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EELE 3331 – Electromagnetic I
Chapter 1
Vector Algebra Islamic University of Gaza
Electrical Engineering Department
Dr. Talal Skaik
2012
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Electromagnetics (EM) → Study of Electric and Magnetic
phenomena.
Applications: Microwave, Antenna, Electric Machines,
Satellite Communications, Radar, Fiber Optics, …etc.
Time-Invariant Conditions:
Electrostatics (Chapters 4, 5 and 6) and Magnetostatics
(Chapters 7 and 8).
Time-Varying Conditions:
Electromagnetics (Chapter 9 to 13)
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Maxwell’s Equations
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Scalars and Vectors
Scalar: Quantity that has only magnitude.
time, distance, temperature, mass, population
Vector: Quantity that ha both magnitude and direction.
Velocity, force, displacement, Electric Field Intensity.
For a vector A, a unit vector aA along A is defined as:
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Unit Vectors
A
A
A
| A |
vector, its magnitude is unity |a |=1
and its direction is along .
| | magnitude of A (scalar)
A= | A |
A
A
a
a
A
A
a
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A=(Ax , Ay , Az) or A=Ax ax + Ay ay + Az az
where: Ax , Ay , Az → Componets of A in the x,y and z directions.
ax , ay , az → Unit Vectors in the x,y and z directions.
A vector in Cartesian (or Rectangular) coordinates
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A vector in Cartesian (or Rectangular) coordinates
2 2 2
2 2 2
The magnitude of vector A is given by:
| A |
and the unit vector along A is given by:
A
| A |
x y z
x x y y z z
A
x y z
A A A
A a A a A aa
A A A
For two vectors A=(Ax , Ay , Az) and B=(Bx , By , Bz)
Addition: C=A+B
C=(Ax+ Bx ) ax + (Ay+ By ) ay + (Az+ Bz ) az
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Vector Addition and Subtraction
Vector addition C A B: (a) parallelogram rule, (b) head-to-tail rule.
Fir two vectors A=(Ax , Ay , Az) and B=(Bx , By , Bz)
Subtraction: D=A-B=A+(-B)
D=(Ax- Bx ) ax + (Ay- By) ay + (Az- Bz ) az
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Vector Addition and Subtraction
Vector subtraction D A - B: (a) parallelogram rule, (b) head-to-tail rule.
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Laws of Algebra
A+B=B+A, kA=Ak
A+(B+C)=(A+B)+C, k(lA)=(kl)A
k(A+B)=kA+kB
k and l are scalars
For Point p in Cartesian Coordinates (x,y,z)
Position Vector : rp (or radius vector) of point p is defined as
the directed distance from origin 0 to p.
rP=0P=xax+yay+zaz
Example:
point P(3,4,5) has position
vector rP=3ax+4ay+5az
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Position and Distance Vectors
The Distance Vector is the displacement from one point to another.
Points → P(xp , yp , zp), Q(xQ , yQ , zQ)
The distance Vector (Separation Vector) : rPQ=rQ-rP
rPQ= (xQ -xp) ax+ (yQ -yp) ay+ (zQ -zp) az
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Position and Distance Vectors
y
2 2 2
( ) =-4
(b) 3A-B=3(10, 4,6) (2,1,0)=(30, 12,18) (2,1,0)
(28, 13,18)
3A-B (28) ( 13) (18) 1277 35.74
(C) Let C=A+2B=(10, 4,6) 2(2,1,0) (14, 2,6)
C (14, 2 a unit vector along C is a =
Cc
a A
- - - -
-
-
- -
-
2 2 2
,6)
(14) ( 2) (6)
a =0.9113a 0.1302a 0.3906a Note: a 1c x y z c
-
- 14
If A=10ax - 4ay +6az , B=2ax + ay , Find
(a) The component of A along ay
(b) The magnitude of 3A-B
(c) A unit Vector along A+2B
Example 1.1
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Points P and Q are located at (0,2,4) and (-3,1,5). Calculate:
(a) The position vector rP
(b) The distance vector from P to Q.
(c) The distance between P and Q.
(d) A vector Parallel to PQ with magnitude of 10.
Example 1.2
2 2 2
2 2 2
( ) r 0a 2a 4a 2a 4a
( ) r r r ( 3,1,5) (0,2,4) ( 3, 1,1)
r 3a a a
( ) = r 3 1 1 3.317
Alternatively: d= 3.317
P x y z y z
PQ Q P
PQ x y z
PQ
Q P Q P Q P
a
b
or
c d
x x y y z z
- - - - -
- -
- -
- - -
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Example 1.2 - continued
( ) Find a vector parallel to with magnitude of 10
Let A be the required vector
A= , but 10
Since A is parallel to , it has the same unit vector
r ( 3, 1,1 a
r
A
PQ
A
PQ
d PQ
A a A
PQ
- -
)
3.317
10( 3, 1,1) A= ( 9.045a 3.015a 3.015a )
3.317x y z
- - - -
When two vectors A and B are multiplied, the result is either a
scalar or a vector.
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Vector Multiplication
(1) Scalar (or dot) product: A B
(2) Vector (or cross) product: A B
Three Vectors A,B,C:
(3) Scalar triple product: A (B C)
(4) Vector triple product: A (B C)
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The product of the magnitudes of A and B and the
cosine of the angle between them.
x y z x y z
x x
A B A B cos
the smaller angle between A and B.
If A=( , , ), B=( , , ), then
A B
AB
AB
y y z z
A A A B B B
A B A B A B
Dot Product A∙B
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2
x y
x y
A and B are orthognal (or perpendicular) if
A B 0
Notes:
A B B A
A (B+C) A B A C
A A A , (cos 0 = 1)
also a a =a a =a a =0
a a =a a =a a =1
y z z x
x y z z
Dot Product A∙B
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A vector quantity whose magnitude is the area of
parallelogram formed by A and B, and its direction
is perpendicular to the plane containing A and B.
Cross Product AxB
Where
an→unit vector normal to plane
containing A and B, and its direction
is found by (right-hand rule), that is
the direction of the right thumb when
the fingers of the right hand rotate
from A to B.
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Cross Product AxB
A×B A B sin aAB n
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Cross Product AxB
x y z x y z
x
x
For A=( , , ), B=( , , ), then
a a a
A×B
= a
a
a
x y z
y z
y z
y z z y x
z x x z y
x y y x z
A A A B B B
A A A
B B B
A B A B
A B A B
A B A B
-
-
-
x
: A B B A, A B B A
A (B C) (A B) C
A (B C) A B A C
A A 0
also a a a
a a a
a a a
Movin
y z
y z x
z x y
Notes -
x
g clockwise positive results
Moving counter-clockwise negative results
a a a , a a a , a a ay x z z y z y x
- - -23
Cross Product AxB
For vectors A, B and C:
A B×C B C×A C A×B
Volume of parallelpiped having A,B and C as edges.
Volume= A×B C= A×B C cos
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Scalar Triple Product
x x x
x
x
x
For vectors A, B and C:
If A=( , , ), B=( , , ), and C=( , , ),
A B×C
y z y z y z
y z
y z
y z
A A A B B B C C C
A A A
then B B B Scalar
C C C
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Scalar Triple Product
For vectors A, B and C:
A B×C B A C C A B
"bac cab" rule
:
A B C A B C
A B C C A B
Notes
but
-
-
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Vector Triple Product
B
B
B
B
Given a vector A, a scalar component of A along B is :
(projection)
A A cos
The vector component A of A along B is,
A A
B B AB
B B
A
A a a
a a
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Components of a vector
Components of A along B: (a) scalar component AB, (b) vector component AB .
2 2 2
2 2 2
1 o
A BA B A B cos cos
A B
A B 3,4,1 0,2, 5 0 8 5 3
A 3 4 1 26
B 0 2 ( 5) 29
A B 3cos 0.1092
A B (26)(29)
cos (0.1092) 83.73
AB AB
AB
AB
-
- -
-
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A=3ax + 4ay +az ,
B=2ay - 5az ,
Find the angle between A and B.
Example 1.4
n
2 2 2
1 o
A BA B A B sin a sin
A B
A B= 3 4 1 20 2 0 15 6 0
0 2 5
A B 22,15,6
A B ( 22) (15) (6) 745
A B 745sin 0.994
A B (26)(29)
sin 0.994 83.73
AB AB
x y z
x y z
AB
AB
Alternatively
a a a
a a a
-
- - -
-
-
-
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Example 1.4 - Continued
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Three field quantities are given by
P=2ax - az , Q=2ax - ay +2 az , R=2ax - 3ay + az
Determine:
(a) (P+Q) x (P-Q)
(b) Q . R x P
(c) P . (Q x R)
(d) sin θQR
(e) P x (Q x R)
(f) A unit vector perpendicular to both Q and R.
(g) The component of P along Q.
Example 1.5
( ) (P + Q) (P Q)
=P (P Q) Q (P Q)
=P P P Q Q P Q Q
= 0 P Q Q P 0
=2 Q P 2 2 1 2 2 12 4
2 0 1
( ) Q R P what makes sense is Q (R P)
Q (R P) (2, 1,2) 2 3 1 (2
2 0 1
x y z
x y z
x y z
a
a a a
a a a
b
a a a
-
- -
- -
- -
-
-
- -
-
, 1, 2) (3,4,6) 6 4 12 14- -
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P=2ax - az , Q=2ax - ay +2 az , R=2ax - 3ay + az
Example 1.5 - Continued
( ) Alternatively Q (R P) (scalar triple product)
2 1 2
Q (R P) 2 3 1 2(3) (1)( 2 2) 2(6) 14
2 0 1
(c) P (Q R)
=Q (R P)=14
Alternatively, P (Q R) (2,0, 1) 2 1 2
2 3 1
x y z
b
a a a
-
- - -
-
- -
-
(2,0, 1) (5,2, 4) 10 0 4 14 - - 32
P=2ax - az , Q=2ax - ay +2 az , R=2ax - 3ay + az
Example 1.5 - Continued
QR
5,2, 4Q R 45( ) sin = = = =0.5976
Q R 2, 1,2 2, 3,1 (3)( 14)
(e) P Q R = 2,0, 1 5,2, 4 (2,3,4)
bac-cab rule: A (B C) B(A C) C(A B)
P Q R =Q(P.R) R(P.Q)
(2, 1,2)(4 0 1) (2, 3,1)(4 0 2)
(6, 3,6) (
d
Alternatively
-
- -
- -
-
-
- - - - -
- -
4, 6,2) (2,3,4)
( ) a unit vector perpendicular to both Q and R
5,2, 4Q R a= 0.745,0.298, 0.596
Q R 45
f
-
- -
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P=2ax - az , Q=2ax - ay +2 az , R=2ax - 3ay + az
Example 1.5 - Continued
x
( ) The component of P along Q
Q (2, 1,2) 2 1 2P P , , ,
Q 3 3 39
2 1 2 4 2 2P. 2,0, 1 , , 0
3 3 3 3 3 3
2 2 1 2P P , ,
3 3 3 3
=0.4444a 0.2222 0.444
Q Q Q Q
Q
Q Q Q
y
g
a a a
a
a a
a a
- -
- - -
-
- y
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P=2ax - az , Q=2ax - ay +2 az , R=2ax - 3ay + az
Example 1.5 - Continued
p1p2 p2 p1
p1p3 p3 p1
p1p2 p1p3
p1p2 p1p3
Distance Vector r =r r = 1,1,2 5,2, 4 = 4, 1,6
Distance Vector r =r r = 3,0,8 5,2, 4 = 8, 2,12
r r 4 1 6 = 0,0,0
8 2 12
shows that the angle between r and r is zero (sin
x y za a a
- - - - -
- - - - - -
- -
- -
1 2 3
=0)
, , lie on a straight line.P P and P
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Example 1.7
Show that points P1(5,2,-4), P2(1,1,2), P3(-3,0,8) all lie on a
straight line.
