Engineering Electromagnetics - 7th Edition - William H.
Hayt Solution Manual
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
(26, 10, 4)
a= = (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. The three vertices of a triangle are located at A(−1, 2, 5), B(−4, −2, −3), and C(1, 3, −2).
a) Find the length of the perimeter of the triangle: Begin with AB = (−3, −4, − 8 ) , √B C = (5, 5, 1),
a√nd CA = (−2, √− 1 , 7). Then the perimeter will be ℓ = |AB| + |BC| + |CA| = 9 + 16+ 64 +
25+ 25+ 1+ 4 + 1 + 49 = 23.9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC: The vector from the origin to the midpoint of AB is MAB = 1 (A +B) = 1 (−5ax + 2az).
The vector from the origin to the midpoint of BC is MBC = 2 (B + C) = 2 (−3ax + ay − 5az).
The vector from midpoint to midpoint is now MAB − MBC = 1 (−2ax − ay + 7az). The unit
vector is therefore
MAB − MBC (−2ax − ay + 7az)
aMM = = = −0.27a − 0.14ay + 0.95az
|MAB − MBC| 7.35
where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC. First we find AC = 2ax + ay − 7az, which we recognize
as −7.35 aMM . The vectors are thus parallel (but oppositely-directed).
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 2A = (6, −2, −4)2 and B = 1 B(2,
1
−2, 1), we use the fact that |B − A| = 10, or
|(6 − 3 B)ax − (2 − 3 B)ay − (4 + 3 B)az| = 10
Expanding, obtain
36 − 8B + 4 B2 + 4 − 8 B + 4 B2 + 16+ 8 B + 1 B2 = 100
√
8± 64−176
or B2 − 8B − 44 = 0. Thus B = = 11.75 (taking positive option) and so
, 2 2 1
B= (11.75)a − − 7.83a + 3.92a
3
,1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determ√ine 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 = √
a φ . Its x and y
components are tx = aφ · ax = − sin φ, and ty = aφ · ay = cos φ. At the point ( 3, 1), φ = 30◦,
and so t = − sin 30◦ax + cos 30◦ay = 0.5(−ax + 3ay).
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 = = ( 0.26, 0.39, 0.88)
|(−48, 72, 162)|
c) a unit vector directed from Q toward P :
P−Q (3, 1, 4)
a = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
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. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz, describe
the surface r · a = B. What is the relation between the the unit vector a and the scalar B to this
surface? (HINT: Consider first a simple example with a = ax and B = 1, and then consider any a
and B.):
We could consider a general unit vector, a = A1ax + A2ay + A3az, where A2 + A2 + A2 = 1.
Then r · a = A1x + A2y + A3z = f (x, y, z) = B. This is the equation of a planar surface, where
f = B. The relation of a to the surface becomes clear in the special case in which a = ax. We
obtain r · a = f (x) = x = B, where it is evident that a is a unit normal vector to the surface
(as a look ahead (Chapter 4), note that taking the gradient of f gives a).
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. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = 3ax 2ay + 4az and B = 2ax + ay 2az. Does this
ambiguity exist when the dot product is used?
We use the relation A × B = |A||B| sin θn. With the given vectors we find
√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
` ˛¸ x
±n
where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really w a √
n t t h√e mag
√ nitude of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two values of θ (75.7◦ and 104.3◦) satisfy this equation, and
hence the real ambiguity.
√
In using the d√o t product, we find A · B = 6 − 2 − 8 = − 4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦. Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
25
G= (xax + yay)
(x2 + y2)
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:
∫ 4∫ 2
G · aydzdx
0 0
∫ 4 ∫ 2 ∫ 4 ∫ 2 ∫ 4
25 25 350
1 4
= 350 × tan−1 − 0 = 26
7 7
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = √b (a x + √ ay + az) and B =
b(ax − ay + az). Now use A · B = |A||B| cos θ, or b (1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
2
1/3 ⇒ θ = 70.53◦. This result (in magnitude) is the same for any two diagonal vectors.
Hayt Solution Manual
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
(26, 10, 4)
a= = (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. The three vertices of a triangle are located at A(−1, 2, 5), B(−4, −2, −3), and C(1, 3, −2).
a) Find the length of the perimeter of the triangle: Begin with AB = (−3, −4, − 8 ) , √B C = (5, 5, 1),
a√nd CA = (−2, √− 1 , 7). Then the perimeter will be ℓ = |AB| + |BC| + |CA| = 9 + 16+ 64 +
25+ 25+ 1+ 4 + 1 + 49 = 23.9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC: The vector from the origin to the midpoint of AB is MAB = 1 (A +B) = 1 (−5ax + 2az).
The vector from the origin to the midpoint of BC is MBC = 2 (B + C) = 2 (−3ax + ay − 5az).
The vector from midpoint to midpoint is now MAB − MBC = 1 (−2ax − ay + 7az). The unit
vector is therefore
MAB − MBC (−2ax − ay + 7az)
aMM = = = −0.27a − 0.14ay + 0.95az
|MAB − MBC| 7.35
where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC. First we find AC = 2ax + ay − 7az, which we recognize
as −7.35 aMM . The vectors are thus parallel (but oppositely-directed).
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 2A = (6, −2, −4)2 and B = 1 B(2,
1
−2, 1), we use the fact that |B − A| = 10, or
|(6 − 3 B)ax − (2 − 3 B)ay − (4 + 3 B)az| = 10
Expanding, obtain
36 − 8B + 4 B2 + 4 − 8 B + 4 B2 + 16+ 8 B + 1 B2 = 100
√
8± 64−176
or B2 − 8B − 44 = 0. Thus B = = 11.75 (taking positive option) and so
, 2 2 1
B= (11.75)a − − 7.83a + 3.92a
3
,1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determ√ine 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 = √
a φ . Its x and y
components are tx = aφ · ax = − sin φ, and ty = aφ · ay = cos φ. At the point ( 3, 1), φ = 30◦,
and so t = − sin 30◦ax + cos 30◦ay = 0.5(−ax + 3ay).
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 = = ( 0.26, 0.39, 0.88)
|(−48, 72, 162)|
c) a unit vector directed from Q toward P :
P−Q (3, 1, 4)
a = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
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. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz, describe
the surface r · a = B. What is the relation between the the unit vector a and the scalar B to this
surface? (HINT: Consider first a simple example with a = ax and B = 1, and then consider any a
and B.):
We could consider a general unit vector, a = A1ax + A2ay + A3az, where A2 + A2 + A2 = 1.
Then r · a = A1x + A2y + A3z = f (x, y, z) = B. This is the equation of a planar surface, where
f = B. The relation of a to the surface becomes clear in the special case in which a = ax. We
obtain r · a = f (x) = x = B, where it is evident that a is a unit normal vector to the surface
(as a look ahead (Chapter 4), note that taking the gradient of f gives a).
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. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = 3ax 2ay + 4az and B = 2ax + ay 2az. Does this
ambiguity exist when the dot product is used?
We use the relation A × B = |A||B| sin θn. With the given vectors we find
√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
` ˛¸ x
±n
where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really w a √
n t t h√e mag
√ nitude of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two values of θ (75.7◦ and 104.3◦) satisfy this equation, and
hence the real ambiguity.
√
In using the d√o t product, we find A · B = 6 − 2 − 8 = − 4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦. Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
25
G= (xax + yay)
(x2 + y2)
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:
∫ 4∫ 2
G · aydzdx
0 0
∫ 4 ∫ 2 ∫ 4 ∫ 2 ∫ 4
25 25 350
1 4
= 350 × tan−1 − 0 = 26
7 7
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = √b (a x + √ ay + az) and B =
b(ax − ay + az). Now use A · B = |A||B| cos θ, or b (1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
2
1/3 ⇒ θ = 70.53◦. This result (in magnitude) is the same for any two diagonal vectors.
