# Edward C. Jordan Memorial Offering of the First Edward C. Jordan Memorial Offering of the First Course under the Indo-US Inter-University Collaborative Initiative in Higher Education and Research: Electromagnetics for Electrical and Computer Engineering by Nannapaneni Narayana Rao Edward C. Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign Urbana, Illinois, USA Amrita Viswa Vidya Peetham, Coimbatore July 10 August 11, 2006 1.1 Vector Algebra 1.1-3

(1) Vectors (A) vs. Magnitude and direction Ex: Velocity, Force Scalars (A) Magnitude only Ex: Mass, Charge (2) Unit Vectors have magnitude unity denoted by symbol a with subscript A A1a1 A2 a 2 A3a 3 aA A A12 A22 A32

Useful for expressing vectors in terms of their components. 1.1-4 (3) Dot Product is a scalar A A A B = AB cos B B Useful for finding angle between two vectors. A B cos AB A A1a1 A2 a 2 A3 a 3

B B1a1 B2 a 2 B3a 3 A1 B1 A2 B2 A3 B3 2 A1 A22 A32 B12 B22 B32 1.1-5 (4) Cross Product is a vector right hand screw A A A B = AB sin an B

B an is perpendicular to both A and B. Useful for finding unit vector perpendicular to two vectors. A B A B an AB sin A B 1.1-6

where a1 A B A1 B1 a2 A2 B2 a3 A3 B3 (5) Triple Cross Product A (B C) is a vector

A (B C) B (C A) C (A B) in general. 1.1-7 (6) Scalar Triple Product A B C B C A C A B A1 A2 A3 B1 B2

B3 C1 C2 C3 is a scalar. 1.1-8 Volume of the parallelepiped Area of base Height A B C a n A B C C A B A B C

AB AB B an C A 1.1-9 D1.2 A = 3a1 + 2a2 + a3 B = a1 + a2 a3 C = a1 + 2a2 + 3a3 (a)

A + B 4C = (3 + 1 4)a1 + (2 + 1 8)a2 + (1 1 12)a3 = 5a2 12a3 A B 4C 25 144 13 1.1-10 (b) A + 2B C = (3 + 2 1)a1 + (2 + 2 2)a2 + (1 2 + 3)a3 = 4a1 + 2a2 4a3 Unit Vector 4a1 2a 2 4a 3 =

4a1 2a 2 4a 3 1 = (2a1 a 2 2a 3 ) 3 1.1-11 (c) A C = 3 1 + 2 2 + 1 3 = 10 a1 a 2 (d) B C 1 1 1 2 a3 1 3 = (3 2)a1 (1 3)a 2 (2 1)a 3

= 5a1 4a2 + a3 1.1-12 3 2 (e) 1 A B C 1 1 1 1 2 3 = 15 8 + 1 = 8 Same as A (B C) = (3a1 + 2a2 + a3) (5a1 4a2 + a3)

= 3 5 + 2 (4) + 1 1 = 15 8 + 1 = 8 1.1-13 P1.5 E D A B C Common

Point D= BA A + D = B) ( E= CB B + E = C) ( D and E lie along a straight line. 1.1-14 D E 0

B A C B 0 B C A C B B A B 0 A B + B C + C A = 0 What is the geometric interpretation of this result? 1.1-15 Another Example Given a1 A a 2 2a3 a 2 A a1 2a3 (1) (2) Find A.

A =C a 2 2a3 a1 2a3 a1 a 2 a3 C 0 1 2 C 2a1 2a 2 a3 1 0 2 1.1-16 To find C, use (1) or (2). a1 C 2a1 2a2 a3 a2 2a3 C 2a3 a2 a2 2a3 C 1 A 2a1 2a2 a3