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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
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)
