6/2027,
stions and
a 1comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success
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 = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
a = (26, 10, 4) = (0.92, 0.36, 0.14)
|(26, 10, 4)|
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. Vector A extends from the origin to (1,2,3) and vector B from the origin to
(2,3,-2).
a) Find the unit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−ax − ay + 5az)
w√hose magnitude is |A − B| = [(−ax − ay + 5az ) · (−ax − ay + 5az )]1/2
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 1
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
,6/2027,
stions and
a 2comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success
√
= 1 + 1 + 25 =
3 3 = 5.20. The unit vector is therefore
aAB = (−ax − ay + 5az)/5.20
b) find the unit vector in the direction of the line extending from the origin to
the midpoint of the line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5,
0.5)
The unit vector is then
a = (1.5ax + 2.5ay + 0.5az)
+ + )/2.96
= (1.5a
2.5a 0.5a
mp p x y z
(1.5)2 + (2.5)2 +
(0.5)2
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 2
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
,6/2027,
stions and Latest update!!!! Engineering electromagnetics solution manuals
a 3comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
best guaranteed success
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 = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or
2 2 31
|(6− 3 B)a−x (2− 3 B)a−y (4 + 3 B)a
| z = 10
Expanding, obtain
36 − 8B + 4 B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B2 − 8B − 44 = 0. Thus B = 8± 64−176
2 = 11.75 (taking positive option) and so
B 2 2 1
= 3 (11.75)ax − 3 (11.75)ay + 3 (11.75)az = 7.83ax − 7.83ay + 3.92az
1.4. circle, centered at the origin with a radius of 2 units, lies in the xy plane.
Determine √the unit
vector in rectangular components that lies in the xy plane, is tangent− to the
circle at ( 3, 1,
0), and is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is
t = −aφ . Its√x and
y components are tx = −aφ · ax = sin φ, and t√y = −aφ · ay = − cos φ. φ
At the point (−
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 3
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
,6/2027,
stions and
a 4comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success
= 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay). 3,
1),
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
(−48, 72, 162)
a =G = ( —0 .26, 0. 39
, . 0 88)
|(−48, 72, 162)|
c) a unit vector directed from Q
toward P :
(3 1 4)
a = P − Q = ,− ,
QP
√ = (0.59, 0.20, −0.78)
26
|P − Q|
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
1.6. Find the acute angle between the two vectors A = 2ax + ay +− 3az and B = ax
3ay + 2az by using the definition of:
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 4
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
stions and
a 1comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success
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 = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
a = (26, 10, 4) = (0.92, 0.36, 0.14)
|(26, 10, 4)|
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. Vector A extends from the origin to (1,2,3) and vector B from the origin to
(2,3,-2).
a) Find the unit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−ax − ay + 5az)
w√hose magnitude is |A − B| = [(−ax − ay + 5az ) · (−ax − ay + 5az )]1/2
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 1
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
,6/2027,
stions and
a 2comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success
√
= 1 + 1 + 25 =
3 3 = 5.20. The unit vector is therefore
aAB = (−ax − ay + 5az)/5.20
b) find the unit vector in the direction of the line extending from the origin to
the midpoint of the line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5,
0.5)
The unit vector is then
a = (1.5ax + 2.5ay + 0.5az)
+ + )/2.96
= (1.5a
2.5a 0.5a
mp p x y z
(1.5)2 + (2.5)2 +
(0.5)2
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 2
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
,6/2027,
stions and Latest update!!!! Engineering electromagnetics solution manuals
a 3comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
best guaranteed success
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 = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or
2 2 31
|(6− 3 B)a−x (2− 3 B)a−y (4 + 3 B)a
| z = 10
Expanding, obtain
36 − 8B + 4 B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B2 − 8B − 44 = 0. Thus B = 8± 64−176
2 = 11.75 (taking positive option) and so
B 2 2 1
= 3 (11.75)ax − 3 (11.75)ay + 3 (11.75)az = 7.83ax − 7.83ay + 3.92az
1.4. circle, centered at the origin with a radius of 2 units, lies in the xy plane.
Determine √the unit
vector in rectangular components that lies in the xy plane, is tangent− to the
circle at ( 3, 1,
0), and is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is
t = −aφ . Its√x and
y components are tx = −aφ · ax = sin φ, and t√y = −aφ · ay = − cos φ. φ
At the point (−
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 3
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
,6/2027,
stions and
a 4comprehensive
answers
of 342Prepare
with complete
collection
with Exam solutions.
ofQ&A
exam-focused
2026/2027,
Designed
questions
aforcomprehensive
students,
and answers
instructors,
collection
with
and
complete
ofprofessional
exam-focused
solutions.
exam
questions
Designed
candidates,
and
for answers
students,
this resource
with
instructors,
complete
supportsand
solutions.
effective
professional
revision,
Designed
exam
practice
for
candidates,
students, ins
th
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success
= 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay). 3,
1),
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
(−48, 72, 162)
a =G = ( —0 .26, 0. 39
, . 0 88)
|(−48, 72, 162)|
c) a unit vector directed from Q
toward P :
(3 1 4)
a = P − Q = ,− ,
QP
√ = (0.59, 0.20, −0.78)
26
|P − Q|
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
1.6. Find the acute angle between the two vectors A = 2ax + ay +− 3az and B = ax
3ay + 2az by using the definition of:
Latest update!!!! Engineering electromagnetics solution manuals
best guaranteed success 4
dprehensive
answersPrepare
with
collection
complete
withof
Exam
exam-focused
solutions.
Q&A 2026/2027,
Designed
questions
for
a comprehensive
students,
and answers
instructors,
with
collection
complete
and of
professional
exam-focused
solutions. exam
Designed
questions
candidates,
for students,
and this
answers
resource
instructors,
with supports
complete
and professional
effective
solutions.
revision,
exam
Designed
candidates,
practice
for students,
testing,
this resource
instructors
a su
