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Solution to the Drill problems of chapter 01
(Engineering Electromagnetics,Hayt,A.Buck 7th ed)
BEE 4A,4B & 4C
~ M N = N (3, −3, 0) − M (−1, 2, 1) = (4, −5, −1) = 4âx − 5ây − âz
D1.1 (a). R
~ M P = P (−2, −3, −4) − M (−1, 2, 1) = (−1, −5, −5)
(b). R
~
~ M P = (4, −5, −1) + (−1, −5, −5) = (3, −10, −6) = 3âx − 10ây − 6âz
RM N + R
(c). ~rM = M (−1, 2, 1) − O(0, 0, 0) = (−1, 2, 1), | ~rM |=
p
(−1)2 + (2)2 + (1)2 = 2.45
~ MP / | R
~ M P |, R
~ M P = (−1, −5, −5) = −âx − 5ây − 5âz , | R
~ M P |=
(d). âM P = R
⇒ âM P = (−âx − 5ây − 5âz )/7.1414 = −0.14âx − 0.7ây − 0.7âz
p
(−1)2 + (−5)2 + (−5)2 = 7.1414
(e). ~rP = P (−2, −3, −4) − O(0, 0, 0) = (−2 − 0, −3 − 0, −4 − 0) = (−2, −3, −4)
⇒ 2~rP = 2 × (−2, −3, −4) = (−4, −6, −8)
~rN = N (3, −3, 0) − O(0, 0, 0) = (3 − 0, −3 − 0, 0 − 0) = (3, −3, 0)
⇒ 3~rN = 3 × (3, −3, 0) = (9, −9, 0)
⇒ 2~rP − 3~rN = (−4,
√ − 9, −6 + 9, −8 − 0) = −13âx + 3ây − 8âz
p −6, −8) − (9, −9, 0) = (−4
2
2
2
⇒| 2~rP − 3~rN |= (−13) + (3) + (−8) = 242 = 15.56
D1.2 (a). S = 125 {(x − 1)âx + (y − 2)ây + (z + 1)âz } / (x − 1)2 + (y − 2)2 + (z + 1)2 , P (2, 4, 3)
⇒ SP (2,4,3) = 125 {(2 − 1)âx + (4 − 2)â
+ (3 + 1)âz } / (2 − 1)2 + (4 − 2)2 + (3 + 1)2
y 2
⇒ SP (2,4,3) = 125 {âx + 2ây + 4âz } / (1) + (2)2 + (4)2 = 125 {âx + 2ây + 4âz } /21
⇒ SP (2,4,3) = 5.95âx + 11.90ây + 23.8âz
p
(b). âS = SP (2,4,3) / | SP (2,4,3) |= (5.95âx + 11.90ây + 23.8âz )/ (5.95)2 + (11.90)2 + (23.8)2
⇒ âS = 0.218âx + 0.436ây + 0.873âz
(c). Wen are given that | S |= 1
o p
⇒ 125
(x − 1)2 + (y − 2)2 + (z + 1)2 / (x − 1)2 + (y − 2)2 + (z + 1)2 = 1
⇒ 125
⇒ 125
⇒
p
np
np
o
(x − 1)2 + (y − 2)2 + (z + 1)2 = (x − 1)2 + (y − 2)2 + (z + 1)2
o
(x − 1)2 + (y − 2)2 + (z + 1)2 =
(x − 1)2 + (y − 2)2 + (z + 1)2 = 125
p
(x − 1)2 + (y − 2)2 + (z + 1)2 ×
p
(x − 1)2 + (y − 2)2 + (z + 1)2
~ AB = B(−2, 3, −4) − A(6, −1, 2) = (−2 − 6, 3 + 1, −4 − 2) = (−8, 4, −6) = −8âx + 4ây − 6âz
D1.3 (a). R
~ AC = C(−3, 1, 5) − A(6, −1, 2) = (−3 − 6, 1 + 1, 5 − 2) = (−9, 2, 3) = −9âx + 2ây + 3âz
(b). R
~ AB · R
~ AC =| R
~ AB || R
~ AC | cos θBAC ⇒ cos θBAC = (R
~ AB · R
~ AC )/(| R
~ AB || R
~ AC |)
(c). R
o
np
p
(−8)2 + (4)2 + (−6)2 × (−9)2 + (2)2 + (3)2
⇒ cos θBAC = {(−8)(−9) + (4)(2) + (−6)(3)} /
⇒ cos θBAC = 62/ (116)(94) ⇒ cos θBAC = 62/104.422 = 0.5937 ⇒ θBAC = arccos(0.5937) = 53.6o
p
~ AB in the direction of R
~ AC , and that scalar component is
(d). First we need to find the scalar component of R
~
~
~
~
~ AC
| RAB | cos θBAC = RAB · RAC / | RAC | ( from part(c)), then we need to find a unit vector in the direction of R
~
~
~
which is given by âR~ AC = RAC / | RAC |, now multiply these two components to find the vector projection of RAB
~
on R
AC
~ AB · R
~ AC / | R
~ AC | R
~ AC / | R
~ AC | , we have already calculated all the values present in the last formula in
⇒ R
√ √ part (c),so using those values we get 62/ 94 (−9âx + 2ây + 3âz ) / 94 = −5.94âx + 1.319ây + 1.979âz
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This document is prepared in LATEX. (Email: ahmadsajjad01@ciit.net.pk)
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~ AB = B(−2, 3, −4) − A(6, −1, 2) = (−8, 4, −6)
D1.4 (a). R
~ AC = C(−3, 1, 5) − A(6, −1, 2) = (−9, 2, 3)
Similarly we have R


âx ây âz
~ AB × R
~ AC =  −8 4 −6  = âx (12 + 12) − ây (−24 − 54) + âz (−16 + 36) = 24âx + 78ây + 20âz
now R
−9 2
3
~ AB | sin θ is the height of the trian(b). Area of a triangle is given by (1/2)(Base)(Height) and in our case | R
~
gle and | RAC | is the base of the triangle,so we get
~ AB | sin θ)(| R
~ AC |) = (1/2) | R
~ AB × R
~ AC |= 42
Area= (1/2)(| R
(c). The required unit vector is given by
√
~ AB × R
~ AC )/(| R
~ AB × R
~ AC |) = (24âx + 78ây + 20âz )/ 7060 = 0.286âx + 0.928ây + 0.238âz
(R
D1.5 (a). we have x = ρ cos φ, by putting the values we get x = (4.4)(cos(−115o ))⇒ x=-1.860
also y=ρ sin φ ⇒ y = (4.4)(sin(−115o )) = −3.99, z parameter is the same.
(b). we have ρ =
p
x2 + y 2 ⇒ ρ =
p
(−3.1)2 + (2.6)2 = 4.05
~ CD = D(−3.1, 2.6, −3) − C(−1.860, −3.99, 2) = (−1.24, 6.59, −5), now the distance from C to D=| R
~ CD |
(c).pR
√
2
2
2
⇒ (−1.24) + (6.59) + (−5) = 69.9657 = 8.36
D1.6 (a). we have P(10,-8,6) and φ = arctan(y/x) ⇒ φ = arctan(−8/10) = −38.66o
now F~ = 10âx − 8ây + 6âz , | F~ρ |= F~ · âρ = (10âx − 8ây + 6âz ) · âρ = 10âx · âρ − 8ây · âρ + 6âz · âρ
⇒ 10 cos φ − 8 sin φ + 6(0) = 10 cos(−38.66o ) − 8 sin(−38.66o ) = 7.80 + 4.99 = 12.8
also | F~φ |= F~ · âφ = (10âx − 8ây + 6âz ) · âφ = 10âx · âφ − 8ây · âφ + 6âz · âφ = 10(− sin φ) − 8 cos φ + 6(0) =
−10 sin(−38.66o ) − 8 cos(−38.66o ) = 6.25 − 6.25 = 0
also | F~z |= 6 now we have F~ =| F~ρ | âρ + | F~φ | âφ + | F~z | âz = 12.8âρ + 6âz
NOTE: Consult table 1.1 of your text book to calculate the dot product of unit vectors.
(b). Same as part (a).
(c). Same as part (a), the only difference is that we have to multiply âx , ây and âz with the given vector and
then consult the table.
p
p
D1.7 (a). we havep
the formula r = x2 + y 2 + z 2p
⇒ r = (−3)2 + 22 + 12 = 3.74
2
2
2
and θ = arccos(z/ x + y + z ⇒ θ = arccos(1/ (−3)2 + 22 + 12 ) = arccos(1/3.74) = 74.5o
and φ = arctan(y/x) = arctan(2/ − 3) = −33.69o (clockwise) or 146.30o (anti clockwise)
(b). use the formula x = r sin θ cos φ ⇒ x = 5 sin 20o cos(−70o ) = 0.585
also y = r sin θ sin φ ⇒ y = 5 sin 20o sin(−70o ) = −1.607
and the formula z = r cos θ ⇒ z = 5 cos 20o = 4.7
~ CD |=| D(0.585, −1.607, 4.70) − C(−3, 2, 1) |=| (3.585, −3.607, 3.70) |= 6.29
(c). The distance fron C to D =| R
p
p
D1.8 (a). We have P(-3,2,4) and θ = arccos(z/ x2 + y 2 + z 2 ⇒ θ = arccos(4/ (−3)2 + 22 + 42 ) = arccos(4/5.385)
⇒ θ = 36.6o also φ = arctan(y/x) = arctan(2/ − 3) = −33.69o (clockwise) or 146.30o (anti clockwise)
~ θ |= A
~ · âθ = 10âx · âθ = 10 cos θ cos φ
~ so we have A
~ = 10âx ,now | A
now the given vector is 10âx , let it be equal to A,
o
o
~
~
⇒| Aθ |= 10 cos 36.6 cos 146.30 = −6.67, similarly | Ar |= 10âx · âr = 10 sin θ cos φ
~ r |= 10 sin 36.6o cos 146.30o = −4.96,similarly | A
~ φ |= 10âx · âφ = −10 sin φ
⇒| A
~ φ |= −10 sin 146.30o = 5.5
⇒| A
~ =| A
~ r | âr + | A
~ θ | âθ + | A
~ φ | âφ = −4.96âr − 6.67âθ + 5.5âφ
so A
(b). Similar to part(a)
(c). Similar to part(a)
NOTE: Consult Table 1.1 & 1.2 of your text book to calculate the dot product of unit vectors.
THE END
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