1
SOLUTIONS MANUAL
, 1
Solutions Manual
1.3. The vector from the origin to the point A is given as −
(6, −2, 4), and the unit vector directed
B is (2, 2, 1)/3. If points A and B are ten units apart, find the
from the origin toward point —
coordinates of point B.
With A = (6, −2, −4) and B = 1 B(2,
3 −2, 1), we use the fact that |B − A| = 10, or
2 2 1
|(6− 3 B)ax − (2− 3 B)ay− (4 + 3 B)a|z = 10
Expanding, 4obtain
36 − 8B + 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
2 2 1
B= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
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
RAM × RAN (350, −200, 340)
a = = = (0.664, −0.379, 0.645)
p
|RAM × RAN | 527.35
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 :
(−10, 8, 15)
aAN = √ = (−0.507, 0.406, 0.761)
389
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
(20, 18,
aAM = = (0.697, 0.627, −0.348)
−10)
|(20, 18, −10)|
Now
1 1
(aAM + aAN ) = [(0.697, 0.627, −0.348) + (−0.507, 0.406, 0.761)] = (0.095, 0.516,
2 2
0.207)
Finally,
(0.095, 0.516,
abis = 0.207) = (0.168, 0.915, 0.367)
|(0.095, 0.516,
0.207)|
, 2
1.27. The surfaces r = 2 and 4, θ = 30◦ and 50◦, and φ = 20◦ and 60◦ identify a closed surface.
a) Find the enclosed volume: This will be
∫ 60◦ ∫ 50◦ ∫ 4
Vol = r2 sin θdrdθdφ = 2.91
20◦ 30◦ 2
where degrees have been converted to
radians.
b) Find the total area of the enclosing surface: ∫ 4 ∫ 60◦
∫ 60 ◦ ∫ 50◦ r(sin 30◦ + sin 50◦)drdφ
Area = (42 + 22) sin θdθdφ + 2 20◦
20◦ 30◦
∫ 50◦ ∫ 4
+2 rdrdθ = 12.61
30◦ 2
c) Find the total length of the twelve edges of the surface:
∫ 4
Length = ∫ 50 ∫ 60
(4+ 2)dθ (4 sin 50◦ + 4 sin 30◦ + 2 sin 50◦ + 2 sin 30◦)dφ
◦ ◦
dr +
4 2 +
30◦ 20◦
2
= 17.49
d) Find the length of the longest straight line that lies entirely within the surface: This will be
from
A(r = 2, θ = 50 ◦ ,φ = 20◦) to B(r = 4, θ = 30 ◦ ,φ = 60◦) or
A(x = 2 sin 50◦ cos 20◦,y = 2 sin 50◦ sin 20◦,x = 2 cos 50◦)
to
B(x = 4 sin 30◦ cos 60◦,y = 4 sin 30◦ sin 60◦,x = 4 cos 30◦)
or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B − A = (−0.44, 1.21, 2.18) and
Length = |B − A| = 2.53
2.5. Let a point charge Q125 nC be located at P1(4, −2, 7) and a charge Q2 = 60 nC be at P2(−3, 4,
−2).
a) If s = s0, find E at P3(1, 2, 3): This field will be
10−9 25R13 60R23
E= +
4πs0 |R 13 |3 |R 23 |3
√
where R13 = −3ax + 4ay − 4az and R23 = 4ax − 2ay + 5az. Also, |R13| = 41 and
√
|R23| = 45.
So
E 10
=
−9
25 × (−3ax + 4ay − 4az)+ 60 × (4ax − 2ay + 5az)
4πs0 (41)1.5 (45)1.5
= 4.58ax − 0.15ay + 5.51az
b) At what point on the y axis is Ex = 0? P3 is now √ at (0, y, 0), so R13 = −4ax + √
( y + 2)ay − 7az
and R23 = 3ax + (y − 4)ay + 2az. Also, 1
|R13| = Ex = + 0
Now the x component of E 4πs [65+
at the new(yP+3 2) 2]1.5
will [13+ (y − 4)2]1.5 −
0 9
be:
SOLUTIONS MANUAL
, 1
Solutions Manual
1.3. The vector from the origin to the point A is given as −
(6, −2, 4), and the unit vector directed
B is (2, 2, 1)/3. If points A and B are ten units apart, find the
from the origin toward point —
coordinates of point B.
With A = (6, −2, −4) and B = 1 B(2,
3 −2, 1), we use the fact that |B − A| = 10, or
2 2 1
|(6− 3 B)ax − (2− 3 B)ay− (4 + 3 B)a|z = 10
Expanding, 4obtain
36 − 8B + 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
2 2 1
B= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
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
RAM × RAN (350, −200, 340)
a = = = (0.664, −0.379, 0.645)
p
|RAM × RAN | 527.35
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 :
(−10, 8, 15)
aAN = √ = (−0.507, 0.406, 0.761)
389
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
(20, 18,
aAM = = (0.697, 0.627, −0.348)
−10)
|(20, 18, −10)|
Now
1 1
(aAM + aAN ) = [(0.697, 0.627, −0.348) + (−0.507, 0.406, 0.761)] = (0.095, 0.516,
2 2
0.207)
Finally,
(0.095, 0.516,
abis = 0.207) = (0.168, 0.915, 0.367)
|(0.095, 0.516,
0.207)|
, 2
1.27. The surfaces r = 2 and 4, θ = 30◦ and 50◦, and φ = 20◦ and 60◦ identify a closed surface.
a) Find the enclosed volume: This will be
∫ 60◦ ∫ 50◦ ∫ 4
Vol = r2 sin θdrdθdφ = 2.91
20◦ 30◦ 2
where degrees have been converted to
radians.
b) Find the total area of the enclosing surface: ∫ 4 ∫ 60◦
∫ 60 ◦ ∫ 50◦ r(sin 30◦ + sin 50◦)drdφ
Area = (42 + 22) sin θdθdφ + 2 20◦
20◦ 30◦
∫ 50◦ ∫ 4
+2 rdrdθ = 12.61
30◦ 2
c) Find the total length of the twelve edges of the surface:
∫ 4
Length = ∫ 50 ∫ 60
(4+ 2)dθ (4 sin 50◦ + 4 sin 30◦ + 2 sin 50◦ + 2 sin 30◦)dφ
◦ ◦
dr +
4 2 +
30◦ 20◦
2
= 17.49
d) Find the length of the longest straight line that lies entirely within the surface: This will be
from
A(r = 2, θ = 50 ◦ ,φ = 20◦) to B(r = 4, θ = 30 ◦ ,φ = 60◦) or
A(x = 2 sin 50◦ cos 20◦,y = 2 sin 50◦ sin 20◦,x = 2 cos 50◦)
to
B(x = 4 sin 30◦ cos 60◦,y = 4 sin 30◦ sin 60◦,x = 4 cos 30◦)
or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B − A = (−0.44, 1.21, 2.18) and
Length = |B − A| = 2.53
2.5. Let a point charge Q125 nC be located at P1(4, −2, 7) and a charge Q2 = 60 nC be at P2(−3, 4,
−2).
a) If s = s0, find E at P3(1, 2, 3): This field will be
10−9 25R13 60R23
E= +
4πs0 |R 13 |3 |R 23 |3
√
where R13 = −3ax + 4ay − 4az and R23 = 4ax − 2ay + 5az. Also, |R13| = 41 and
√
|R23| = 45.
So
E 10
=
−9
25 × (−3ax + 4ay − 4az)+ 60 × (4ax − 2ay + 5az)
4πs0 (41)1.5 (45)1.5
= 4.58ax − 0.15ay + 5.51az
b) At what point on the y axis is Ex = 0? P3 is now √ at (0, y, 0), so R13 = −4ax + √
( y + 2)ay − 7az
and R23 = 3ax + (y − 4)ay + 2az. Also, 1
|R13| = Ex = + 0
Now the x component of E 4πs [65+
at the new(yP+3 2) 2]1.5
will [13+ (y − 4)2]1.5 −
0 9
be:
