01A Vector Calculus Review Lecture Slides.pdf

Vector calculus review

1. Introduction to Vectors
Dr. Steven Weiss
Intermediate Electromagnetics

2. 2
Introduction to Vectors
§ At the end of this presentation, you will be able to:
o Recall the fundamental definitions of the dot and cross
product operations as they pertain to vectors
o Compute a unit vector normal to a plane defined by 3 points
o Calculate the magnitude of a vector using the dot product
operation

3. 3
Introduction to Vectors (cont.)
§ We describe Electric, Magnetic, and electromagnetic fields
using vectors. Accordingly, it is appropriate to introduce some
notation used in this course. At this point in your education,
you should already be familier with basic concepts such as
the dot and cross products, so only a brief review will be
given. We review vector calculus more extensively.

4. 4
The Dot Product – A Scalar Product

The dot product is a scalar operation:
!
A⋅
!
B=ABcosθ
AB
. Two vectors dotted
together produce a scalar. The magnitude
of a vector is properly understood as dot
producting the vector with itself and taking
the square root. That is:
A=
!
A=
!
A⋅
!
A B=
!
B=
!
B⋅
!
B
This holds for real values of
!
A and
!
B. If a vector is complex, the
magnitude is understood of dot producting the vector with its
complex conjugate and taking the squre root of the result.
!A!BθABAcosθAB
AsinθAB

5. 5
Properties of Vectors

1) Commutative property
!
A+
!
B=
!
B+
!
A,
!
A⋅
!
B=
!
B⋅
!
A.
2) Associative law (
!
A+
!
B)+
!
C=
!
A+(
!
B+
!
C)
3) 0 is the additive identity element
4)
!
A+(−
!
A)=0
5) Distributive law r(
!
A+
!
B)=r
!
A+r
!
B,
!
A⋅(
!
B+
!
C)=
!
A⋅
!
B+
!
A⋅
!
C
6) (r+s)
!
A=r
!
A+s
!
A
7) (rs)
!
A=r(s
!
A)
8) 1
!
A=
!
A 1 is a scalar called the multiplicative identity element

6. 6
Properties of Vectors (cont.) A vector space is a set containing elements calledvectors. Vectors in three dimensional coordinatespace are called three tuples (e.g., a1, a2, a3).Relativity theory uses four tuples. Vectors can also befunctions in a function space called a Hilbert space.These concepts come from linear algebra.

7. 7
The Cross Product – A Vector Product (1) The cross product is also commonly used andis defined by the sin between two vectors withthe result being a vector orthogonal to bothvectors having a magnitude of: ABsinθAB. Asthe cross product produces a vector from twovectors, it gives a vector product. That is: !A×!B=ˆanABsinθAB.Here, ˆan is a unit vector perpendicular to the plane containing thevectors !A and !B. The direction of ˆan follows the right-hand rule.
!A
!BθAB ˆan

8. 8
The Cross Product – A Vector Product (2) Note that: !A×!B=−!B×!A. Therefore, the cross product is notcommutative. !A×(!B+!C)= !A×!B+ !A×!C. Therefore, the crossproduct is distributive. But, !A×(!B×!C)≠ (!A×!B)×!C, so thecross product is not associative. Some comments:1) The vector product of two parallel vectors is always zero:(i.e., sinθAB=sin0=0.)2) The vector product with itself is always zero:(i.e., sinθAA=sin0=0.)

9. 9
The Cross Product – A Vector Product (3) It can be shown that: !A×(!B×!C)=!B(!A⋅!C)−!C(!A⋅!B). This usefulrelationship is called the BAC-CAB rule. If !A and !B each contain3 components (e.g.,vectors in the Cartesian coordinate system),we would have: !A=ˆaxAx+ˆayAy+ˆazAz and !B=ˆaxBx+ˆayBy+ˆazBz. The cross product may be written in matrix form as:!A×!B=ˆaxˆayˆazAxAyAzBxByBz=ˆaxAyBz−AzBy( )−ˆayAxBz−AzBx( )+ˆazAxBy−AyBx( )

10. 10
Example Problem Example problem: Find a unit vector normal to a plane that contains P1(0,1,0),P2(1,0,1), and P3(0,0,1).Let !A=P1P2=!P2−!P1=(ˆax1+ˆay0+ˆaz1)−(ˆax0+ˆay1+ˆaz0)=ˆax−ˆay+ˆazLet !B =P1P3=!P3−!P1=(ˆax0+ˆay0+ˆaz1)−(ˆax0+ˆay1+ˆaz0)=−ˆay+ˆazThe cross product gives a vector that normal to the plane containing !A and !B.Therefore: !C=!A×!B=(ˆax−ˆay+ˆaz)×(−ˆay+ˆaz)=−(ˆay+ˆaz). This vector isnormal to the plane containing !A and !B but it is not the unit vector. Note that!C=ˆac!C so that: ˆac=!C/!C where!C=!C⋅!C=2. Therefore: ˆac=−(ˆay+ˆaz)2.

11. 11
Example Problem (cont.) Remember, to obtain the magnitude of a vector, dot product thevector with itself and then take the square root of the result:!A=!A⋅!A. If the vector is complex (e.g., !A=ˆax3+jˆay) then wedot product the vector with its complex conjugate.!A=!A⋅!A*=(ˆax3+jˆay)⋅(ˆax3−jˆay)=9+1=10.Note: (+j)(−j)=+1, where j=−1.

12. 12
Conclusion
The cross product follows the right-hand rule!

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