CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N =
... [Show More] 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
a =
(26, 10, 4)
|(26, 10, 4)|
= (0.92, 0.36, 0.14)
b) the magnitude of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7, −2, 1):
a) Specify the vector A extending from the origin to the point A.
A = (4, 3, 2) = 4ax + 3ay + 2az
b) Give a unit vector extending from the origin to the midpoint of line AB.
The vector from the origin to the midpoint is given by
M = (1/2)(A + B) = (1/2)(4 − 2, 3 + 0, 2 + 5) = (1, 1.5, 3.5)
The unit vector will be
m =
(1, 1.5, 3.5)
|(1, 1.5, 3.5)|
= (0.25, 0.38, 0.89)
c) Calculate the length of the perimeter of triangle ABC:
Begin with AB = (−6, −3, 3), BC = (9, −2, −4), CA = (3, −5, −1).
Then
|AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.32
1.3. The vector from the origin to the point A is given as (6, −2, −4), and the unit vector directed from the
origin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, find the coordinates of point
B.
With A = (6, −2, −4) and B = 13B(2, −2, 1), we use the fact that |B − A| = 10, or
|(6 − 23B)ax − (2 − 23B)ay − (4 + 13B)az| = 10
Expanding, obtain
36 − 8B + 4
9B2 + 4 − 8 3B + 49B2 + 16 + 83B + 19B2 = 100
or B2 − 8B − 44 = 0. Thus B = 8±√64−176
2 = 11.75 (taking positive option) and so
B =
2 3
(11.75)ax −
2 3
(11.75)ay +
1 3
(11.75)az = 7.83ax − 7.83ay + 3.92az
1
SOLUTION MANUAL ENGINEERING ELECTROMAGNETICS BY WILLIAM H. HAYT 8TH
EDITION COMPLETE CHAPTERS1.4. given points A(8, −5, 4) and B(−2, 3, 2), find:
a) the distance from A to B.
|B − A| = |(−10, 8, −2)| = 12.96
b) a unit vector directed from A towards B. This is found through
aAB =
B − A
|B − A|
= (−0.77, 0.62, −0.15)
c) a unit vector directed from the origin to the midpoint of the line AB.
a0M =
(A + B)/2
|(A + B)/2|
=
(3, −1, 3)
√19 = (0.69, −0.23, 0.69)
d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3.
Note that the midpoint, (3, −1, 3), as determined from part c happens to have z coordinate of 3. This
is the point we are looking for.
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az. Given two points, P(1, 2, −1) and
Q(−2, 1, 3), find:
a) G at P: G(1, 2, −1) = (48, 36, 18)
b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so
aG =
(−48, 72, 162)
|(−48, 72, 162)|
= (−0.26, 0.39, 0.88)
c) a unit vector directed from Q toward P:
aQP =
P − Q
|P − Q|
=
(3, −1, 4)
√26 = (0.59, 0.20, −0.78)
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
21.6. For the G field in Problem 1.5, make sketches of Gx, Gy, Gz and |G| along the line y = 1, z = 1, for
0 ≤ x ≤ 2. We find G(x, 1, 1) = (24x, 12x2 + 24, 18), from which Gx = 24x, Gy = 12x2 + 24,
G
z = 18, and |G| = 6√4x4 + 32x2 + 25. Plots are shown below.
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region |x|, |y|, and |z| less
than 2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with
|x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2;
4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez: This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y, with
|x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
1.8. Two vector fields are F = −10ax +20x(y−1)ay and G = 2x2yax −4ay +zaz. For the point P(2, 3, −4),
find:
a) |F|: F at (2, 3, −4) = (−10, 80, 0), so |F| = 80.6.
b) |G|: G at (2, 3, −4) = (24, −4, −4), so |G| = 24.7.
c) a unit vector in the direction of F − G: F − G = (−10, 80, 0) − (24, −4, −4) = (−34, 84, 4). So
a =
F − G
|F − G|
=
(−34, 84, 4)
90.7
= (−0.37, 0.92, 0.04)
d) a unit vector in the direction of F + G: F + G = (−10, 80, 0) + (24, −4, −4) = (14, 76, −4). So
a =
F + G
|F + G|
=
(14, 76, −4)
77.4
= (0.18, 0.98, −0.05)
31.9. A field is given as
G =
25
(x2 + y2)(xax + yay)
Find:
a) a unit vector in the direction of G at P(3, 4, −2): Have Gp = 25/(9 + 16) × (3, 4, 0) = 3ax + 4ay,
and |Gp| = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G and ax at P: The angle is found through aG · ax = cos θ. So cos θ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus θ = 53◦.
c) the value of the following double integral on the plane y = 7:
04 02 G · aydzdx
04 02 x2 25 + y2 (xax + yay) · aydzdx = 04 02 x2 25 + 49 × 7 dzdx = 04 x2350 + 49dx
= 350 ×
1 7
tan−1 47 − 0 = 26
1.10. Use the definition of the dot product to find the interior angles at A and B of the triangle defined by the
three points A(1, 3, −2), B(−2, 4, 5), and C(0, −2, 1):
a) Use RAB = (−3, 1, 7) and RAC = (−1, −5, 3) to form RAB · RAC = |RAB||RAC| cos θA. Obtain
3 + 5 + 21 = √59√35 cos θA. Solve to find θA = 65.3◦.
b) Use RBA = (3, −1, −7) and RBC = (2, −6, −4) to form RBA · RBC = |RBA||RBC| cos θB. Obtain
6 + 6 + 28 = √59√56 cos θB. Solve to find θB = 45.9◦.
1.11. Given the points M(0.1, −0.2, −0.1), N(−0.2, 0.1, 0.3), and P(0.4, 0, 0.1), find:
a) the vector RMN: RMN = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4).
b) the dot product RMN · RMP : RMP = (0.4, 0, 0.1) − (0.1, −0.2, −0.1) = (0.3, 0.2, 0.2). RMN ·
RMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP :
RMN · aRMP = (−0.3, 0.3, 0.4) ·
(0.3, 0.2, 0.2)
√0.09 + 0.04 + 0.04 =
0.05
√0.17 = 0.12
d) the angle between RMN and RMP :
θM = cos−1 |RRMN MN|| ·RRMP MP| = cos−1 √0.34 0.05 √0.17 = 78◦
41.12. Given points A(10, 12, −6), B(16, 8, −2), C(8, 1, −4), and D(−2, −5, 8), determine:
a) the vector projection of RAB + RBC on RAD: RAB + RBC = RAC = (8, 1, 4) − (10, 12, −6) =
(−2, −11, 10) Then RAD = (−2, −5, 8) − (10, 12, −6) = (−12, −17, 14). So the projection will
be:
(RAC · aRAD)aRAD = (−2, −11, 10) · (−12√, −629 17, 14) (−12√, −629 17, 14) = (−6.7, −9.5, 7.8)
b) the vector projection of RAB + RBC on RDC: RDC = (8, −1, 4) − (−2, −5, 8) = (10, 6, −4). The
projection is:
(RAC · aRDC)aRDC = (−2, −11, 10) · (10√, 6152 , −4) (10√, 6152 , −4) = (−8.3, −5.0, 3.3)
c) the angle between RDA and RDC: Use RDA = −RAD = (12, 17, −14) and RDC = (10, 6, −4).
The angle is found through the dot product of the associated unit vectors, or:
θD = cos−1(aRDA · aRDC) = cos−1 (12, 17,√−629 14)√· 152 (10, 6, −4) = 26◦
1.13. a) Find the vector component of F = (10, −6, 5) that is parallel to G = (0.1, 0.2, 0.3):
F||G =
F · G
|G|2 G =
(10, −6, 5) · (0.1, 0.2, 0.3)
0.01 + 0.04 + 0.09
(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
b) Find the vector component of F that is perpendicular to G:
FpG = F − F||G = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21)
c) Find the vector component of G that is perpendicular to F:
GpF = G−G||F = G−
G · F
|F|2 F = (0.1, 0.2, 0.3)−
1.3
100 + 36 + 25
(10, −6, 5) = (0.02, 0.25, 0.26)
1.14. The four vertices of a regular tetrahedron are located at O(0, 0, 0), A(0, 1, 0), B(0.5√3, 0.5, 0), and
C(√3/6, 0.5, √2/3).
a) Find a unit vector perpendicular (outward) to the face ABC: First find
RBA × RBC = [(0, 1, 0) − (0.5√3, 0.5, 0)] × [(√3/6, 0.5, 2/3) − (0.5√3, 0.5, 0)]
= (−0.5√3, 0.5, 0) × (−√3/3, 0, 2/3) = (0.41, 0.71, 0.29)
The required unit vector will then be:
RBA × RBC
|RBA × RBC|
= (0.47, 0.82, 0.33)
b) Find the area of the face ABC:
Area =
1 2
|RBA × RBC| = 0.43
51.15. Three vectors extending from the origin are given as r1 = (7, 3, −2), r2 = (−2, 7, −3), and r3 = (0, 2, 3).
Find:
a) a unit vector perpendicular to both r1 and r2:
a
p12 =
r1 × r2
|r1 × r2|
=
(5, 25, 55)
60.6
= (0.08, 0.41, 0.91)
b) a unit vector perpendicular to the vectors r1 − r2 and r2 − r3: r1 − r2 = (9, −4, 1) and r2 − r3 =
(−2, 5, −6). So r1 − r2 × r2 − r3 = (19, 52, 32). Then
a
p =
(19, 52, 32)
|(19, 52, 32)|
=
(19, 52, 32)
63.95
= (0.30, 0.81, 0.50)
c) the area of the triangle defined by r1 and r2:
Area =
1 2
|r1 × r2| = 30.3
d) the area of the triangle defined by the heads of r1, r2, and r3:
Area =
1 2
|(r2 − r1) × (r2 − r3)| =
1 2
|(−9, 4, −1) × (−2, 5, −6)| = 32.0
1.16. Describe the surfaces defined by the equations:
a) r · ax = 2, where r = (x, y, z): This will be the plane x = 2.
b) |r × ax| = 2: r × ax = (0, z, −y), and |r × ax| = z2 + y2 = 2. This is the equation of a cylinder,
centered on the x axis, and of radius 2.
1.17. Point A(−4, 2, 5) and the two vectors, RAM = (20, 18, −10) and RAN = (−10, 8, 15), define a triangle.
a) Find a unit vector perpendicular to the triangle: Use
a
p =
RAM × RAN
|RAM × RAN|
=
(350, −200, 340)
527.35
= (0.664, −0.379, 0.645)
The vector in the opposite direction to this one is also a valid answer.
b) Find a unit vector in the plane of the triangle and perpendicular to RAN:
aAN =
(−10, 8, 15)
√389 = (−0.507, 0.406, 0.761)
Then
apAN = ap ×aAN = (0.664, −0.379, 0.645)×(−0.507, 0.406, 0.761) = (−0.550, −0.832, 0.077)
The vector in the opposite direction to this one is also a valid answer.
c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit vector
in the required direction is (1/2)(aAM + aAN), where
aAM =
(20, 18, −10)
|(20, 18, −10)|
= (0.697, 0.627, −0.348)
61.17c. (continued) Now
1 2
(aAM + aAN) =
1 2
[(0.697, 0.627, −0.348) + (−0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207)
Finally,
abis =
(0.095, 0.516, 0.207)
|(0.095, 0.516, 0.207)|
= (0.168, 0.915, 0.367)
1.18. Given points A(ρ = 5, φ = 70◦, z = −3) and B(ρ = 2, φ = −30◦, z = 1), find:
a) unit vector in cartesian coordinates at A toward B: A(5 cos 70◦, 5 sin 70◦, −3) = A(1.71, 4.70, −3), In
the same manner, B(1.73, −1, 1). So RAB = (1.73, −1, 1) − (1.71, 4.70, −3) = (0.02, −5.70, 4) and
therefore
aAB =
(0.02, −5.70, 4)
|(0.02, −5.70, 4)|
= (0.003, −0.82, 0.57)
b) a vector in cylindrical coordinates at A directed toward B: aAB · aρ = 0.003 cos 70◦ − 0.82 sin 70◦ =
−0.77. aAB · aφ = −0.003 sin 70◦ − 0.82 cos 70◦ = −0.28. Thus
aAB = −0.77aρ − 0.28aφ + 0.57az
.
c) a unit vector in cylindrical coordinates at B directed toward A:
Use aBA = (−0, 003, 0.82, −0.57). Then aBA ·aρ = −0.003 cos(−30◦)+0.82 sin(−30◦) = −0.43, and
aBA · aφ = 0.003 sin(−30◦) + 0.82 cos(−30◦) = 0.71. Finally,
aBA = −0.43aρ + 0.71aφ − 0.57az
1.19 a) Express the field D = (x2 + y2)−1(xax + yay) in cylindrical components and cylindrical variables:
Have x = ρ cos φ, y = ρ sin φ, and x2 + y2 = ρ2. Therefore
D =
1 ρ
(cos φax + sin φay)
Then
D
ρ = D · aρ =
1 ρ
cos φ(ax · aρ) + sin φ(ay · aρ) = 1
ρ
cos2 φ + sin2 φ = 1
ρ
and
Dφ = D · aφ =
1 ρ
cos φ(ax · aφ) + sin φ(ay · aφ) = 1
ρ
[cos φ(− sin φ) + sin φ cos φ] = 0
Therefore
D =
1 ρ
a
ρ
71.19b. Evaluate D at the point where ρ = 2, φ = 0.2π, and z = 5, expressing the result in cylindrical and
cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5aρ. To express this in
cartesian, we use
D = 0.5(aρ · ax)ax + 0.5(aρ · ay)ay = 0.5 cos 36◦ax + 0.5 sin 36◦ay = 0.41ax + 0.29ay
1.20. Express in cartesian components:
a) the vector at A(ρ = 4, φ = 40◦, z = −2) that extends to B(ρ = 5, φ = −110◦, z = 2): We
have A(4 cos 40◦, 4 sin 40◦, −2) = A(3.06, 2.57, −2), and B(5 cos(−110◦), 5 sin(−110◦), 2) =
B(−1.71, −4.70, 2) in cartesian. Thus RAB = (−4.77, −7.30, 4).
b) a unit vector at B directed toward A: Have RBA = (4.77, 7.30, −4), and so
aBA =
(4.77, 7.30, −4)
|(4.77, 7.30, −4)|
= (0.50, 0.76, −0.42)
c) a unit vector at B directed toward the origin: Have rB = (−1.71, −4.70, 2), and so −rB =
(1.71, 4.70, −2). Thus
a =
(1.71, 4.70, −2)
|(1.71, 4.70, −2)|
= (0.32, 0.87, −0.37)
1.21. Express in cylindrical components:
a) the vector from C(3, 2, −7) to D(−1, −4, 2):
C(3, 2, −7) → C(ρ = 3.61, φ = 33.7◦, z = −7) and
D(−1, −4, 2) → D(ρ = 4.12, φ = −104.0◦, z = 2).
Now RCD = (−4, −6, 9) and Rρ = RCD · aρ = −4 cos(33.7) − 6 sin(33.7) = −6.66. Then
Rφ = RCD · aφ = 4 sin(33.7) − 6 cos(33.7) = −2.77. So RCD = −6.66aρ − 2.77aφ + 9az
b) a unit vector at D directed toward C:
RCD = (4, 6, −9) and Rρ = RDC · aρ = 4 cos(−104.0) + 6 sin(−104.0) = −6.79. Then Rφ =
RDC · aφ = 4[− sin(−104.0)] + 6 cos(−104.0) = 2.43. So RDC = −6.79aρ + 2.43aφ − 9az
Thus aDC = −0.59aρ + 0.21aφ − 0.78az
c) a unit vector at D directed toward the origin: Start with rD = (−1, −4, 2), and so the vector toward
the origin will be −rD = (1, 4, −2). Thus in cartesian the unit vector is a = (0.22, 0.87, −0.44).
Convert to cylindrical:
a
ρ = (0.22, 0.87, −0.44) · aρ = 0.22 cos(−104.0) + 0.87 sin(−104.0) = −0.90, and
aφ = (0.22, 0.87, −0.44) · aφ = 0.22[− sin(−104.0)] + 0.87 cos(−104.0) = 0, so that finally,
a = −0.90a
ρ − 0.44az.
1.22. A field is given in cylindrical coordinates as
F = ρ240 + 1 + 3(cos φ + sin φ) aρ + 3(cos φ − sin φ)aφ − 2az
where the magnitude of F is found to be:
|F| = √F · F = (ρ1600 2 + 1)2 + ρ2240 + 1(cos φ + sin φ) + 221/2 [Show Less]