Engineering Electromagnetics - Solution Manual- 7th Edition - Hayt & Buck

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1、CHAPTER11.1.GiventhevectorsM=−10ax+4ay−8azandN=8ax+7ay−2az,find:a)aunitvectorinthedirectionof−M+2N.−M+2N=10ax−4ay+8az+16ax+14ay−4az=(26,10,4)Thus(26,10,4)a==(0.92,0.36,0.14)

2、(26,10,4)

3、b)themagnitudeof5ax+N−3M:(5,0,0)+(8,7,−2)−(−30,12,−24)=(43,−5,22),and

4、(43,−5

5、,22)

6、=48.6.c)

7、M

8、

9、2N

10、(M+N):

11、(−10,4,−8)

12、

13、(16,14,−4)

14、(−2,11,−10)=(13.4)(21.6)(−2,11,−10)=(−580.5,3193,−2902)1.2.Giventhreepoints,A(4,3,2),B(−2,0,5),andC(7,−2,1):a)SpecifythevectorAextendingfromtheorigintothepointA.A=(4,3,2)=4ax+3ay+2azb)Giveaunitvectorextendingf

15、romtheorigintothemidpointoflineAB.ThevectorfromtheorigintothemidpointisgivenbyM=(1/2)(A+B)=(1/2)(4−2,3+0,2+5)=(1,1.5,3.5)Theunitvectorwillbe(1,1.5,3.5)m==(0.25,0.38,0.89)

16、(1,1.5,3.5)

17、c)CalculatethelengthoftheperimeteroftriangleABC:BeginwithAB=(−6,−3,3),BC=(9,

18、−2,−4),CA=(3,−5,−1).Then

19、AB

20、+

21、BC

22、+

23、CA

24、=7.35+10.05+5.91=23.321.3.ThevectorfromtheorigintothepointAisgivenas(6,−2,−4),andtheunitvectordirectedfromtheorigintowardpointBis(2,−2,1)/3.IfpointsAandBaretenunitsapart,findthecoordinatesofpointB.WithA=(6,−2,−4)andB=1B(2,

25、−2,1),weusethefactthat

26、B−A

27、=10,or32B)a21

28、(6−x−(2−B)ay−(4+B)az

29、=10333Expanding,obtain36−8B+4B2+4−8B+4B2+16+8B+1B2=100939√39orB2−8B−44=0.ThusB=8±64−176=11.75(takingpositiveoption)andso2221B=(11.75)ax−(11.75)ay+(11.75)az=7.83ax−7.83ay+3.92az33311.4.givenpointsA(

30、8,−5,4)andB(−2,3,2),find:a)thedistancefromAtoB.

31、B−A

32、=

33、(−10,8,−2)

34、=12.96b)aunitvectordirectedfromAtowardsB.ThisisfoundthroughB−AaAB==(−0.77,0.62,−0.15)

35、B−A

36、c)aunitvectordirectedfromtheorigintothemidpointofthelineAB.(A+B)/2(3,−1,3)a0M==√=(0.69,−0.23,0.69)

37、(A+B)/

38、2

39、19d)thecoordinatesofthepointonthelineconnectingAtoBatwhichthelineintersectstheplanez=3.Notethatthemidpoint,(3,−1,3),asdeterminedfrompartchappenstohavezcoordinateof3.Thisisthepointwearelookingfor.1.5.AvectorfieldisspecifiedasG=24xyax+12(x2+2)ay+18z2az.Giventwo

40、points,P(1,2,−1)andQ(−2,1,3),find:a)GatP:G(1,2,−1)=(48,36,18)b)aunitvectorinthedirectionofGatQ:G(−2,1,3)=(−48,72,162),so(−48,72,162)aG==(−0.26,0.39,0.88)

41、(−48,72,162)

42、c)aunitvectordirected