Solutions Manual Engineering Management Meeting the Global Challenges 2nd Edition By C. M. Chang

[Solutions Manual] Engineerin
e e



g Electromagnetics
e



Engineering
DaffodileInternationaleUniversity
25epag.




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,D1.1 Given points M(-1,2,1),N(3,-3,0) and P(-2,-3,-4),find (a) RMN
e e e e e e e


(b) RMN + RMP (c) |rM| (d) aMP (e) |2rP-3rN|.
e e e e e e e e e




SOLUTION:
(a) RMN RMN = rN - rM e e e e e


RMN = (3x-3ay+0az)-(-1ax+2ay+1az) M(-1,2,1) N(3,-
e e e e


3,0) rM rN = (3+1)ax +(-3-2)ay +(0-1)az
e e e e e e


RMN = 4ax-5ay-az e e


(b) RMN + RMP = ? e e e e e


RMP
=
rP -rM M(-1,2,1) rM
e e e


= (-2,-3,-4) –(-
e e


1,2,1) P(-2,-3,-4) e


= (-2+1)ax +(-3-2)ay+(-4-1)az
e e


= -ax-5ay-5az
e


We know RMN = 4ax-5ay-az
e e e e


RMN + RMP =(4ax-5ay-az)+ (-ax-5ay-5az)
e e e e


= 3ax-10ay-6az
e


(c) |rM| =? e


RM = (-ax+2ay+az)
e e


|rM| = (-1)2 + (2)2+(1)2
e e e e


= √6
e


|rM| = 2.45 e e


(d) aMP = ? e e e


aMP = RM e e


P
|RMP|
RMP = -ax-5ay-5az e e




|RMP| = (-1)2 + (-5)2+(-5)2 e e e e




= √51
e


= 7.14
e




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,aMP = -ax-5ay-5az/7.14 e e


=-0.14ax-0.7ay-0.7az
(e) |2rP-3rN|=?
Professor Aatif Saeed Preston University Islamabad rP Page 1 rP = -2ax –
e e e e e e e e e e e e


e3ay-4az
2rP = -4ax-6ay-az e e


rN = 3ax - e e e


3ay+0az 3rN = 9ax- e e e


9ay+0az
2rP-3rN = (-4ax-6ay-az) – (9ax-9ay+0az) e e e e


= -13ax+3ay-8az
e




|2rP-3rN|= (-13)2 + (3)2+(8)2 e e e


=√ 242 e


|2rP-3rN| = 15.56 ANSWER e e e




D1.2. A vector field S is expressed in rectangular coordinates as S = {125/ [(
e e e e e e e e e e e e e e


x
−1)2+(y−2)2+(z+1)2]}{(x −1)ax +(y−2)ay +(z+1)az}. (a) EvaluateS at P(2, 4, 3 e e e e e e e e e


). (b)
e


Determine a unit vector that gives the direction of S at P.
e e e e e e e e e e e




SOLUTION:
(a) S = {125/[(x −1)2+(y−2)2+(z+1)2]}{(x −1)ax +(y−2)ay +(z+1)az}
e e e e e e


S = {125/[(2 −1)2+(4−2)2+(3+1)2]}{(2 −1)ax +(4−2)ay +(3+1)az} S
e e e e e e e


e= {125/[1+4+16]}{(ax +2ay +4az}
e e e


S = {125/21}{(ax +2ay +4az}
e e e e e


S =5.95{(ax +2ay +4az} S=5.
e e e e


95 ax +11.9ay +23.8
e e e


(b) a e


S= e e


S
│S│
25
S 2 5 = (5.95)2+(11.9)2+(23.8)2
e
e
e e
e
0e2e e 0 e2

=35.40+141.61+556.96
=733.97




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