Solutions Manual – Engineering Electromagnetics, 9th Edition by Hayt and Buck

,
,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),√ BC = (5, 5, 1),
and
√ 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 = 12 (A + B) = 12 (−5ax + 2az ).
The vector from the origin to the midpoint of BC is MBC = 21 (B + C) = 12 (−3ax + ay − 5az ).
The vector from midpoint to midpoint is now MAB − MBC = 12 (−2ax − ay + 7az ). The unit
vector is therefore

MAB − MBC (−2ax − ay + 7az )
aM M = = = −0.27ax − 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 aM M . 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 A = (6, −2, −4) and B = 31 B(2, −2, 1), we use the fact that |B − A| = 10, or
|(6 − 23 B)ax − (2 − 23 B)ay − (4 + 31 B)az | = 10
Expanding, obtain
36 − 8B + 94 B 2 + 4 − 83 B + 49 B 2 + 16 + 38 B + 19 B 2 = 100
√
8± 64−176
or B 2 − 8B − 44 = 0. Thus B = 2 = 11.75 (taking positive option) and so

2 2 1
B= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3

1

, 1.4. A 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 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 + 18z 2 az . 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)
aG = = (−0.26, 0.39, 0.88)
|(−48, 72, 162)|

c) a unit vector directed from Q toward P :

P−Q (3, −1, 4)
aQP = = √ = (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), 18z 2 )|, or
10 = |(4xy, 2x2 + 4, 3z 2 )|, so the equation is

100 = 16x2 y 2 + 4x4 + 16x2 + 16 + 9z 4


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 = A1 ax + A2 ay + A3 az , where A21 + A22 + A23 = 1.
Then r · a = A1 x + A2 y + A3 z = 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 = 4zy 2 cos 2xax + 2zy sin 2xay + y 2 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 = y 2 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 zy 2 cos 2x = zy sin 2x =
y 2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.



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