax + yay + zaz (1.14)

Fig. 1.4 Illustration of position vector

rp = 3ax + 4ay + 5az

13

The distance vector is the displacement from one point to another.

Fig. 1.5 Distance vector 𝐫𝐏𝐐.

)15.1(a)zz(

a)yy(

a)xx(

rrr

zPQ

yPQ

xPQ

PQPQ

O

PQ

𝐫𝐏𝐐

𝐫𝐐 𝐫𝐏

14

)2,2(

)1,3()3,1(

rrr

)3,1(Q),1,3(P

PQPQ

O

P(3,1)Q(1,3)

𝐫𝐏𝐐

𝐫𝐐 𝐫𝐏

𝐫𝐏𝐐=(-2,2)

15

예제 1.1 A = 10,−4,6 , B = 2,1,0 일 경우에

(a) ay 방향의 A의 성분

(b) 3A − B

(c) A + 2B 방향의 unit vector

(a) ay ∙ A = ay ∙ (10ax − 4ay + 6az) = −4

(b) 3A − B=3(10, -4, 6)-(2, 1, 0)

=(30, -12, 18)-(2, 1, 0)

=(28, -13, 18)

(c) C = A + 2B =(10, -4, 6)+2(2, 1, 0)=(14, -2, 6)

aC =C

C=

(14, −2, 6)

142 + (−2)2+62= (0.9113, −0.130, 0.3906)

16

예제 1.2 Point P(0,2,4), Point Q(-3,1,5) 일 때 다음을 구하라

(a) OP

(b) PQ

(c) PQ

(d) 10aPQ

(a) OP = (0,2,4)

(b) PQ =(-3,1,5)-(0,2,4)=(-3,-1,1)

(c) PQ = (−3)2+(−1)2+12 = 3.317

(d) 10aPQ = (−9.045, −3.015, 3.015)

17

예제 1.3 Boat가 강물을 따라 남동쪽으로 10 km/hr로 움직이고 그 위에서 사람이 Boat의 가는 방향의 오른 쪽 수직으로 2 km/hr로 움직일 때 그 사람의 절대 속도와 가는 방향은?

북 y

동 x

배 10km/hr

사람 2 km/hr

45°=π/4

10.2 km/hr

hr/km2.10uu

)485.8,657.5(

)414.1,414.1()071.7,071.7(uu

)414.1,414.1()4

sin,4

cos(2u

)071.7,071.7()4

sin,4

(cos10u

Boat

mb

mb

m

b

절대속도사람의

상대속도사람의

속도의

θ

18

4/

1

1

2

4/

2/

21

32/

3/

6/

2

3

3sin

2

1

3cos

2

1

6sin

2

3

6cos

2

1

4sin

2

1

4cos

19

북 y

동 x

θ

10.2 km/hr

예제 1.3 Boat가 강물을 따라 남동쪽으로 10 km/hr로 움직이고 그 위에서 사람이 Boat의 가는 방향의 오른 쪽 수직으로 2 km/hr로 움직일 때 그 사람의 절대 속도와 가는 방향은?

o

1

1

mb

mb

3.56

)499.1(tan

)499.1(tan

499.1

657.5

485.8tan

?

hr/km2.10uu

)485.8,657.5(

)414.1,414.1()071.7,071.7(uu

방향가는실제

절대속도사람의

20

1.7 Vector Multiplication

1. Scalar (or dot) product : A ∙ B

2. Vector (or cross) product : A × B

3. Scalar triple product : A ∙ (A × C)

4. Vector triple product : A × (A × C)

21

A. Dot Product

)15.1(cosABBA AB

)16.1(BABABABA)B,B,B(B

)A,A,A(A

zzyyxxzyx

zyx

θAB

aABcosθAB

B

A

22

12

0343

BABABA

12

2

1423

cosABBA

2/1)45cos(

4B

23A

(4,0)B(3,3),A

yyxx

AB

o

2

1

3

1 2 3 4

y

B

A

x

45o

23

)b20.1(1aaaaaa

)a20.1(0aaaaaa

:note (iii)

)19.1(A|A|AA

(1.18)CABA)CB(A

:law veDistributi (ii)

(1.17)ABBA

:law eCommutativ (i)

zzyyxx

xzzyyx

22

24

B. Cross Product

nasinJBBJ

BJam

수직

수직

자장

전류

힘

an θ

25

Fig. 1.7 The cross product of 𝐀 and 𝐁.

)21.1(asinABBA nAB

수직

수직

B~자장

A~전류

A × B~힘

an θAB

26

zxyyx

yxzzx

xyzzy

zyx

zyx

zyx

zyx

zyx

a)BABA(

)b22.1(a)BABA(

a)BABA(

)a22.1(

BBB

AAA

aaa

BA

)B,B,B(B

)A,A,A(A

27

Fig. 1.8 Direction of 𝐀, 𝐁 and 𝐚𝐧 using

(a) the right-hand rule and

(b) the right-handed-screw rule.

28

)12,0,0(

a)12(a)0(a)0(

040

033

aaa

BBB

AAA

aaa

BA

(0,0,12)BA

12

45sin423BA

45

4B

23A

asinABBA

zyx

zyx

zyx

zyx

zyx

o

oAB

nAB

x

an

θAB = 45°

A = (3,3,0)

B = (0,4,0)

A × B = (0,0,12)

y

z

29

zA × B

B × A

B

A

y

x

i It is not commutative ∶ A × B ≠ B × A (1.23a)

It is anti − commutative: A × B = −B × A (1.23b)

30

* A, B, C가 동일 평면에 있을 경우

𝐀 × (𝐁 × 𝐂)

BA

B × C

C

A × B

(𝐀 × 𝐁) × 𝐂

BC

A

ii It is not associative: A × (B × C) ≠ (A × B) × C (1.24)

31

)27.1(

aaa

aaa

aaa

)26.1(0AA

: Note (iv)

(1.25)CABA)CB(A

:vedistributi isIt (iii)

yxz

xzy

zyx

z

ax

ay

a𝑧

x

y

32

Fig. 1.9 Cross product using cyclic permutation

(a) Moving clockwise leads to positive results.

(b) Moving counterclockwise leads to negative results.

33

C. Scalar Triple Product

)29.1(

CCC

BBB

AAA

CCC

BBB

aaa

)aAa.Aa.A(

CCC

BBB

aaa

A)CB(A

)C,C,C(C

)B,B,B(B

)A,AA(A

(1.28))BA(C)AC(B)CB(A

zyx

Zyx

zyx

zyx

zyx

zyx

zzyyxy

zyx

zyx

zyx

zyx

zyx

zy,x

34

zyx

zyx

zyx

zyx

zyx

zyx

zyx

zyx

zyx

BBB

AAA

CCC

AAA

CCC

BBB

CCC

BBB

AAA

)BA(C)AC(B)CB(A

35

D. Vector Triple Product

(1.32))BA(CC)BA(

(1.31))CB(AC)BA(

(1.30))BA(C)CA(B)CB(A

36

1.8 Component of a Vector

Figure 1.10 Components of A along B:

(a) scalar component AB,

(b) vector component AB .

)34.1(a)aA(aAA

)33.1(aAA

cos|a||A|

cosAA

BBBBB

BB

ABB

ABB

θAB

A

BAB

(a)

B

A

A − B

θAB

AB = ABaB

(b)

37

예제 1.4 Vector 𝐀=(3,4,1), 𝐁=(0,2,-5) 일 때 Vector 사이의 각도를 구하라.

o1

222222

AB

73.83)1092.0(cos

1092.0

)5(20143

)5,2,0()1,4,3(

AB

BAcos

38

예제 1.5 Vector P=(2,0,-1), Q=(2,-1,2), R=(2,-3,1) 일 때 다음을 구하라.

(a) (P + Q) × (P − Q) (e) P × (Q × R)

(b) Q ∙ R × P (f) Q와 R에 수직인 unit vector

(c) P ∙ Q × R (g) PQ

(d) sin θQR

)4,12,2(

)2,6,1(2

)2102,2212,0211(2

102

212

aaa

2PQ2

0PQPQ0

QQPQQPPP

)QP(Q)QP(P

)QP()QP()a(

zyx

39

예제 1.5 Vector P=(2,0,-1), Q=(2,-1,2), R=(2,-3,1) 일 때 다음을 구하라.

(b) Q ∙ R × P

14

1246

)6()2(

)4()1(

)3()2(

)2302()2(

)2112()1(

)0113()2(

102

132

212

PRQ

14

)2012()206(

132

212

102

132

212

102

132

212

PRQ

-12 -

0 -

2 -

+ 6

+ 0

+ -2

40

예제 1.5 Vector P=(2,0,-1), Q=(2,-1,2), R=(2,-3,1) 일 때 다음을 구하라.

(c) P ∙ Q × R

)BA(C

)AC(B

)CB(A(1.28)

(cf)

12464010

14

102

132

212

132

212

102

RQP

14

102

132

212

PRQ

41

예제 1.5 Vector P=(2,0,-1), Q=(2,-1,2), R=(2,-3,1) 일 때 다음을 구하라.

(d) sin θQR

(e) P × (Q × R)

)4,3,2(

)204)(1,3,2()104)(2,1,2(

)QP(R)RP(Q)RQ(P

)BA(C)CA(B)CB(A)30.1(Eq

)e(

5976.0149/

132

212

aaa

RQ

RQsin)d(

zyx

RQ

42

예제 1.5 Vector P=(2,0,-1), Q=(2,-1,2), R=(2,-3,1) 일 때 다음을 구하라.

(f) Q와 R에 수직인 unit vector

)596.0,298.0,745.0(

)4(25

)4,2,5(

)4,2,5(

)4,2,5(

RQ

RQa)f(

222

43

예제 1.5 Vector P=(2,0,-1), Q=(2,-1,2), R=(2,-3,1) 일 때 다음을 구하라.

(g) PQ

)2,1,2(9

2

)2)1(2(

)2,1,2)(204(

Q

Q)QP(

Q

Q

Q

QP

a)aP(

acosPP)g(

222

2

QQ

QPQQ

Q

P

θPQ

PQ = PQaQ

44

예제 1.6 다음을 유도 하라(a) cosine 공식: a2 = b2 + c2 − 2bcosA

b sine 공식:sinA

a=

sinB

b=

sinC

c

c

Csin

b

Bsin

a

Asin

Bsinca2

1Asinbc

2

1Csinab

2

1

ac2

1cb

2

1ba

2

1

)b(

면적삼각형

Acosbc2cba

cb2ccbb

)cb()cb(aa

acb

0cba)a(

222

A

b

c

bsinAb

c

a

A

C

B

45

예제 1.6 다음을 유도 하라

(a) cosine 공식: a2 = b2 + c2 − 2bcosA

Acosbc2cb

Acosbc2c)AcosA(sinb

Acosbc2cAcosbAsinb

AcosbAcosbc2cAsinb

)Acosbc()Asinb(a

Pythagoras *

22

2222

22222

22222

222

정리삼각형의빨간

b

c

a

A

C

B

A

ba

bsinA

b cosA c-b cosA

46

예제 1.7 (a) Point P1, P2, P3 가 동일 직선에 있음을 보여라.(b) P4 (3,-1,0)와 직선 사이의 최단 거리.

직선동일

)0,0,0(

1228

614

aaa

rr

)4,3,2(

)4,2,5()8,0,3(

rrr

)6,1,4(

)4,2,5()2,1,1(

rrr)a(

zyx

1312

PP13

PP12

13

12

P1(5,2,-4)

P2(1,1,2)

P3(-3,0,8) r13

r12

47

예제 1.7 (b) P4 (3,-1,0)와 직선 사이의 최단 거리.

426.2

)6,1,4(

)6,1,4()4,3,2(

arsinrd214141 PPPPPP

d=?

P

θP4(3,-1,0)

P2(1,1,2)

P1(5,2,-4) rP1P4

rP1P2

aP1P2

P3(-3,0,8)

48

P1, P2를 잊는 직선의 방정식

방정식개의

때일

2

64z

2y

45x

P2

)64,2,45(

)rr(rr

)rr(r

rrr

3

PPPP

PPPP

PPPP

121

1221

11

P1(5,2,-4)

P2(1,1,2)P

P3(-3,0,8)

rP1P

rP1P2

P4(3,-1,0)