CHAPTER 7
Applications of Vectors
m 2 5 52 1 122 2 2(5)(12) cos 135
2 !2
5 25 1 144 2 120a
b
2
5 169 1 84.85
5 253.85
Review of Prerequisite Skills, p. 350
1. The velocity relative to the ground has a
magnitude equivalent to the hypotenuse of a
triangle with sides 800 and 100. So, by the
Pythagorean theorem we can find the magnitude of
the velocity.
v 2 5 8002 1 1002
5 640 000 1 10 000
5 650 000
v 5 "650 000
8 806 km>h
100
800
m 5 "253.85
8 15.93 units
122 5 15.932 1 52 2 2(15.93)(5) cos u
144 5 253.76 1 25 2 159.3 cos u
2134.76 5 2159.3 cos u
134.76
cos u 5
159.3
134.76
u 5 cos21 a
b
159.3
8 32.2°
So the displacement is 15.93 units, W 32.2° N.
3.
z
u
C(–2, 0, 1)
B(–3, 2, 0)
y
A(0, 1, 0)
tan u 5
u 5 tan
100
800
21
x
100
a
b
800
u 8 7.1°
The velocity of the airplane relative to the ground is
about 806 km> h N 7.1° E.
2.
displacement
12
u
5
The angle between the two displacements is 135°.
The magnitude, m, and the angle, u, of the
displacement can be found using the cosine law.
Calculus and Vectors Solutions Manual
D(0, 2, –3)
4. a. (3, 22, 7)
l 5 magnitude
5 "32 1 (22)2 1 72
5 "9 1 4 1 49
5 "62
8 7.87
b. (29, 3, 14)
l 5 magnitude
5 "(29)2 1 32 1 142
5 !81 1 9 1 196
5 !286
8 16.91
7-1
c. (1, 1, 0)
l 5 magnitude
5 "12 1 12 1 02
5 "2
8 1.41
d. (2, 0, 29)
l 5 magnitude
5 "22 1 02 1 (29)2
5 "4 1 0 1 81
5 "85
8 9.22
5. a. A(x, y, 0)
In the xy-plane at the point (x, y).
b. B(x, 0, z)
In the xz-plane at the point (x, z).
c. C(0, y, z)
In the yz-plane at the point (y, z).
6. a. (26, 0) 1 7(1, 21)
>
>
>
>
5 (26i 1 0j ) 1 7(i 2 j )
>
>
>
>
5 (26i 1 0j ) 1 (7i 2 7j )
>
>
5 i 2 7j
b. (4, 21, 3) 2 (22, 1, 3)
>
>
>
>
>
>
5 (4i 2 j 1 3k ) 2 (22i 1 j 1 3k )
>
>
5 6i 2 2j
c. 2(21, 1, 3) 1 3(22, 3, 21)
>
>
>
>
>
>
5 2(2i 1 j 1 3k ) 1 3(22i 1 3j 2 k )
>
>
>
>
>
>
5 (22i 1 2j 1 6k ) 1 (26i 1 9j 2 3k )
>
>
>
5 28i 1 11j 1 3k
1
3
d. 2 (4, 26, 8) 1 (4, 26, 8)
2
2
>
>
>
>
3 >
1 >
5 2 (4i 2 6j 1 8k ) 1 (4i 2 6j 1 8k )
2 >
2>
>
>
>
>
5 (22i 1 3j 2 4k ) 1 (6i 2 9j 1 12k )
>
>
>
5 4i 2 6j 1 8k
>
>
7. a. a> 1 b >
>
>
>
5 (3i
1
2j
2
k
)
1
(22i
1
j
)
>
>
>
5 i 1 3j 2 k
>
>
b. a 2> b >
>
>
>
1
j
)
5 (3i> 1 2j> 2 k>) 2 (22i
>
>
5 (3i
1
2j
2
k
)
1
(2i
2
j
)
>
>
>
5 5i 1 j 2 k
>
>
c. 2a 2> 3b >
>
>
>
1 2j
2 k>) 2 3(22i
1> j )
5 2(3i
>
>
>
5 (6i> 1 4j
2 2k
) 1 (6i 2 3j )
>
>
5 12i 1 j 2 2k
7-2
7.1 Vectors as Forces, pp. 362–364
1. a. 10 N is a melon, 50 N is a chair, 100 N is a
computer
b. Answers will vary.
2. a.
10 N
20 N
30 N
b. 180°
3.
10 N
20 N
The forces should be placed in a line along the
same direction.
4. For three forces to be in equilibrium, they must
form a triangle, which is a planar figure.
5.
equilibrant
f2
resultant
f1
a. The resultant is equivalent in magnitude to the
hypotenuse, h, of the triangle with 5 and 12 as sides
12
and is directed northeast at an angle of sin21 h .
Thus, the resultant is "52 1 122 5 13 N at an angle
of sin21 12
13 5 N 22.6° E. The equlibrant is equal in
magnitude and opposite in direction of the resultant.
Thus, the equilibrant is 13 N at an angle of S 22.6° W.
b. The resultant is "92 1 122 5 15 N at an angle of
sin21 12
15 5 S 36.9° W. The equilibrant, then, is 15 N
at N 36.9° E.
6. For three forces to form equilibrium, they must
be able to form a triangle or a balanced line, so
a. Yes, since 3 1 4 . 7 these can form a triangle.
b. Yes, since 9 1 40 . 41 these can form a triangle.
c. No, since "5 1 6 , 9 these cannot form a
triangle.
d. Yes, since 9 1 10 5 19, placing the 9 N and 10 N
force in a line directly opposing the 19 N force
achieves equilibrium.
Chapter 7: Applications of Vectors
7. Arms 90 cm apart will yield a resultant with a
smaller magnitude than at 30 cm apart. A resultant
with a smaller magnitude means less force to
counter your weight, hence a harder chin-up.
8. Using the cosine law, the resultant has a magnitude,
r, of
>
>
> >
r 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos 120°
Now we look at x1 and x2. We know
>
x1 5 @ f1 @ sin 15
r 5 "148
8 12.17 N
Using the sine law, the resultant’s angle, u, can be
found by
sin u
sin 120°
5
8
12.17
@ f2 @ (1.035) 5 10
1
5 62 1 82 2 2(6)(8)a2 b
2
5 36 1 64 1 48
5 148
sin u 5 8
"3
2
12.17
u 5 sin21 8
"3
2
12.17
8 34.7° from the 6 N force toward the 8 N
force. The equilibrant, then, would be 12.17 N at
180° 2 34.7° 5 145.3° from the 6 N force away
from the 8 N force.
9.
10 N
f1
f2
x2 5 @ f2 @ sin 75
x1 1 x2 5 10
>
So @ f1 @ sin 15 1 @ f2 @ sin 75 5 10
>
Substituting then solving for f2 yields
> cos 75
>
@ f2 @
sin 15 1 @ f2 @ sin 75 5 10
cos 15
> cos 75
@ f2 @ a
sin 15 1 sin 75b 5 10
cos 15
>
>
>
@ f2 @ 5 9.66 N
>
Now we solve for f1 :
>
@ f1 @ 5 @ f2 @
cos 75
cos 15
>
cos 75
@ f1 @ 5 (9.66)
cos 15
>
@ f1 @ 5 (9.66)(0.268)
>
>
@ f1 @ 5 2.59 N
So the force 15° from the 10 N force is 9.66 N and
the force perpendicular to it is 2.59 N.
10. The force of the block is
(10 kg)(9.8 N>kg) 5 98 N. The component of this
force parallel to the ramp is
(98) sin 30° 5 (98)A 12 B 5 49 N, directed down the
ramp. So the force preventing this block from
moving would be 49 N directed up the ramp.
11. a.
>
7N
13 N
>
f1 5 force 15° from the 10 N force
>
>
f2 5 force perpendicular to f1
>
x1 5 component of f1 parallel to the 10 N force
>
x2 5 component of f2 parallel to the 10 N force
>
>
We know that the components of f1 and f2
perpendicular to the 10 N force must be equal, so we
can write
>
>
@ f1 @ cos 15 5 @ f2 @ cos 75
>
> cos 75
@ f1 @ 5 @ f2 @
cos 15
Calculus and Vectors Solutions Manual
8N
b. Using the cosine law for the angle, u, we have
132 5 82 1 72 2 2(8)(7) cos u
169 5 64 1 49 2 112 cos u
56 5 2112 cos u
256
cos u 5
112
21
u 5 cos21
2
5 120
7-3
This is the angle between the vectors when placed
head to tail. So the angle between the vectors when
placed tail to tail is 180° 2 120° 5 60°.
12. The 10 N force and the 5 N force result in a 5 N
force east. The 9 N force and the 14 N force result
in a 5 N force south. The resultant of these is now
equivalent to the hypotenuse of the right triangle
with 5 N as both bases and is directed 45° south of
east. So the resultant is "52 1 52 5 "50 8 7.1 N
45° south of east.
13.
of an equilateral triangle are 60°, the resultant will be
at a 60° angle with the other two vectors.
b. Since the equilibrant is directed opposite the
resultant, the angle between the equilibrant and the
other two vectors
>
>is 180° 2 60° 5 120°.
15. Since f1 and f2 act opposite one another,
they
>
>
net a 10 N force directed west. Since f3 and f4 act
opposite one another, they net a 10 N force directed
45° north of east.
So using the cosine law to find
>
the resultant, fr,
>
@ fr @ 2 5 102 1 102 2 2(10)(10) cos 45°
5 200 2 200 cos 45°
5 200 2 200 a
f1 = 24 N
@ fr @ 5
>
resultant = 25 N
equlibrant
f2
a. Using the Pythagorean theorem,
>
>
@ f1 @ 2 1 @ f2 @ 2 5 252
>
>
@ f2 @ 2 5 252 2 @ f1 @ 2
5 252 2 242
5 49
>
@ f2 @ 5 7
>
b. The angle, u, between f1 and the resultant is
given by
>
@ f2 @
sin u 5
25
7
sin u 5
25
7
u 5 sin21
25
8 16.3°
>
So the angle between f1 and the equilibrant is
180° 2 16.3° 5 163.7°.
14. a.
60°
1N
1N
60°
60°
1N
For these three equal forces to be in equilibrium, they
must form an equilateral triangle. Since the resultant
will lie along one of these lines, and since all angles
7-4
Å
"2 b
2
200 2 200a
"2
b
2
8 7.65 N
Since our net forces are equal at 10 N, the angle of
the resultant is directed halfway
between
the two, or
>
>
1
at 2 (135°) 5 67.5° from f2 toward f3.
16.
T2
T1
20 kg
Let T1 be the tension in the 30° rope and T2 be the
tension in the 45° rope.
Since this system is in equilibrium, we know that
the horizontal components of T1 and T2 are equal
and opposite and the vertical components add to be
opposite the action of the mass. Also, the force
produced by the mass is (20 kg)(9.8 N>kg) 5 196 N.
So we have a system of two equations: the first,
(T1 ) cos 30° 5 (T2 ) cos 45° represents the balance
of the horizontal components, and the second,
(T1 ) sin 30° 1 (T2 ) sin 45° 5 196 represents the
balance of the vertical components with the mass.
So solving this system of two equations with two
variable gives the desired tensions.
T1 cos 30° 5 T2 cos 45°
cos 45°
T1 5 T2
cos 30°
T1 sin 30° 1 T2 sin 45° 5 196
cos 45°
b sin 30° 1 T2 sin 45° 5 196
aT2
cos 30°
cos 45°
T2 aa
b sin 30° 1 sin 45°b 5 196
cos 30°
Chapter 7: Applications of Vectors
T2 (1.12) 5 196
T2 8 175.73 N
cos 45°
T1 5 (175.73)
cos 30°
8 143.48 N
Thus the tension in the 45° rope is 175.73 N and the
tension in the 30° rope is 143.48 N.
17.
40 cm
24 cm
Thus the tension in the 24 cm string is 39.2 N and
the tension in the 32 cm string is 29.4 N.
18.
resultant
x
35°
u
2x
32 cm
5 kg
First, use the Cosine Law to find the angles the
strings make at the point of suspension. Let u1 be the
angle made by the 32 cm string and u2 be the angle
made by the 24 cm string.
242 5 322 1 402 2 2(32)(40) cos u1
22048 5 22560 cos u1
2048
u1 5 cos21
2560
8 36.9°
322 5 242 1 402 2 2(24)(40) cos u2
21152 5 21920 cos u2
1152
u2 5 cos21
1920
8 53.1°
A keen eye could have recognized this triangle as a
3-4-5 right triangle and simply used the Pythagorean
theorem as well. Now we set up the same system of
equations as in problem 16, with T1 being the tension
in the 32 cm string and T2 being the tension in the
24 cm string, and the force of the mass being
(5 kg)(9.8 N>kg) 5 49 N.
T1 cos 36.9° 5 T2 cos 53.1°
cos 53.1°
T1 5 T2
cos 36.9°
T1 sin 36.9° 1 T2 sin 53.1° 5 49
cos 53.1°
aT2
b sin 36.9° 1 T2 sin 53.1° 5 49
cos 36.9°
cos 53.1°
T2 aa
b sin 36.9° 1 sin 53.1°b 5 49
cos 36.9°
T2 (1.25) 5 49
T2 8 39.2 N
cos 53.1°
T1 5 (39.2)
cos 36.9°
8 29.4 N
Calculus and Vectors Solutions Manual
N
W
E
S
(Port means left and starboard means right.) We are
looking for the resultant of these two force vectors
that are 35° apart. We don’t know the exact value
of the force, so we will call it x. So the small tug
is pulling with a force of x and the large tug is
pulling with a force of 2x. To find the magnitude
of the resultant, r, in terms of x, we use the cosine
law.
r 2 5 x 2 1 (2x)2 2 2(x)(2x) cos 145°
5 x 2 1 4x 2 2 4x 2 cos 145°
8 5x 2 2 4x 2 (20.8192)
8 5x 2 1 3.2768x 2
8 8.2768x 2
r 8 "8.2768x 2
8 2.8769x
Now we use the cosine law again to find the angle,
u, made by the resultant.
x 2 5 r 2 1 (2x)2 2 2(2.8769x)(2x) cos u
x 2 5 8.2768x 2 1 4x 2 2 11.5076x 2 cos u
x 2 5 12.2768x 2 2 11.5076x 2 cos u
211.2768x 2 5 211.5076x 2 cos u
11.2768
cos u 5
11.5076
11.2768
u 5 cos21 a
b
11.5076
8 11.5° from the large tug toward the
small tug, for a net of 8.5° to the starboard side.
7-5
19.
u10N 5 cos21 a
5N
8N
10N
a. First we will find the resultant of the 5 N and
8 N forces. Use the Pythagorean theorem to find the
magnitude, m.
m 2 5 52 1 82
5 25 1 64
5 89
m 5 "89 8 9.4
Next we use the Pythagorean theorem again to find
the magnitude, M, of the resultant of this net force
and the 10 N force.
M 2 5 m 2 1 102
5 89 1 100
5 189
M 5 "189 8 13.75
Since the equilibrant is equal in magnitude to the
resultant, we have the magnitude of the equilibrant
equal to approximately 13.75 N.
b. To find each angle, use the definition of cosine
with respect each force as a leg and the resultant as
the hypotenuse. Let u5N be the angle from the
5 N force to the resultant, u8N be the angle from the
8 N force to the resultant, and u10N be the angle
from the 10 N force to the resultant.
Let the sign of the resultant be negative, since it is
in a direction away from the head of each of the
given forces.
5
cos u5N 5
213.75
5
u5N 5 cos21 a
b
213.75
8 111.3°
8
cos u8N 5
213.75
8
u8N 5 cos21 a
b
213.75
8 125.6°
10
cos u10N 5
213.75
7-6
10
b
213.75
8 136.7°
20. We know that the resultant of these two forces is
equal in magnitude and angle to the> diagonal
line
>
of the parallelogram formed with f1 and f2 as legs
>
>
and has diagonal length @ f1 1 f2 @ . We also know
from the cosine law that
>
>
>
>
>
>
@ f1 1 f2 @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos f
where f is the supplement to u in our parallelogram.
Since we know f 5 180 2 u, then
cos f 5 cos (180 2 u) 5 2cos u.
Thus
we> have >
>
>
>
>
@ f1 1 f2 @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos f
>
>
>
>
5 @ f1 @ 2 1 @ f2 @ 2 1 2 @ f1 @ @ f2 @ cos u
@ f1 1 f2 @ 5 " @ f1 @ 2 1 @ f2 @ 2 1 2 @ f1 @ @ f2 @ cos u
>
>
>
>
>
>
7.2 Velocity, pp. 367–370
1. a. Both the woman and the train’s velocities are
in the same direction, so we add them.
80 km>h 1 4 km>h 5 84 km>h
b. The woman’s velocity is directed opposite that of
train, so we subtract her velocity from the train’s.
80 km>h 2 4 km>h 5 76 km>h. The resultant is in
the same direction as the train’s movement.
2. a. The velocity of the wind is directed opposite that
of the airplane, so we subtract the wind’s velocity
from the airplane’s.
600 km>h 2 100 km>h 5 500 km>h north.
b. Both the wind and the airplane’s velocities are in
the same direction, so we add them.
600 km>h 1 100 km>h 5 700 km>h north.
3. We use the Pythagorean theorem to find the
magnitude, m, of the resultant velocity and we use
the definition of sine to find the angle, u, made.
m 2 5 3002 1 502
5 90 000 1 2500
5 92 500
m 5 "92 500
8 304.14 km>h
50
tan u 5
300
50
u 5 tan21
300
8 9.5°. The resultant is 304.14 km> h, W 9.5° S.
Chapter 7: Applications of Vectors
4. Adam must swim at an angle, u, upstream so as
to counter the 1 km> h velocity of the stream. This is
equivalent to Adam swimming along the hypotenuse
of a right traingle with 1 km> h leg and a 2 km> h
hypotenuse. So the angle is found using the definition
of cosine.
1
cos u 5
2
1
u 5 cos21
2
5 60° upstream
5. a. 2 m> s forward
b. 20 m> s 1 2 m> s 5 22 m> s in the direction of the car
6. Since the two velocities are at right angles we
can use the Pythagorean theorem to find the
magnitude, m, of the resultant velocity and we use
the definition of sine to find the angle, u, made.
m 2 5 122 1 52
5 144 1 25
5 169
m 5 "169
5 13 m>s
5
sin u 5
13
5
u 5 sin21
13
8 22.6° from the direction of the boat toward
the direction of the current. This results in a net of
22.6° 1 15° 5 37.6°, or N 37.6° W.
7. a. First we find the components of the resultant
directed north and directed west. The component
directed north is the velocity of the airplane, 800,
minus 100 sin 45°, since the wind forms a 45°
angle south of west. The western component of
the resultant is simply 100 cos 45°. So we use the
Pythagorean theorem to find the magnitude, m, of
the resultant and the definition of sine to find the
angle, u, of the resultant.
m 2 5 (800 2 100 sin 45°)2 1 (100 cos 45°)2
8 (729.29)2 1 (71.71)2
8 536 863.8082
m 8 732.71 km>h
Use the sine law to determine the direction.
sin u
sin 45°
5
100
732.71
u 8 5.5°
The direction is N 5.5° W.
Calculus and Vectors Solutions Manual
b. The airplane is travelling at approximately
732.71 km> h, so in 1 hour the airplane will travel
about 732.71 km.
8. a. First we find the velocity of the airplane. We
use the Pythagorean theorem to find the magnitude,
m, of the resultant.
m 2 5 4502 1 1002
5 202 500 1 10 000
5 212 500
m 5 "212 500
8 461 km>h
So in 3 hours, the airplane will travel about
(461 km>h)(3 h) 5 1383 km.
b. To find the angle, u, the airplane travels, we use
the definition of sine.
100
sin u 5
461
100
u 5 sin21
461
8 12.5° east of north.
9. a. To find the angle, u, at which to fly is the
equivalent of the angle of a right triangle with 44 as
the opposite leg and 244 as the hypotenuse. So we
use the definition of sine to find this angle.
44
sin u 5
244
44
u 5 sin21
244
8 10.4° south of west.
b. By the Pythagorean Theorem, the resultant ground
speed of the airplane is "(2442 2 442 ) 5 240 km>h.
Since time 5 distance>rate, the duration of the
flight is simply (480 km)> (240 km>h) 5 2 h.
10. a. Since Judy is swimming perpendicular to
the flow of the river, her resultant velocity is simply
the hypotenuse of a right triangle with 3 and 4 as
bases, which is a 3-4-5 right triangle. Thus, Judy’s
resultant velocity is 5 km> h. The direction is
determined by tan u 5 43. u 8 53.1° downstream
b. Judy’s distance traveled down the river would be the
“4” leg of the 3-4-5 triangle formed by the vectors, but
scaled down so that 1m (the width of the river) is
equivalent to the “3” leg. So her distance traveled is
4
3 8 1.33 km. This makes her about 0.67 km from
Helen’s cottage.
c. While in the river, Judy is swimming at
5 km> h for a distance of 53 km. Since
time 5 distance>rate, her time taken is
5
3 km
5 1 hours 5 20 minutes.
5 km>h 3
7-7
11.
h
m/
v
5k
20 30°
u
212 km/h
a. and b. Here, 205 km> h directed 30° north of east
is the resultant of 212 km> h directed east, and the
wind speed, v, directed at some angle. This problem
is more easily approached finding the wind speed,
v, first. So we will do that using the cosine law.
v 2 5 2052 1 2122 2 2(205)(212) cos 30°
5 42 025 1 44 944 2 86 920 cos 30°
5 86 969 2 75 275
5 11 694
v 5 "11 694
8 108 km>h
Now to find the wind’s direction, we simply find the
angle supplementary to the lesser angle, u, formed
by the parallelogram of these three velocities. We
can use the sine law for this.
sin u
sin 30°
5
205
108
sin 30°
sin u 5 205a
b
108
sin 30°
u 5 sin21 a205a
bb
108
8 71.6°
Thus, the direction of v is the angle supplementary
to u in the parallelogram:
180° 2 71.6° 5 108.4° 5 18.4° west of north.
12.
4
5
Since her swimming speed is a maximum of 4 km> h,
this is her maximum resultant magnitude, which is
also the hypotenuse of the triangle formed by her and
the river’s velocity vector. Since one of these legs is
5 km> h, we have a triangle with a leg larger than its
hypotenuse, which is impossible.
13. a. First we need to find Mary’s resultant
velocity, v. Since this resultant is the diagonal of the
parallelogram formed by hers and the river’s
velocity, we can use the cosine law with the angle, u,
of the parallelogram adjacent 30°.
7-8
v 2 5 32 1 42 2 2(3)(4) cos 150°
5 9 1 16 2 24 cos 150°
5 25 1 20.8
5 45.8
v 5 "45.8
8 6.8 m>s
So in 10 seconds, Mary travels about
(6.8 m>s)(10 s) 5 68 m.
b. Since Mary is travelling at 3 m>s at an angle of 30°,
to find the component of her velocity, v, perpendicular
to the current, we use the definition of sine.
v 5 3 sin 30
1
5 3a b
2
5 1.5 m>s perpendicular to the current.
So since time 5 distance>rate, the time taken is
(150 m)> (1.5 m>s) 5 100 s.
14. a. So we have a 5.5 m> s vector and a 4 m> s
vector with a resultant vector that is directed 45°
south of west. Letting u be the angle between the
4 km> h vector and the resultant, we can construct
a parallelogram using these three vectors and a
subsequent triangle with u opposite the 5.5 m> s
vector and 45° opposite the 4 m> s vector. We now
use the sine law to find u.
sin u
sin 45°
5
5.5
4
sin 45°
sin u 5 5.5a
b
4
sin 45°
u 5 sin21 a5.5a
bb
4
8 76.5° from the resultant.
Since the resultant is 45° west of south, Dave’s
direction is 76.5° 1 45° 5 121.5° west of south,
which is equivalent to about 180° 2 121.5° 5 58.5°
upstream.
b. First, we find the magnitude, m, of Dave’s 4 m> s
velocity in the direction perpendicular to the river.
This is done using the definition of sine.
m 5 4 sin 58.5°
8 3.41 m>s perpendicular to the river.
Since time is distance>rate, we have
(200 m)> (3.41 m>s) 8 58.6 s.
15. Let b represent the speed of the steamboat and c
represent the speed of the current. On the way
downstream, the effective speed is b 1 c, and
upstream is b 2 c. The distance upstream and
downstream is the same, so 5(b 1 c) 5 7(b 2 c).
So, b 5 6c. This means that the speed of the boat
is 6 times the speed of the current. So, (6c 1 c) ? 5
Chapter 7: Applications of Vectors
or 35c is the distance. This means that it would take
a raft 35 hours moving with the speed of the current
to get from A to B.
7.3 The Dot Product of Two Geometric
Vectors, pp. 377–378
> >
> >
>
1. a ? b 5 0 a 0 @ b @ cos u 5 0. This means 0 a 0 5 0,
>
or @ b @ 5 0, or cos u 5 0. To be guaranteed that the
two vectors are perpendicular, the vectors must be
nonzero.
> >
2. a ? b is a scalar, and
a dot product is only defined
> >
>
for vectors, so (a ? b ) ? c is meaningless.
> >
> >
>
>
3. Answers
may vary. Let a 5 i, b 5 j, c 5 2i.
>
>
>
>
>
>
a ? b 5 0, b ? c 5 0, but a 5 2c .
>
>
>
>
>
>
>
>
4. a ? b 5 b ? a 5 b ? c because c 5 a
>
>
>
>
5. Since a and b are unit vectors, 0 a 0 5 @ b @ 5 1 and
since they are pointing in opposite directions
then
> >
u 5 180° so cos u 5 21. Therefore a ? b 5 21.
> >
> >
6. a. p ? q 5 0 p 0 0 q 0cos u
5 (4)(8) cos (60°)
5 (32)(.5)
5 16
> >
> >
b. x ? y 5 0 x 0 0 y 0cos u
5 (2)(4) cos (150°)
"3
5 (8)a2
b
2
8 26.93
> >
> >
c. a ? b 5 0 a 0 @ b @ cos u
5 (0)(8) cos (100°)
50
> >
> >
d. p ? q 5 0 p 0 0 q 0cos u
5 (1)(1) cos (180°)
5 (1)(21)
5 21
> >
> >
e. m ? n 5 0 m 0 0 n 0cos u
5 (2)(5) cos (90°)
5 (10)(0)
50
> >
> >
f. u ? v 5 0 u 0 0 v 0cos u
5 (4)(8) cos 145°
8 226.2
> >
> >
7. a. x ? y 5 0 x 0 0 y 0cos u
12"3 5 (8)(3) cos u
"3
5 cos u
2
u 5 30°
Calculus and Vectors Solutions Manual
> >
> >
b. m ? n 5 0 m 0 0 n 0cos u
(6) 5 (6)(6) cos u
1
5 cos u
6
u 8 80°
> >
> >
c. p ? q 5 0 p 0 0 q 0cos u
3 5 (5)(1) cos u
3
5 cos u
5
u 8 53°
> >
> >
d. p ? q 5 0 p 0 0 q 0cos u
23 5 (5)(1) cos u
3
2 5 cos u
5
u 8 127°>
> >
>
e. a ? b 5 0 a 0 @ b @ cos u
10.5 5 (7)(3) cos u
1
5 cos u
2
u 5 60°
> >
> >
f. u ? v 5 0 u 0 0 v 0cos u
250 5 (10)(10) cos u
1
2 5 cos u
2
u 5 120°>
> >
>
8. a ? b 5 0 a 0 @ b @ cos u
5 (7.5)(6) cos (180° 2 120°)
1
5 (45)a b
2
5 22.5
Note: u is the angle between the two vectors when
they are tail to> tail, so u 2> 120°.
>
>
>
>
>
>
9. a. (a 1 5b ) ? (2a 2 3b ) 5 a ? 2a 2 a ? 3b
>
>
>
>
1 5b ? 2a 2> 5b ? 3b
>2
5 2 0 a 0 2 15 @ b @ 2
> >
> >
2 3a ? b 1 10a
?b
>
>
2
@
@
5 2 0 a 0 2 2 15
b
> >
1 7a ? b
>
>
>
>
>
>
>
b. 3x ? (x 2 3y ) 2 (x 2 3y ) ? (23x 1 y )
>
>
>
>
> >
>
>
5 3 0 x 0 2 2 3x ? 3y 1 3 0 x 0 2 2 x ? y 2 (23y ? 23x )
>2
1 30 y 0
>
> >
> >
> >
>
5 6 0 x 0 2 2 9x ? y 2 x ? y 2 9x ? y 1 3 0 y 0 2
>2
> >
>2
5 6 0 x 0 2 19x ? y 1 30 y 0
>
10. @ 0 @ 5 0 so the dot product of any vector with
>
0 is 0.
7-9
>
>
> >
>
>
>
>
11. (a 2 5b ) ? (a 2 b ) 5 @ a 2 5b @ @ a 2 b @ cos (90°)
>
>
>
>
>
>
0 a 0 2 2 a ? b 2 5b ? a 1 5 @ b @ 2 5 0
>
>
> >
0 a 0 2 1 5 @ b @ 2 5 6a ? b
>
> > 1 >
a ? b 5 ( 0 a 0 2 1 5 @ b @ 2)
6
5
1
>
>
>
>
> >
> >
12. a. (a 1 b ) ? (a 1 b ) 5 a ? >a 1 a ? b
>
>
>
1 b ? a 1 b> ? b
>
>
> >
5 0 a 02 1 a ? b 1 a ? b
>
1 @b@ 2
>
>2
> >
@
@2
1
2a
1
b
5
0
a
0
?
b
>
>
>
>
>
> >
> >
>
b. (a 1 b ) ? (a 2 b ) 5 a ? a 2 a ? b 1 b ? a
>
>
2b?b
>
>
> >
> >
5 0 a 02 2 a ? b 1 a ? b 2 @b@ 2
>
>
5 0 a 02 2 @b@ 2
>
> >
13. a. 0 a 0 2 5 a ?> a
>
>
>
5 (b 1 c ) ? (b 1 c )
>
>
>
>
5 @ b @ 2 1 2b ? c 1 0 c 0 2
>
>
>
>
b. b ? c 5 @ b @ 0 c 0cos (90°) 5 0
>
>
>
Therefore 0 a 0 2 5 @ b @ 2 1 0 c 0 2.
This is just what the Pythagorean theorem says,
>
>
where b and c are the legs of the right triangle.
>
>
>
>
>
>
14. (u 1 v 1 w ) ? (u 1 v 1 w )
> >
> >
>
>
> >
> >
5u?u1u?v1u?w1v?u1v?v
>
>
> >
> >
>
>
1v?w1w?u1w?v1w?w
>2
>2
>2
> >
5 0 u 0 1 0 v 0 1 0 w 0 1 2 0 u 0 0 v 0cos (90°)
> >
> >
1 2 0 u 0 0 w 0cos (90°) 1 2 0 v 0 0 w 0cos (90°)
5 (1)2 1 (2)2 1 (3)2
5 14
>
>
>
>
15. 0 u 1 v 0 2 1 0 u 2 v 0 2
>
>
>
>
>
>
>
>
5 (u 1 v ) ? (u 1 v ) 1 (u 2 v ) ? (u 2 v )
>
> >
>
>
> >
>
5 0 u 0 2 1 2u ? v 1 0 v 0 2 1 0 u 0 2 2 2u ? v 1 0 v 0 2
>
>
5 20 u 0 2 1 2 0 v 0 2
>
>
>
>
>
16. (a 1 b ) ? (a 1 b 1 c )
>
>
>
>
> >
> >
>
>
5 0 a 02 1 a ? b 1 a ? c 1 b ? a 1 @b@ 2 1 b ? c
> >
> >
5 1 1 20 a 0 @ b @ cos (60°) 1 0 a 0 0 c 0cos (60°) 1 1
> >
1 @ b @ 0 c 0cos (120°)
1
1
1
5 2 1 2a b 1 2
2
2
2
53
>
>
>
>
17. a 1 b >1 c 5 0 >
>
>
>
>
>
>
a ? (a 1 b 1 c ) 1 b ? (a 1 b 1 c )
>
>
>
>
1 c ? (a 1 b 1 c ) 5 0
>
>
>
>
>
>
> >
>
>
0 a 02 1 a ? b 1 a ? c 1 b ? a 1 @b@ 2 1 b ? c
>
> >
>
>
1 c ? a 1 c ? b 1 0 c 02 5 0
7-10
>
> >
> >
>
1 1 4 1 9 1 2(a ? b 1 a ? c 1 b ? c ) 5 0
>
>
>
> >
>
2(a ? b 1 a ? c 1 b ? c ) 5 214
>
>
>
> >
>
a ? b 1 a ? c 1 b ? c 5 27
>
>
>
18. d> 5 b> 2 c
>
b 5 d 1> c
> >
> >
>
c ? a 5 ((b ? a ) a ) ? a
>
>
> >
> > >
>
c ? a 5 (b ? a )(a ? a ) because b ? a is a scalar
>
> >
> >
c ? a 5 (b ? a ) 0 a 0 2
>
> >
>
>
>
c ? a 5 (d 1 c ) ? a because 0 a 0 5 1
>
> >
>
> >
c> ? a 5 d ? a 1 c ? a
>
d?a50
7.4 The Dot Product for Algebraic
Vectors, pp. 385–387
> >
a?b50
1.
(21)b1 1 b2 5 0
b2 5 b1
Any vector of the form (c, c) is perpendicular
>
to a . Therefore there are infinitely many vectors
>
perpendicular to a . Answers may vary. For example:
(1, 1), (2, > 2), (3, 3).
>
2. a. a ? b 5 (22)(1) 1 (1)(2)
50
u
5 90°
> >
b. a ? b 5 (2)(4) 1 (3)(3) 1 (21)(217)
5 8 1 9 1 17
5 34 . 0
cos u . 0
u is acute
> >
c. a ? b 5 (1)(3) 1 (22)(22) 1 (5)(22)
5 3 1 4 2 10
5 23 , 0
cos u , 0
u is obtuse
3. Any vector in the xy-plane
is of the form
>
>
a 5 (a1, a2, 0). Let b 5 (0, 0, 1).
> >
a ? b 5 (0)(a1 ) 1 (0)(a2 ) 1 (0)(1)
50
Therefore (0, 0, 1) is perpendicular to every vector
in the xy-plane.
Any vector in the xz-plane
is of the form
>
>
c 5 (c1, 0, c3 ). Let d 5 (0, 1, 0).
> >
c ? d 5 (0)(c1 ) 1 (0)(1) 1 (0)(c3 )
50
Therefore (0, 1, 0) is perpendicular to every vector
in the xz-plane.
Chapter 7: Applications of Vectors
Any vector in the yz-plane
is of the form
>
>
e 5 (0, e2, e3 ). Let f 5 (1, 0, 0).
> >
e ? f 5 (1)(0) 1 (0)(e2 ) 1 (0)(e3 )
50
Therefore (1, 0, 0) is perpendicular to every vector
in the yz-plane.
4. a.
(1, 2, 21) ? (4, 3, 10) 5 4 1 6 2 10
50
5
(24, 25, 26) ? a5, 23, 2 b 5 220 1 15 1 5
6
50
b. If any of the vectors were collinear then one
would be a scalar multiple of the other. Comparing
the signs of the individual components of each
vector eliminates (1, 2, 21) and ( 5, 23, 2 56) . All of
the components of (24, 25, 26) have the same
sign and the same is true for (4, 3, 10), but (4, 3, 10)
is not a scalar multiple of (24, 25, 26). Therefore
none of the vectors are collinear.
5. a. Using the strategy of Example 5 yields
(x, y) ? (1, 22) 5 0 and (x, y) ? (1, 1) 5 0
x 2 2y 5 0 and x 1 y 5 0
3y 5 0
Therefore the only result is x 5 y 5 0, or (0, 0).
This is because (1, 22) and (1, 1) both lie on the
xy-plane and are not collinear, so any vector that is
perpendicular to both vectors must be in R 3 which
does not exist in R 2.
b. If we select any two vectors that are not collinear
in R 2, then any vector that is perpendicular to both
cannot be in R 2 and must be in R 3. This is not
possible since R 3 does
not exist in R 2.
> >
a?b
6. a. cos u 5 > >
0 a 0 @b@
(5)(21) 1 (3)(22)
5
"25 1 9"1 1 4
211
5
"(34)(5)
211
5
"170
u 8 148°
> >
a?b
b. cos u 5 > >
0 a 0 @b@
(21)(6) 1 (4)(22)
5
"1 1 16"36 1 4
214
5
"680
u 8 123°
Calculus and Vectors Solutions Manual
> >
a?b
c. cos u 5 > >
0 a 0 @b@
(2)(2) 1 (2)(1) 1 (1)(22)
5
"4 1 4 1 1"4 1 1 1 4
4
5
(3)(3)
4
5
9
u 8 64° >
>
a?b
d. cos u 5 > >
0 a 0 @b@
(2)(25) 1 (3)(0) 1 (26)(12)
5
"4 1 9 1 36"25 1 144
282
5
(7)(13)
282
5
91
u 8 154°
> >
> >
7. a.
a ? b 5 0 a 0 @ b @ cos u
> >
(21)(26k) 1 (2)(21) 1 (23)(k) 5 0 a 0 @ b @
cos (90°)
6k 2 2 2 3k 5 0
3k 5 2
2
k5
3
> >
> >
a ? b 5 0 a 0 @ b @ cos u
b.
(1)(0) 1 (1)(k) 5 "1 1 1"k 2 cos (45°)
1
k 5 "2 0 k 0
"2
k 5 0k0
k$0
8. a.
y
2
(0, 1)
1
(1, 0) x
–2 –1 0 1 2
–1
–2
b.
y
2
(0, 1)
1
(1, 0) x
0
–2 –1
1 2
–1
–2
7-11
The diagonals are (1, 0) 1 (0, 1) 5 (1, 1) and
(1, 0) 2 (0, 1) 5 (1, 21) or
(1, 0) 1 (0, 1) 5 (1, 1) and
(0, 1) 2 (1, 0) 5 (21, 0).
c. (1, 1) ? (1, 21)
5121
50
or (1, 1) ? (21, 1)
5 21 1 1
50
> >
a?b
9. a. cos u 5 > >
0 a 0 @b@
(1 2 "2)(1) 1 ("2 2 1)(1)
5
> >
0 a 0 @b@
50
u 5> 90°>
a?b
b. cos u 5 > >
0 a 0 @b@
5
5
"2 2 1 1 "2 1 1 1 "2
" (2 2 2"2 1 1) 1 (2 1 2"2 1 1) 1 2 "1 1 1 1 1
3"2
"8"3
"3
5
2
u 5 30°
>
>
10. a. i. a 5 kb
8 5 12k
2
k5
3
2
p 5 4a b
3
8
p5
3
2
25 q
3
q53
ii. Answers may
vary. For example:
> >
a?b50
2q 1 4p 1 96 5 0
q 5 22p 2 48
Let p 5 1
q 5 250
b. In part a., the values are unique because both
vectors have their third component specified, and >
the ratios must be the same for each component b.
In part b. the values are not unique; any value of
p could have been chosen, each resulting in a
different value of q.
7-12
>
>
>
11. AB 5 (2, 6), BC 5 (25, 25), CA 5 (3, 21)
>
>
AB ? CA
cos (180° 2 uA ) 5
>
>
@ AB @ @ CA @
626
5
>
>
@ AB @ @ CA @
50
180° 2 uA 5 90°
uA 5 90° >
>
AB ? BC
cos (180° 2 uB ) 5
>
>
@ AB @ @ BC @
210 2 30
5
"4 1 36"25 1 25
240
5
"(40)(50)
4
52
Å5
180° 2 uB 8 153.4°
uB 8 26.6°
uC 5 180° 2 uA 2 uB
uC 8 63.4°
12. a. O 5 (0, 0, 0), A 5 (7, 0, 0), B 5 (7, 4, 0),
C 5 (0, 4, 0), D 5 (7, 0, 5), E 5 (0, 4, 5),
F 5 (0, 0, 5)
>
>
>
>
b.
AE ? BF 5 @ AE @ @ BF @ cos u
(27, 4, 5) ? (27, 24, 5) 5 "49 1 16 1 25
3 "49 1 16 1 25 cos u
49 2 16 1 25 5 90 cos u
58
5 cos u
90
u 8 50°
13. a. Answers may vary. For example:
(x, y, z) ? (21, 3, 0) 5 0
2x 1 3y 5 0
x 5 3y
(x, y, z) ? (1, 25, 2) 5 0
x 2 5y 1 2z 5 0
22y 1 2z 5 0
y5z
Let y 5 1.
(3, 1, 1) is perpendicular to (21, 3, 0) and
(1, 25, 2).
b. Answers may vary. For example:
(x, y, z) ? (1, 3, 24) 5 0
x 1 3y 2 4z 5 0
x 5 4z 2 3y
(x, y, z) ? (21, 22, 3) 5 0
Chapter 7: Applications of Vectors
2x 2 2y 1 3z 5 0
3y 2 4z 2 2y 1 3z 5 0
y5z
Let y 5 1.
(1, 1, 1) is perpendicular to (1, 3, 24) and
(21, 22, 3).
14. (p, p, 1) ? (p, 22, 23) 5 0
p 2 2 2p 2 3 5 0
2 6 "22 2 4(23)
2
p5162
p 5 3 or 21
15. a. (23, p, 21) ? (1, 24, q) 5 0
23 2 4p 2 q 5 0
3 1 4p 1 q 5 0
b. 3 1 4p 2 3 5 0
p50
16. Answers may vary. For example: Note that
>
>
s 5 22r , so they are collinear. Therefore any
>
vector that is perpendicular to s is also
>
perpendicular to r .
(x, y, z) ? (1, 2, 21) 5 0
x 1 2y 2 z 5 0
Let x 5 z 5 1.
(1, 0, 1) is perpendicular to (1, 2, 21) and
(22, 24, 2).
Let x 5 y 5 1.
(1, 1, 3) is perpendicular to (1, 2 2 1) and
(22, 24, 2).
> >
> >
17. x ? y 5 0 x 0 0 y 0cos u
(24, p, 22) ? (22, 3, 6)
p5
5 "16 1 p 2 1 4"4 1 9 1 36 cos u
8 1 3p 2 12 5 "20 1 p 2 (7) cos u
(3p 2 4)2 5 a7"20 1 p 2 cos ub
9p 2 2 24p 1 16 5 49(20 1 p 2 )a
2
4 2
b
21
320
16
1 p2
9
9
65p 2 2 216p 2 176 5 0
9p 2 2 24p 1 16 5
p5
216 6 "(2216)2 2 4(65)(2176)
2(65)
p 5 4 or 2
44
65
> >
18. a. a ? b 5 23 1 3
50
Therefore, since the two diagonals are perpendicular,
all the sides must be the same length.
Calculus and Vectors Solutions Manual
>
>
1 >
b. AB 5 (a 1 b )
2
5 (1, 2, 21)
>
>
1 >
BC 5 (a 2 b )
2
5 (2, 1, 1)
>
>
@ AB @ 5 @ BC @ 5 "6
>
>
>
>
c. AB ? BC 5 @ AB @ @ BC @ cos u1
2 1 2 2 1 5 6 cos u1
1
5 cos u1
2
u1 5 60°
2u1 1 2u2 5 360°
u2 5 120°
>
>
19. a. AB 5 (3, 4, 212), DA 5 (24, 2 2 q, 25)
>
>
AB ? DA 5 0
212 1 8 2 4q 1 60 5 0
21 2 q 1 15 5 0
q 5 14
>
>
DA 5 CB
(24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)
x 5 6, y 5 18, z 5 24
The coordinates of vertex C are (6, 18, 24).
>
>
>
>
b.
AC ? BD 5 @ AC @ @ BD @ cos u
(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 49
3 "1 1 64 1 289 cos u
7 1 128 2 119 5 354 cos u
16
5 cos u
354
u 8 87.4°
20. The two vectors representing the body diagonals
are (0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1) and
(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)
(21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u
1 2 1 1 1 5 3 cos u
1
5 cos u
3
u 8 70.5°
a 5 180° 2 u
a 8 109.5°
Mid-Chapter Review, pp. 388–389
> >
1. a. a ? b 5 (3)(2) cos (60°)
1
5 (6)
2
53
7-13
>
>
>
>
>
> >
b. (3a 1 2b ) ? (4a 2 3b ) 5 12 0 a 0 2> 2 9a ? b >
>
1 8b ? a 2 6 @ b @ 2
5 12(3) 2 2 3 2 6(2) 2
5 81
2.
20 cm
u2
15 cm
25 cm
u1
Let T1 be the tension in the 15 cm cord and T2 be
the tension in the 20 cm cord. Let u1 be the angle
the 15 cm cord makes with the ceiling and u2 be the
angle the 20 cm cord makes with the ceiling. By the
cosine law:
(15)2 5 (20)2 1 (25)2 2 2(20)(25) cos (u2 )
cos (u2 ) 5 0.8
sin (u2 ) 5 "1 2 cos2 (u2 )
sin (u2 ) 5 0.6
(20)2 5 (15)2 1 (25)2 2 (2)(15)(25) cos (u1 )
cos (u1 ) 5 0.6
sin (u1 ) 5 0.8
Horizontal Components:
2T1 cos (u1 ) 1 T2 cos (u2 ) 5 0
(0.8)T2 5 (0.6)T1
T2 5 (0.75)T1
Vertical Components:
T1 sin (u1 ) 1 T2 sin (u2 ) 2 (15)(9.8) 5 0
(0.8)T1 1 (0.6)(0.75)T1 5 147
(1.25)T1 5 147
T1 5 117.6 N
T2 5 (0.75)T1
T2 5 88.2 N
Therefore the tension in the 15 cm cord is 117.60 N
and the tension in the 20 cm cord is 88.20 N.
3. The diagonals of a square are perpendicular, so
the dot product is 0.
4. a.
a
v
v +w
135°
w>
>
0 v 0 5 500, 0 w 0 5 100
By the cosine law:
>
>
0 v 1 w 0 2 5 (500)2 1 (100)2
2 2(500)(100) cos (135°)
>
>
0 v 1 w 0 8 575.1
7-14
By the cosine law:
sin (a)
sin (135°)
5
100
575.1
sin (a) 8 0.123
a 8 7.06°
The resultant velocity of the airplane is 575.1 km>h
at S7.06°E
b. (distance) 5 (rate)(time)
1000
km
t8
?
575.1 (km/h)
t 8 1.74 hours
5. a.
E
F
@ E ' @ 5 @ E @ cos (40°)
>
@ E ' @ 5 (9.8)(15)cos (40°)
>
@ E ' @ 8 112.61 N
>
>
b. @ F @ 5 @ E @ sin (40°)
>
@ F @ 8 94.49 N
6. 6u 5 360°
>
>
u 5 60° >
> >
>
a ? b 5 0 a 0 @ b @ cos (60°)
5 (3)(3)(0.5)
5 4.5
> >
7. a. a ? b 5 (4)(1) 1 (25)(2) 1 (20)(2)
5 34>
> >
>
b. a ? b 5 0 a 0 @ b @ cos (u)
34 5 "16 1 25 1 400 "1 1 4 1 4 cos (u)
34
cos (u) 5
63 >
>
>
>
>
>
> >
8. a. a ? b 5 (i 1 2j 1 k ) ? (2i 2 3j 1 4k )
522614
5 0>
>
>
>
>
>
>
>
b. b ? c 5 (2i 2 3j 1 4k ) ? (3i 2 j 2 k )
561324
55 >
>
>
>
>
>
>
>
c. b 1 c 5 (2i 2 3j 1 4k ) 1 (3i 2 j 2 k )
>
>
>
5 5i 2 4j >1 3k>
>
>
>
>
>
>
>
d. a ? (b 1 c ) 5 (i 1 2j 1 k ) ? (5i 2 4j 1 3k )
552813
5
>
> 0>
>
>
>
>
e. (a 1 b ) ? (b 1 c ) 5 (3i 2 j 1 5k )
>
>
>
? (5j 2 4j 1 3k )
5 15 1 4 1 15
5 34
Chapter 7: Applications of Vectors
>
>
>
>
>
>
>
f. (2a 2 3b ) ? (2a 1 c ) 5 ((2i 1 4j 1 2k )
>
>
>
2 (6i 2 9j 1 12k ))
>
>
>
? ((2i 1 4j 1 2k )
>
>
>
1 (3i 2 j 1 k ))
>
>
>
5 (24i 1 13j 2 10k )
>
>
>
? (5i 1 3j 1 k )
5 220 1 39 2 10
59
> >
9. a.
?
q
5
0
p
>
>
>
>
>
>
(xi 1 j 1 3k ) ? (3xi 1 10xj 1 k ) 5 0
3x 2 1 10x 1 3 5 0
210 6 "(10)2 2 4(3)(3)
x5
2(3)
210 6 8
x5
6
1
x 5 23 or x 5 2
3
>
>
b. If p and q are parallel then one is a scalar
multiple of the other.
>
>
p 5 nq where n is a constant
>
>
>
>
>
>
xi 1 j 1 3k 5 n(3xi 1 10xj 1 k )
>
n 5 3 by the k > component
x 5 9x by the i component
x50
>
1 5 30(0) by the j component
120
Therefore there is no value of x that will make these
two vectors parallel.
>
>
>
>
>
>
>
>
10. a. 3x 2 2y 5 (3i 2 6j 2 3k ) 2 (2i 2 2j 2 2k )
>
>
>
5 i 2 4j 2 k
>
>
>
>
>
>
>
>
b. 3x ? 2y 5 (3i 2 6j 2 3k ) ? (2i 2 2j 2 2k )
5 6 1 12 1 6
5 24 >
>
>
>
>
>
>
>
c. 0 x 2 2y 0 5 @ (i 2 2j 2 k ) 2 (2i 2 2j 2 2k ) @
>
>
5 @ 2i 1 k @
5 " (2i 1 k ) ? (2i 1 k )
>
>
>
>
5 "2 or 1.41
>
>
>
>
>
>
>
d. (2x 2 3y ) ? (x 1 4y ) 5 ((2i 2 4j 2 2k )
>
>
>
2 (3i 2 3j 2 3k )) ?
>
>
>
1 (( i 2 2j 2 k )
>
>
>
1 (4i 2 4j 2 4k )
>
>
>
5 (2i 2 j 1 k )
>
>
>
? (5i 2 6j 2 5k )
5 25 1 6 2 5
5 24
Calculus and Vectors Solutions Manual
> >
> >
> >
> >
e. 2x ? y 2 5y ? x 5 2x ? y 2 5x ? y
> >
5 23x ? y
>
>
>
>
>
5 23(i 2 2j 2 k) ? (i 2 j 2 k )
5 23(1 1 2 1 1)
5 212
11.
5N
4N
180° - u
3N
2
2
(4) 5 (5) 1 (3)2 2 2(3)(5) cos (180° 2 u)
0.6 5 cos (180° 2 u)
180° 2 u 8 53.1
u 8 126.9°
12. (F)2 5 (3)2 1 (4)2 2 2(3)(4) cos (180° 2 60°)
(F)2 5 25 2 24 cos (120°)
(F)2 5 37
F 8 6.08 N
(3)2 5 (4)2 1 "37
(
cos u 5
44
)2 2 2(4)("37) cos u
8"37
u 8 25.3°
>
F 8 6.08 N, 25.3° from the 4 N force towards the
3 >N force.
E 8 6.08 N, 180° 2 25.3° 5 154.7° from the
4 N force away from the 3 N force.
>
>
>
>
13. a. The diagonals are m 1 n and m 2 n
>
>
m 1 n 5 (1, 4, 10)
>
>
m 2 n 5 (3, 210, 0)
>
>
>
>
>
> >
>
(m 1 n ) ? (m 2 n ) 5 0 m 1 n 0 0 m 2 n 0cos u
3 2 40 5 "1 1 16 1 100 "9 1 100 cos u
cos u 8 20.3276
u 8 109.1°
>
>
>
>
> >
b. 0 m 2 n 0 2 5 0 m 0 2 1 0 n 0 2 2 2 0 m 0 0 n 0cos u
(9 1 100) 5 (4 1 9 1 25) 1 (1 1 49 1 25)
2 2"38 "75 cos u
cos u 8 0.0374
u 8 87.9°
7-15
14. a. 45 sin (150°) 5 500 sin u
u 8 N 2.6° E
b. v 5 500 cos (2.6°) 2 45 cos (30°)
8 460.5 km> h
1000
t8
460.5
t 8 2.17 hours
> >
a?x50
15.
2x1 1 2x2 1 5x3 5 0
x 5 2x2 1 5x3
> 1>
b?x50
x1 1 3x2 1 5x3 5 0
2x2 1 5x3 1 3x2 1 5x3 5 0
x2 1 2x3 5 0
choose x3 5 1
x2 5 22
x1 5 1
1
>
x5
(1, 22, 1)
!6
1 2
1
2 1
1
>
x5a
,2
,
b or a2
,
,2
b
!6 !6 !6
!6 !6
!6
16. a. v 5 4 1 3 cos (45°)
8 6.12 m> s
d 8 (6.12)(10)
8 61.2 m
b. w 5 3 sin (45°)
8 2.12 m> s
180
t8
2.12
t 8 84.9 seconds
>
>
>
>
17. a. (x 1 y ) ? (x 2 y ) 5 0
>2
> >
> >
>2
0x0 2 x ? y 1 y ? x 2 0y0 5 0
>
>
0 x 02 5 0 y 02
>
>
>
>
>
>
(x 1 y ) ? (x 2 y ) 5 0 when x and y have
the same length. >
>
b. Vectors >a and b determine a parallelogram. Their
>
sum a 1 b is one diagonal of the parallelogram
formed, with
its tail in the same location
as the tails
>
>
>
>
of a and b. Their difference a 2 b is the other
diagonal
of the parallelogram.
>
18. @ F @ 5 350 cos (40°)
8 268.12 N
7-16
7.5 Scalar and Vector Projections,
pp. 398–400
>
>
>
>
a ?b
1. a. Scalar projection of a on b is > where
@b@
>
>
a 5 (2, 3) and b is the positive x-axis (X, 0).
> >
a ? b 5 (2X) 1 (3 3 0)
5 2X 1 0
5 2X
@ b @ 5 "X 2 1 02
>
5X
2X
> 5
X
@b@
5 2;
The vector projection
is the
scalar projection
>
>
> >
a?b
multiplied by
b
@ b@
>
where
b
@ b@
>
is the x-axis divided by
>
the magnitude of the x-axis which is equal
> to i.
>
The scalar projection of 2 multiplied> by i equals 2i.
>
>
>
a ?b
b. Scalar projection of a on b is > where
@b@
>
>
a 5 (2, 3) and b is now the positive y-axis (0, Y).
> >
a ? b 5 (2 3 0) 1 (3Y)
5 0 1 3Y
> 5 3Y
@ b @ 5 "02 1 Y 2
5Y
> >
a @b@
3Y
> 5
Y
@b@
5 3;
The vector projection is the scalar projection
multiplied by
>
>
b
b
@ b@
>
where
@ b@
>
is the y-axis divided
>
by the magnitude of the y-axis which is equal
to j. >
>
The scalar projection of 3 multiplied by j equals 3j.
2. Using the formula
would cause a division by 0.
>
Generally the 0 has any direction and 0 magnitude.
You can not project onto nothing.
>
>
3. You are projecting a onto the tail of b which
>
is a point with magnitude
0. Therefore it is 0; the
>
>
projections
of b onto the tail of a are also 0
>
and 0.
Chapter 7: Applications of Vectors
>
>
4. Answers
may
vary.
For
example:
p
,
5
AE
>
>
q 5 AB
D
p
A
E
C
q
B
>
>
>
Scalar projection p on q 5 @ AC @ ;
>
>
>
Vector projection p on q 5 AC;
>
>
>
Scalar projection q on p 5 @ AD @ ;
>
>
>
Vector projection q on p 5 AD
>
>
5. When a 5 (21, 2, 5) and b 5 (1, 0, 0) then
> >
a ? b 5 (21 3 1 1 2 3 0 1 5 3 0)
5 21
>
@ b @ 5 "12 1 02 1 02
51
> >
a ?b
21
Therefore the scalar projection is > 5
1
@b@
5 21;
>
b
(1, 0, 0)
The vector equation is 21 3 > 5 21 3
1
@b@
5 21;
>
Under> the same approach, when a 5 (21, 2, 5)
and b 5 (0, 1, 0), then
> >
a ? b 5 (21 3 0 1 2 3 1 1 5 3 0)
52
>
@ b @ 5 "02 1 1 1 02
51
> >
2
a ?b
Therefore the scalar projection is > 5
1
@b@
5 2,
>
b
(0, 1, 0)
The vector equation is 2 3 > 5 2 3
1
@b@
5 2;
>
The
same is also true when a 5 (21, 2, 5) and
>
b 5 (0, 0, 1) then
> >
a ? b 5 (21 3 0 1 2 3 0 1 5 3 1)
55
>
@ b @ 5 "02 1 02 1 12
51
> >
a ?b
5
Therefore the scalar projection is > 5
1
@b@
5 5,
>
b
(0, 0, 1)
The vector equation is 5 3 > 5 5 3
1
@b@
5 5;
Calculus and Vectors Solutions Manual
Without having to use formulae, a projection of
> >
>
(21, 2, 5) on i, j, or k is the same as a projection
>
>
>
of (21, 0, 0) on i, (0, 2, 0) on j, and (0, 0, 5) on k
which intuitively yields the same result.
> >
6. a. p ? q 5 (3 3 24) 1 (6 3 5)
1 (222 3 220)
5 212 1 30 1 440
5 458
>
0 q 0 5 "(24)2 1 52 1 (220)2
5 "16 1 25 1 400
5 "441
5 21
Therefore the scalar projection is
> >
p?q
458
,
> 5
0q0
21
>
458
q
The vector equation 5
3 >
21
0q0
458 (24, 5, 220)
5
.
21
21
458
(24, 5, 20).
5
441
>
>
b. Direction angles for p where p 5 (a, b, c)
a
include a, b, and g. cos a 5
"a 2 1 b 2 1 c 2
3
5
"32 1 62 1 (222)2
3
5
"9 1 36 1 484
3
5
"529
3
5 ,
23
3
Therefore a 5 cos21 a b
23
8 82.5°;
b
cos b 5
"a 2 1 b 2 1 c 2
6
5
2
2
"3 1 6 1 (222)2
6
5
"9 1 36 1 484
6
5
"529
6
5 ,
23
6
Therefore b 5 cos21 a b
23
8 74.9°;
7-17
cos g 5
5
5
5
>
8. a. The scalar projection of a on the x-axis
c
"a 2 1 b 2 1 c 2
222
(X, 0, 0) is
"3 1 6 1 (222)
222
2
2
2
"9 1 36 1 484
222
"529
222
5
,
23
Therefore g 5 cos21 a
222
b
23
8 163.0°
> >
7. a. x ? y 5 (1 3 1) 1 (1 3 21)
5 1 1 (21)
50
>
0 y 0 5 "12 1 (21)2
5 "2
> >
x ?y
0
> 5
0y0
"2
5 0;
>
>
y
The vector projection is 0 3 > 5 0
0y0
> >
b. x ? y 5 (2 3 1) 1 (2"3 3 0)
52
>
0 y 0 5 "12 1 02
51
> >
x ?y
2
The scalar projection is > 5
0y0
1
5 2;
>
y
(1, 0)
The vector projection is 2 3 > 5 2 3
0y0
1
>
5 2i
> >
c. x ? y 5 (2 3 25) 1 (5 3 12)
5 210 1 60
5 50
>
0 y 0 5 "(25)2 1 122
The scalar projection is
5 "25 1 144
5 "169
5 13
> >
x ?y
50
The scalar projection is > 5 .
0y0
13
>
50
y
(25, 12)
50
The vector projection is
3 > 5
3
13
0y0
13
13
50
(25, 12)
5
169
7-18
>
a ? (X, 0, 0)
0 (X, 0, 0) 0
>
a ? (X, 0, 0)
(21 3 X) 1 (2 3 0) 1 (4 3 0)
5
0 (X, 0, 0) 0
"X 2 1 02 1 02
2X
5
X
5 21;
>
The vector projection of a on the x-axis is
(X, 0, 0)
(X, 0, 0)
5 21 3
21 3
2
2
2
X
"X 1 0 1 0
>
5 2i;
>
The scalar projection of a on the y-axis (0, Y, 0) is
>
(21 3 0) 1 (2 3 Y) 1 (4 3 0)
a ? (0, Y, 0)
5
0 (0, Y, 0) 0
"02 1 Y 2 1 02
2Y
5
Y
52
>
The vector projection of a on the y-axis is
(0, Y, 0)
(0, Y, 0)
23
523
2
2
2
Y
"0 1 Y 1 0
>
5 2j;
>
The scalar projection of a on the z-axis (0, 0, Z) is
>
(21 3 0) 1 (2 3 0) 1 (4 3 Z)
a ? (0, 0, Z)
5
0 (0, 0, Z) 0
"02 1 02 1 Z 2
4Z
5
Z
5 4;
>
The vector projection of a on the z-axis is
(0, 0, Z)
(0, 0, Z)
43
543
2
2
2
Z
"0 1 0 1 Z
>
5 4k.
>
b. The scalar projection of m a on the x-axis
(X, 0, 0) is
>
ma ? (X, 0, 0)
(2m 3 X) 1 (2m 3 0)
5
0 (X, 0, 0) 0
"X 2 1 02 1 02
(4m 3 0)
1
"X 2 1 02 1 02
2mX
5
X
5 2m
>
The vector projection of ma on the x-axis is
(X, 0, 0)
(X, 0, 0)
5 2m 3
2m 3
2
2
2
X
"X 1 0 1 0
>
5 2mi;
Chapter 7: Applications of Vectors
>
The scalar projection of ma on the y-axis (0, Y, 0) is
>
ma ? (0, Y, 0)
(2m 3 0) 1 (2m 3 Y)
5
0 (0, Y, 0) 0
"0 2 1 Y 2 1 02
(4m 3 0)
1
"0 2 1 y2 1 02
2mY
5
Y
5 2m;
>
The vector projection of ma on the y-axis is
(0, Y, 0)
(0, Y, 0)
5 2m 3
2m 3
2
2
2
Y
"0 1 Y 1 0
>
5 2mj;
>
The scalar projection of ma on the z-axis (0, 0, Z) is
>
ma ? (0, 0, Z)
(2m 3 0) 1 (2m 3 0)
5
0 (0, 0, Z) 0
"0 2 1 02 1 Z2
(4m 3 Z)
1
"0 2 1 02 1 Z2
4mZ
5
Z
5 4m;
>
The vector projection of ma on the z-axis is
(0, 0, Z)
(0, 0, Z)
4m 3
5 4m 3
2
2
2
Z
"0 1 0 1 Z
>
5 4mk.
9. a.
a
>
a projected onto itself will yield itself. The scalar
projection will be the magnitude of itself.
b. Using the formula for the scalar projection
>
>
0 a 0cos u 5 0 a 0cos 0
>
5 0 a 0 (1)
>
5 0 a 0.
The vector projection is the scalar projection
>
>
>
>
a
a
multiplied by 0 > 0 , 0 a 0 3 0 > 0 5 a .
a
a
10. a. B –a
O
a
>
>
>
(2a ) ? a
2 0 a 02
b.
5
>
>
0a0
0a0
>
5 20a 0
A
> 0a0
>
So the vector projection is 2 0 a 0 a > b 5 2 a .
0a0
>
Calculus and Vectors Solutions Manual
>
11. a. AB 5 Point B 2 Point A
5 (21, 3, 4) 2 (1, 2, 2)
5 (22, 1, 2)
>
The scalar projection of AB on the x-axis (X, 0, 0) is
>
(22 3 X) 1 (1 3 0) 1 (2 3 0)
a ? (X, 0, 0)
5
0 (X, 0, 0) 0
"X 2 1 02 1 02
22X
5
X
5 22;
>
The vector projection of AB on the x-axis is
(X, 0, 0)
(X, 0, 0)
22 3
5 22 3
X
"X 2 1 02 1 02
>
5 22i;
>
The scalar projection of AB on the y-axis (0, Y, 0) is
>
(22 3 0) 1 (1 3 Y) 1 (2 3 0)
a ? (0, Y, 0)
5
0 (0, Y, 0) 0
"02 1 Y 2 1 02
Y
5
Y
5 1;
>
The vector projection of AB on the y-axis is
(0, Y, 0)
(0, Y, 0)
13
513
2
2
2
Y
"0 1 Y 1 0
>
5 j;
>
The scalar projection of AB on the z-axis (0, 0, Z) is
>
a ? (0, 0, Z)
(22 3 0) 1 (1 3 0) 1 (2 3 Z)
5
0 (0, 0, Z) 0
"02 1 02 1 Z 2
2Z
5
Z
5 2;
>
The vector projection of AB on the z-axis is
(0, 0, Z)
(0, 0, Z)
23
523
2
2
2
Z
"0 1 0 1 Z
>
5 2k
b. The angle made with the y-axis is b
b
cos b 5
2
"a 1 b 2 1 c 2
1
5
2
"(22) 1 12 1 22
1
5
"4 1 1 1 4
1
5
"9
1
5 ,
3
7-19
1
Therefore b 5 cos21 a b
3
8 70.5°
>
12. a. @ BD @
C
a
u
B
b. @ BD @
D
>
b
B
c
b
u
A
C
a
u
5
D
c
b
u
A
c. In an isosceles triangle, CD is a median
and a
>
>
right bisector of BA. Therefore a and b have the
>
same magnitude projected on c .
d. Yes, not only do they have the same magnitude,
but they are in the same direction as well which
makes them have equivalent vector projections.
>
13.
the formula for the scalar projection of a on
> a. Use
>
b 5 0 a 0cos u
5 10 cos 135°
5 27.07
>
And the> formula for the scalar projection of b on
>
a 5 @ b @ cos u
5 12 cos 135°
5 28.49
b.
b
12
135° 10
Q
O
a
P
>
>
>
OQ> is the vector projection of b on a>
>
OP is the vector projection of a on b
>
14. a. AB 5 Point B 2 Point A
5 (1, 3, 3) 2 (22, 1, 4)
5 (3, 2, 21)
>
>
The scalar projection of AB on OD is
>
>
AB ? OD
(3 3 21) 1 (2 3 2) 1 (21 3 2)
>
5
@ OD @
"(21)2 1 22 1 22
(23) 1 4 1 (22)
5
"1 1 4 1 4
7-20
21
"9
1
52
3
>
b. BC 5 Point C 2 Point B
5 (26, 7, 5) 2 (1, 3, 3)
5 (27, 4, 2)
>
>
The scalar projection of BC on OD is
>
>
BC ? OD
(27 3 21) 1 (4 3 2) 1 (2 3 2)
>
5
@ OD @
"(21)2 1 22 1 22
71814
5
"1 1 4 1 4
19
5
"9
19
5
3
>
>
>
>
AB ? OD
BC ? OD
1
19
>
>
1
52 1
3
3
@ OD @
@ OD @
18
5
3
5
6
>
AC 5 Point C 2 Point A
5 (26, 7, 5) 2 (22, 1, 4)
5 (24, 6, 1)
>
>
The> scalar
> projection of AC on OD is
AC ? OD
(24 3 21) 1 (6 3 2) 1 (1 3 2)
>
5
@ OD @
"(21)2 1 22 1 22
4 1 12 1 2
5
"1 1 4 1 4
18
5
"9
18
5
3
56
>
c. Same lengths and both are in the direction of OD.
Add to get one vector.
15. a. 1 5 cos2 a 1 cos2 b 1 cos2 g
2
2
a
b
5a
b 1a
b
"a 2 1 b 2 1 c 2
"a 2 1 b 2 1 c 2
2
c
1a
b
2
"a 1 b 2 1 c 2
a2
b2
5 2
1
a 1 b2 1 c2
a2 1 b2 1 c2
2
c
1 2
a 1 b2 1 c2
Chapter 7: Applications of Vectors
a2 1 b2 1 c2
a2 1 b2 1 c2
51
b. a 5 90°, b 5 30°, g 5 60°
cos a 5 cos 90°
5 0,
x50
cos b 5 cos 30°
"3
5
,
2
y is a multiple of "3
2 .
cos g 5 cos 60°
1
5 ,
2
1
z is a multiple of .
2
1
Answers include Q 0, "3
2 , 2 R , Q 0, "3, 1 R , etc.
c. If two angles add to 90°, then all three will add to
180°.
16. a. a 5 b 5 g
cos a 5 cos b 5 cos g
cos2 a 5 cos2 b 5 cos2 g
1 5 cos2 a 1 cos2 b 1 cos2 g
1 5 3 cos2 x
1
5 cos2 x
3
1
5 cos x
Å3
1
x 5 cos21
Å3
x 8 54.7°
1
.
b. For obtuse, use cos x 5 2
Å3
1
x 5 cos21 a2
b
Å3
x 8 125.3°
17. cos2 x 1 sin2 x 5 1
cos2 x 5 1 2 sin2 x
1 5 cos2 a 1 cos2 b 1 cos2 g
1 5 (1 2 sin2 a) 1 (1 2 sin2 b) 1 (1 2 sin2 g)
1 5 3 2 (sin2 a 1 sin2 b 1 sin2 g)
sin2 a 1 sin2 b 1 sin2 g 5 2
5
Calculus and Vectors Solutions Manual
18. Answers may vary. For example:
z
B (0, c, d)
y
x
A (a, b, 0)
7.6 The Cross Product of Two
Vectors, pp. 407–408
z
1. a.
b
axb
y
a
x
>
>
>
a 3 b is perpendicular to a . Thus, their dot product
must equal 0. The same applies to the second case.
z
a3b
b
a1b
y
a
x >
>
>
b.> a 1 b is still
> in the same plane formed
> by a and
>
>
b, thus a 1 b is perpendicular to a 3 b making the
dot product 0 again. >
>
c. Once again, a 2> b is still in> the same plane
>
>
formed
> by a and b, thus a 2 b is perpendicular to
>
the dot product 0 again.
a 3 b making
>
>
2. a 3 b produces a vector, not a scalar. Thus, the
equality is meaningless.
3. a. It’s possible because there is a vector crossed
with a vector, then dotted with another vector,
producing a scalar.
> >
b. This is meaningless because a ? b produces a
scalar. This results in a scalar crossed with a vector,
which is meaningless.
7-21
>
>
c. This> is possible. a 3 b produces a vector, and
>
c 1 d also produces a vector. The result is a vector
dotted with a vector producing
a scalar.
> >
d. This
is
possible.
produces
a scalar, and
a
?
b
>
>
c 3 d produces a vector. The product of a scalar
and vector produces a vector.
>
>
e. This> is possible. a 3 b produces a vector, and
>
c 3 d produces a vector. The cross product of a
vector and vector produces
> a vector.
>
f. This is possible. a 3 b produces a vector. When
added to another vector, it produces another vector.
4. a. (2, 23, 5) 3 (0, 21, 4)
5 (23(4) 2 5(21), 5(0) 2 2(4),
2(21) 2 (23)(0))
5 (27, 28, 22)
(2, 23, 5) ? (27, 28, 22) 5 0
(0, 21, 4) ? (27, 28, 22) 5 0
b. (2, 21, 3) 3 (3, 21, 2)
5 (21(2) 2 3(21), 3(3) 2 2(2),
2(21) 2 (21)(3))
5 (1, 5, 1)
(2, 21, 3) ? (1, 5, 1) 5 0
(3, 21, 2) ? (1, 5, 1) 5 0
c. (5, 21, 1) 3 (2, 4, 7)
5 (21(7) 2 1(4), 1(2) 2 5(7),
5(4) 2 (21)(2))
5 (211, 233, 22)
(5, 21, 1) ? (211, 233, 22) 5 0
(2, 4, 7) ? (211, 233, 22) 5 0
d. (1, 2, 9) 3 (22, 3, 4)
5 (2(4) 2 9(3), 9(22) 2 1(4),
1(3) 2 2(22))
5 (219, 222, 7)
(1, 2, 9) ? (219, 222, 7) 5 0
(22, 3, 4) ? (219, 222, 7) 5 0
e. (22, 3, 3) 3 (1, 21, 0)
5 (3(0) 2 3(21), 3(1) 2 (22)(0),
22(21) 2 3(1))
5 (3, 3, 21)
(22, 3, 3) ? (3, 3, 21) 5 0
(1, 21, 0) ? (3, 3, 21) 5 0
f. (5, 1, 6) 3 (21, 2, 4)
5 (1(4) 2 6(2), 6(21) 2 5(4),
5(2) 2 1(21))
5 (28, 226, 11)
(5, 1, 6) ? (28, 226, 11) 5 0
(21, 2, 4) ? (28, 226, 11) 5 0
5. (21, 3, 5) 3 (0, a, 1)
5 (3(1) 2 5(a), 5(0) 2 (21)(1),
21(a) 2 3(0))
7-22
If we look at the x component, we know that:
3(1) 2 5(a) 5 22
25(a) 5 25
>a 5 1
>
6. a. a 3 b 5 (1(1) 2 1(5), 1(0) 2 0(1),
0(5) 2 0(1))
5 (24, 0, 0)
b. Vectors of the form (0, b, c) are in the
yz-plane. Thus, the only vectors perpendicular to the
yz-plane are those of the form (a, 0, 0) because they
are parallel to the x-axis.
7. a. (1, 2, 1) 3 (2, 4, 2)
5 (2(2) 2 1(4), 1(2) 2 1(2), 1(4) 2 2(2))
5 (0, 0, 0)
b. (a, b, c) 3 (ka, kb, kc)
5 (b(kc) 2 c(kb), c(ka) 2 a(kc),
a(kb) 2 b(ka))
Using the commutative law of multiplication we
can rearrange this:
5 (bck 2 bck, ack 2 ack, abk 2 abk)
5 (0, 0, 0)
>
>
>
8. a. p 3 (q 1 r ) 5 (1, 22, 4) 3 3(1, 2, 7)
1 (21, 1, 0)4
5 (1, 22, 4) 3 (1 2 1, 2 1 1, 7 1 0)
5 (1, 22, 4) 3 (0, 3, 7)
5 (22(7) 2 4(3), 4(0) 2 1(7),
1(3) 1 2(0))
5 (226, 27, 3)
>
>
>
>
p 3 q 1 p 3 r 5 (22(7) 2 4(2),
4(1) 2 1(7), 1(2) 1 2(1))
1 (22(0) 2 4(1),
4(21) 2 1(0), 1(1) 1 2(21))
5 (222, 23, 4) 1 (24, 24, 21)
5 (226, 27, 3)
>
>
>
b. p 3 (q 1 r ) 5 (4, 1, 2) 3 3(3, 1, 21)
1 (0, 1, 2)4
5 (4, 1, 2) 3 (3, 1 1 1, 21 1 2)
5 (4, 1, 2) 3 (3, 2, 1)
5 (1(1) 2 2(2), 3(2) 2 4(1),
4(2) 2 1(3))
5 (23, 2, 5)
>
>
>
>
p 3 q 1 p 3 r 5 (1(21) 2 2(1), 2(3) 2 4(21),
4(1) 2 1(3)) 1 (1(2) 2 2(1),
2(0) 2 4(2), 4(1) 2 1(0))
5 (23, 10, 1) 1 (0, 28, 4)
5 (23, 2, 5)
>
>
9. a. i 3 j 5 (1, 0, 0) 3 (0, 1, 0)
5 (0 2 0, 0 2 0, 1 2 0)
5 (0, 0, 1)
>
5k
Chapter 7: Applications of Vectors
>
>
2j 3 i 5 (0, 21, 0) 3 (1, 0, 0)
5 (0 2 0, 0 2 0, 0 2 (21))
5 (0, 0, 1)
>
>
> 5k
b. j 3 k 5 (0, 1, 0) 3 (0, 0, 1)
5 (1 2 0, 0 2 0, 0 2 0)
5 (1, 0, 0)
>
>
>5i
2k 3 j 5 (0, 0, 21) 3 (0, 1, 0)
5 (0 2 (21), 0 2 0, 0 2 0)
5 (1, 0, 0)
>
>
>5i
c. k 3 i 5 (0, 0, 1) 3 (1, 0, 0)
5 (0 2 0, 1 2 0, 0 2 0)
5 (0, 1, 0)
>
>
>5j
2i 3 k 5 (21, 0, 0) 3 (0, 0, 1)
5 (0 2 0, 0 2 (21), 0 2 0)
5 (0, 1, 0)
>
5j
10. k(a2b3 2 a3b2, a3b1 2 a1b3, a1b2 2 a2b1 )
? (a1, a2, a3 )
5 k(a1a2b3 2 a1a3b2 1 a2a3b1 2 a2a1b3
1 a3a1b2 2 a3a2b1 )
5 k(0)
50
>
>
>
k(a
a is perpendicular
to
).
3
b
>
>
11. a. a 3 b 5 (2, 0, 0) 3 (0, 3, 0)
5 (0 2 0, 0 2 0, 6 2 0)
5 (0, 0, 6)
>
>
c 3 d 5 (2, 3, 0) 3 (4, 3, 0)
5 (0 2 0, 0 2 0, 6 2 12)
5
(0, 0,> 26)
>
>
>
b. (a 3 b ) 3 (c 3 d ) 5 (0, 0, 6) 3 (0, 0, 26)
(by part a.)
5 (0 2 0, 0 2 0, 0 2 0)
5 (0, 0, 0)
c. All the vectors are in the xy-plane. Thus, the cross
product in part b. is between vectors parallel to the
z-axis and so parallel to each other.
The cross
>
product of parallel vectors is 0.
>
12. Let x 5 (1, 0, 1)
>
y 5 (1, 1, 1)
>
z 5 (1, 2, 3)
>
>
Then x 3 y 5 (0 2 1, 1 2 1, 1 2 0)
5 (21, 0, 1)
>
>
>
(x 3 y ) 3 z 5 (0 2 2, 1 2 (23), 23 2 0)
5 (22, 4, 23)
Calculus and Vectors Solutions Manual
>
>
y 3 z 5 (3 2 2, 1 2 3, 2 2 1)
5 (1, 22, 1)
>
>
>
x 3 (y 3 z ) 5 (0 1 2, 1 2 1, 22 2 0)
5 (2, 0, 22)
>
>
>
>
>
>
Thus (x 3> y ) 3 z 2 x> 3 (y 3 z ).
>
>
13. (a 2 b ) 3 (a 1 b )
By the distributive property of cross product:
>
>
>
>
>
>
5 (a 2 b ) 3 a 1 (a 2 b ) 3 b
By the distributive
property >again:
>
>
>
>
>
>
>
5 a 3 a 2 b 3 a 1 a 3 b 2 b 3 >b
A vector
with> itself equals 0, thus:
> crossed
>
>
5 2b 3> a 1> a 3 b
>
>
5 a 3 b> 2 b 3 a >
>
>
5 a 3 b 2 (2a 3 b )
>
>
5 2a 3 b
7.7 Applications of the Dot Product
and Cross Product, pp. 414–415
1. By pushing as far away from the hinge as
>
possible, 0 r 0 is increased making the cross product
bigger. By pushing at right angles, sine is its largest
value, 1, making
the cross product larger.
>
>
2. a. a 3 b 5 (1, 2, 1) 3 (2, 4, 2)
5 (2(2) 2 1(4), 1(2)
2 1(2), 1(4) 2 2(2))
5 (0, 0, 0)
>
>
@a 3 b @ 5 0
b. This makes sense because the vectors lie on the
same line. Thus, the parallelogram would just be a
line making
its area 0.
> >
3. a. f ? s 5 3 ? 150 5 450 J
b.
x
y
50°
392 N
40 m
The axes are tilted to illustrate the force of gravity
can be split up into components to find the part in
the direction of the motion. Let x be the component
of force going in the motion’s direction.
x
cos (50°) 5
392
x 5 (392) cos (50°)
Now we have our force, so:
(392) cos 50° N ? 40 m 8 10 078.91 J
7-23
c.
140 N
20°
250 m
First find the x component of the force:
(140) cos (20°) 5 x
Calculate work:
140 cos 20° N ? 250 m 8 32 889.24 J
d.
100 N
45°
500 m
First calculate the x component of the force:
x 5 (100) cos (45°)
Calculate work:
100 cos
m 5 35 355.34 J
> 45°
> ? 500
>
4. a. i 3 j 5 k
The square formed by the 2 vectors has an area of 1.
The 2 vectors are >on the xy-plane, thus, the cross
product must be k by the right hand rule.
>
>
>
b. 2i 3 j 5 2k
Once again, the area is 1, making the possible vector
have a magnitude of 1. Also, the 2 vectors are on the
xy-plane again so the cross product must lie on the
z axis. However, because
of the right hand rule, the
>
product
must
be
this
time.
2k
>
>
>
c. i 3 k 5 2j
The square has an area of 1, so the magnitude of the
vector produced must be 1. The 2 vectors are on the
xz-plane. The> new vector must be on the y axis
making it – j because of the right hand rule.
>
>
>
d. 2i 3 k 5 2j
The square has an area of 1. The 2 vectors
> are on
the xz-plane. So the new vector must be j because
of the right> hand rule.
>
5. a. a 3 b 5 (1, 1, 0) 3 (1, 0, 1)
5 (1 2 0, 0 2 1, 0 2 1)
5 (1, 21, 21)
>
>
@ a 3 b @ 5 "1 1 1 1 1 5 "3
So the area
> of the parallelogram is "3 square units.
>
b. a 3 b 5 (1, 22, 3) 3 (1, 2, 4)
5 (28 2 6, 3 2 4, 2 1 2)
5 (214, 21, 4)
>
>
@ a 3 b @ 5 "196 1 1 1 16 5 "213
So the area of the parallelogram is "213 square units.
7-24
>
>
6. p 3 q 5 (a, 1, 21) 3 (1, 1, 2)
5 (2 1 1, 22a 2 1, a 2 1)
5 (3, 2a 1 1, a 2 1)
>
>
0 p 3 q 0 5 "9 1 (2a 1 1)2 1 (a 2 1)2 5 "35
9 1 (2a 1 1)2 1 (a 2 1)2 5 35
2
9 1 4a 1 4a 1 1 1 a 2 2 2a 1 1 5 35
5a 2 1 2a 2 24 5 0
22 6 "22 2 4(5)(224)
a5
2(5)
22 6 22
5
10
212
5 2,
5
7. a.
B
AB
A
AC
C
As we see from the picture, the area of the triangle
ABC is just half the area of> the parallelogram
>
determined by vectors AB and AC. Thus, we use the
magnitude of the cross product to calculate the area.
>
AB> 5 (1 1 2, 0 2 1, 1 2 3) 5 (3, 21, 22)
AC> 5 (2 1
> 2, 3 2 1, 2 2 3) 5 (4, 2, 21)
AB 3 AC 5 (1 1 4, 23 1 8, 6 1 4) 5 (5, 5, 10)
>
>
@ AB 3 AC @ 5 "25 1 25 1 100 5 5"6
Since triangle ABC is half the area of the
parallelogram, its area is 5"6
2 square units.
b. This is just a different way of describing the first
triangle, thus the area is 5"6
2 square units.
c. Any two sides of a triangle can be used to
calculate> its area.
>
>
>
8. @ r 3 f @ 5 ( 0 r 0sin (u)) @ f @
5 (0.14) sin (45°) ? 10
8 0.99 J
9.
A
BN
OA
C
O
OB
B
We know that the area of a parallelogram is equal to
its height multiplied with its base. Its height is BN
>
>
and its base is AC 5 OB as can be seen from the
picture. We can calculate the area using the given
vectors,
then
>
> use the area to find BN.
OA 3 OB 5 (8 2 4, 12 2 16, 4 2 6)
5 (4, 24, 22)
>
>
@ OA 3 OB @ 5 "16 1 16 1 4 5 "36 5 6
Chapter 7: Applications of Vectors
Now we need to calculate @ OB @ to know the length
of the base.
>
>
AC 5 @ OB @ 5 "9 1 1 1 16 5 "26
Substituting these results into the equation for area:
>
@ OB @ ? BN 5 6
>
"26 BN 5 6
BN 5
6
or about 1.18
"26
>
>
10. a.
p 3 q 5 (26 2 3, 6 2 3, 1 1 4)
5 (29, 3, 5)
>
>
>
(p 3 q ) 3 r 5 (0 2 5, 5 1 0, 29 2 3)
5 (25, 5, 212)
a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)
Looking at x components:
a 1 2b 5 25; a 5 25 2 2b
y components:
22a 1 b 5 5
Substitute in a:
10 1 4b 1 b 5 5
5b 5 25
b 5 21
Substitute b back into the x components:
a 5 25 1 2; a 5 23
Check in z components:
3a 1 3b 5 212
29 2 3 5 212
> >
b. p ? r 5 1 2 2 1 0 5 21
> >
q?r52111053
> > >
> > >
(p ? r )q 2 (q ? r )p 5 21(2, 1, 3) 2 3(1, 22, 3)
5 (2, 21, 23) 2 (3, 26, 9)
5 (22 2 3, 21 1 6, 23 2 9)
5 (25, 5, 212)
Review Exercise, pp. 418–421
>
>
1. a. a 3 b 5 (2 2 0, 21 1 1, 0 1 2)
5 (2, 0, 2)
>
b. b 3 c 5 (0 2 4, 25 1 5, 24 2 0)
5 (24, 0, 24)
c. 16
d. The cross products are parallel, so the original
vectors are in the same plane.
>
2. a.> 0 a 0 5 "22 1 (21)2 1 22 5 3
b. @ b @ 5> "62 1 32 1 (22)2 5 7
>
c. a 2 b 5 (2 2 6, 21 2 3, 2 1 2)
5 (24, 24, 4)
>
>
@ a 2 b @ 5 "(24)2 1 (24)2 1 42 5 4"3
>
Calculus and Vectors Solutions Manual
>
>
d. a 1 b 5 (2 1 6, 21 1 3, 2 2 2)
5 (8, 2, 0)
>
>
@ a 1 b> @ 5 "82 1 22 1 02 5 2"17
>
e. a ? b 5 2(6)
2 1(3) 1 2(22) 5 5
>
>
a 2 2b 5 (2 2 12, 21 2 6, 2 1 4)
f.
> 5 (210, 27, 6)
>
>
a ? (a 2 2b ) 5 2(210) 2 1(27) 1 2(6) 5 21
>
>
3. a. If a 5> 6, then
y will be twice x , thus collinear.
>
x 3 y 5 (3, a, 9) ? (a, 12, 18) 5 0
b.
3a 1 12a 1 162 5 0
15a 5 2162
254
a5
5
> >
a?b
4. cos (u) 5 > >
0 a 0 @b@
> >
a ? b 5 4(23) 1 5(6) 1 20(22) 5 458
>
0 a 0 5 "42 1 52 1 202 5 21
>
@ b @ 5 "(23)2 1 62 1 222 5 23
458
u 5 cos21 a
b
483
u 8 18.52°
5. a.
y
4
2 OA
OB
x
0
–4 –2
2 4
–2
–4
b. We can use the dot product of the 2 diagonals to
calculate
the
angle. >The diagonals
are the vectors
>
>
>
OA> 1 OB> and OA 2 OB.
OA 1 OB 5 (5 2 1, 1 1 4) 5 (4, 5)
>
>
OA 2 OB 5 (5 1 1, 1 2 4) 5 (6, 23)
>
>
>
>
(OA 1 OB ) ? (OA 2 OB )
>
>
>
>
cos (u) 5
@ OA 1 OB @ @ OA 2 OB @
>
>
>
>
(OA 1 OB ) ? (OA 2 OB ) 5 4(6) 1 5(23) 5 9
>
>
@ OA 1 OB @ 5 "42 1 52 5 "41
@ OA 2 OB @ 5 "62 1 (23)2 5 3"5
>
>
u 5 cos21 a
9
3"205
b
u 8 77.9°
7-25
6.
>
T1
30°
5 (250, 220, 18)
>
>
0 x 3 y 0 5 "502 1 202 1 182 5 "3224 8 56.78
9. (0, 3, 25) 3 (2, 3, 1)
5 (3 1 15, 210 2 0, 0 2 6) 5 (18, 210, 26)
The cross product is perpendicular to the given
vectors, but its magnitude is
T2
45°
x
98 N
The vertical components of the tensions must equal
the downward force:
T1 sin (30°) 1 T2 sin (45°) 5 98 N
1
1
T 1
T 5 98
2 1 "2 2
T1 5 196 2 "2T2
The horizontal components:
T1 cos (30°) 1 T2 cos (45°) 5 0 N
"3
1
T 2
T 50
2 1 "2 2
Substitute in T1:
"6
98"3 2
T 5 298"3
2 2
2"6 2 "2
T2 5 298"3
2
T2 8 87.86N
Substitute this back in to get T1:
T1 8 71.74N
7.
x
50 km/h
300 km/h
x 5 "50 1 3002 8 304.14
50
tan21 a
b 8 9.46°
300
The resultant velocity is 304.14 km> h, W 9.46° N.
8. a.
z
2
y
x
x
7-26
>
b. x 3 y 5 (215 2 35, 25 2 15, 21 2 3)
y
y
"18 2 1 (210)2 1 (26)2, or 2"115. A unit vector
perpendicular to the given vectors is
a !115, 2 !115, 2 !115 b.
9
5
3
>
10. a.cos (a) 5
>
>
AB ? AC
@ AB @ @ AC @
>
>
AB> 5 (0, 23, 4) 2 (2, 3, 7) 5 (22, 26, 23)
> AC> 5 (5, 2, 24) 2 (2, 3, 7) 5 (3, 21, 211)
AB ? AC 5 22(3) 2 6(21) 2 3(211) 5 33
>
@ AB @ 5 "(22)2 1 (26)2 1 (23)2 5 7
>
2
@ AC @ 5 "32 1 (21)
1
(211)2 5 "131
>
>
AB ? AC
>
>
a 5 cos21
@ AB @ @ AC @
33
5 cos21
7"131
8 65.68° >
>
21 BA ? BC
>
>
b 5 cos
@
@
@
@
BA
BC
>
>
BA> 5 2AB 5 (2, 6, 3)
> BC> 5 (5 2 0, 2 1 3, 24 2 4, ) 5 (5, 5, 28)
BA ? BC 5 2(5) 1 6(5) 1 3(28) 5 16
>
@ BA @ 5 "22 1 62 1 32 5 7
>
@ BC @ 5 "52 1 52 1 (28)2 5 "144
16
b 5 cos21
7"114
8 77.64°
g 5 180 2 a 2 b 8 36.68°
So b 8 77.64° is the largest angle.
b. The area is >half the magnitude
of the cross
>
product of AB and AC.
>
>
1
1
AB 3 AC 5 0 (63, 231, 20) 0 8 36.50
2
2
11. The triangle formed by the two strings and the
ceiling is similar to a 3-4-5 right triangle, with the
30 cm and 40 cm strings as legs. So the angle
adjacent to the 30 cm leg satisfies
3
cos u 5
5
Chapter 7: Applications of Vectors
The angle adjacent to the 40 cm leg satisfies
4
cos f 5
5
Also,
4
3
sin u 5 and sin f 5 .
5
5
Let T1 be the tension in the 30 cm string, and T2 be
the tension in the 40 cm string. Then
T1 cos u 2 T2 cos f 5 0
3
4
T1 2 T2 5 0
5
5
4
T1 5 T2
3
Also,
T1 sin u 1 T2 sin f 5 (10)(9.8) 5 98
4
3
T1 2 T2 5 98
5
5
4
3
4
a T2 b 1 T2 5 98
3
5
5
5
T 5 98
3 2
T2 5 58.8 N
4
T1 5 (58.8)
3
5 78.4 N
So the tension in the 30 cm string is 78.4 N and the
tension in the 40 cm string is 58.8 N.
12. a.
54 N
30 N
25 N
42 N
b. The east- and west-pulling forces result in a force
of 5 N west. The north- and south-pulling forces
result in a force of 12 N north. The 5 N west and
12 N north forces result in a force pulling in the
north-westerly direction with a force of
"52 1 122 5 13 N,
by using the Pythagorean theorem. To find the exact
direction of this force, use the definition of sine.
Calculus and Vectors Solutions Manual
If u is the angle west of north, then
5
sin u 5
13
u 8 22.6°
So the resultant is 13 N in a direction
N22.6°W. The equilibrant is 13 N in a direction
S22.6°E.
13. a. Let D be the origin, then:
A 5 (2, 0, 0), B 5 (2, 4, 0), C 5 (0, 4, 0),
D 5 (0, 0, 0), E 5 (2, 0, 3), F 5 (2, 4, 3),
G 5 (0, 4, 3) H 5 (0, 0, 3)
>
b. AF> 5 (0, 4, 3)
AC 5 (22, 4, 0)
>
>
AF ? AC> 5 0 1 16 1 0 5 16
@ AF @ 5 "02 1 42 1 32 5 5
>
@ AC @ 5 "(22)2 1 42 1 02 5 2"5
>
>
AF ? AC
>
>
cos (u) 5
@ AF @ @ AC @
16
b
u 5 cos21 a
10"5
u 8 44.31°
>
c. Scalar projection 5 @ AF @ cos (u)
By part b.:
5 (5) cos (44.31°)
8 3.58 >
>
> >
14. a ? b 5 0 a 0 @ b @ cos (u) 5 cos (u)
1
cos (u) 5 2 (cosine law)
2 >
>
>
>
(2a 2 5b ) ? (b 1 3a )
>
>
> >
> >
5 213a ? b 1 6a ? a 2 5b ? b
>
>
5 213a ? b 1 1
5 213 cos (u) 1 1
5 7.5
15. a. The angle to the bank, u, will satisfy
2
sin (90° 2 u) 5 3
90° 2 u 8 41.8°
u 8 48.2°
b. By the Pythagorean theorem, Kayla’s net
swimming speed will be
"32 2 22 5 "5 km> h.
So since distance 5 rate 3 time, it will take her
0.3
t5
"5
8 0.13 h
8 8 min 3 sec
to swim across.
7-27
c. Such a situation would have resulted in a right
triangle where one of the legs is longer than the
hypotenuse, which is impossible.
>
>
16. > a. The > diagonals are OA 1 OB and
OA> 2 OB>.
OA 1 OB 5 (3 2 6, 2 1 6, 26 2 2)
5 (23, 8, 28)
>
>
OA 2 OB 5 (3 1 6, 2 2 6, 26 1 2)
5 (9, 24, 24)
>
>
b. OA ? OB> 5 3(26) 1 2(6) 2 6(22) 5 6
@ OA @ 5 "32 1 22 1 (26)2 5 7
>
2
@ OB @ 5 "(26)
1
62 1 (22)2 5 2"19
>
>
OA ? OB
>
>
cos (u) 5
@ OA @ @ OB @
6
b
u 5 cos21 5 a
14"19
8 84.36°
17. a. The z value is >double, so if a 5 4 and
b 5 24, the vector q will be collinear.
>
>
b. If p and q are perpendicular, then their dot
product will equal 0.
> >
p ? q 5 2a 2 2b 2 18 5 0
c. Let a 5 9, and b> 5 0, then we have a vector
perpendicular to p . Now it must be divided by its
magnitude to make it a unit vector:
>
0 p 0 5 "81 1 0 1 324 5 9"5
So the unit vector is:
1
2
a
, 0,
b
!5 !5
> >
18. a. m ? n 5 2"3 2 2"3 1 3 5 3
>
0 m 0 5 "3 1 4 1 9 5 4
>
0 n 0 5 "4 1 3 1 1 5 2"2
> >
m?n
cos (u) 5 > >
0m00n0
3
u 5 cos21 a
b
8"2
8 74.62°
>
b. Scalar projection 5 0 n 0cos (u)
5 2"2 cos (74.62°)
8 0.75
with the unit vector
c. Scalar projection multiplied
>
in the direction of m :
>
m
5 (0.75) >
0m0
("3, 22, 23)
5 (0.75)
4
5 (0.1875)("3, 22, 23)
7-28
>
>
d. m ? k 5 23
u 5 cos21 a
23
b
4
8 138.59°
19. a. If the dot product is 0, then the vectors are
perpendicular:
(1, 0, 0) ? (0, 0, 21) 5 0 1 0 1 0 5 0
(1, 0, 0) ? (0, 1, 0) 5 0 1 0 1 0 5 0
(0, 0, 21) ? (0, 1, 0) 5 0 1 0 1 0 5 0 special
1
1
21 1
1
b. a ,
, 0b ? a
,
,
b
!2 !2
!3 !3 !3
1
1
52
1
10
!6
!6
50
1
1
a
,
, 0b ? (0, 0, 21) 5 0 1 0 1 0 5 0
!2 !2
21 1
1
a
,
,
b ? (0, 0, 21)
!3 !3 !3
1
1
52
not special
501012
!3
>
> !3
20. a. p 3 q
5 (22(1) 2 1(21), 1(2) 2 1(1), 1(21) 1 2(2))
5 (21, 1, 3)
>
>
b. p> 2 q> 5 (21, 21, 0)
p 1 q 5 (3, 23, 2)
>
>
>
>
(p 2 q ) 3 (p 1 q ) 5 (22 2 0, 0 1 2, 3 2 (23))
5 (22, 2, 6)
>
>
c. p 3 r 5 (4 2 1, 0 1 2, 1 2 0)
5 (3, 2, 1)
>
>
>
(p 3 r ) ? r 5 0 1 2 2 2 5 0
>
>
d. p 3 q 5 (22 1 1, 2 2 1, 21 1 4)
5 (21, 1, 3)
21. Since the angle between the two vectors is 60°,
the angle formed when they are placed head-to-tail
is 120°. So the resultant, along with these two
vectors, forms an isosceles triangle with top angle
120° and two equal angles 30°. By the cosine law,
the two equal forces satisfy
202 5 2F 2 2 2F 2 cos 120°
400
F2 5
3
20
F5
"3
8 11.55
> N
>
22. a 3 b 5 (2 2 0, 25 2 3, 0 2 10)
5 (2, 28, 210)
Chapter 7: Applications of Vectors
23.
First we need to determine the dot product of
>
>
x and y :
> >
> >
x ? y 5 0 x 0 0 y 0cos u
5 (10) cos (60°)
55
>
>
>
>
(x 2 2y ) ? (x 1 3y )
By the distributive property:
> >
> >
> >
> >
5 x ? x 1 3x ? y 2 2x ? y 2 6y ? y
5 4 1 15 2 10 2 150
5 2141
24. 0 (2, 2, 1) 0 5 "22 1 22 1 12 5 3
Since the magnitude of the scalar projection is 4,
the scalar projection itself has value 4 or 24.
If it is 4, we get
(1, m, 0) ?(2, 2, 1)
54
3
2 1 2m 5 12
m 5 5
If it is 24, we get
(1, m, 0) ?(2, 2, 1)
5 24
3
2 1 2m 5 212
m 5 27
So the two possible values for m are 5 and 27.
> >
25. a ? j 5 23
>
0 a 0 5 "144 1 9 1 16 5 13
23
u 5 cos21 a
b
13
8 103.34°
26. a.> C 5 (3, 0, 5), F 5 (0, 4, 0)
b. CF 5 (0, 4, 0) 2 (3, 0, 5) 5 (23, 4, 25)
c. @ CF @ > 5 "9 1 16 1 25 5 5"2
>
OP 5 (3, 4, 5)
>
@
@ > 5 "9 1 16 1 25 5 5"2
OP
>
CF ? OP 5 29 1 16 2 25 5 218
218
u 5 cos21 a
b
50
8 111.1°
27.
d
50°
130°
e
a. Using properties of parallelograms, we know that
the other angle is 130° (Angles must add up to
360°, opposite angles are congruent).
Using the cosine law,
Calculus and Vectors Solutions Manual
@ d 1 e @ 2 5 32 1 52 2 2(3)(5) cos 130°
>
>
@ d 1 e @ 8 7.30
b. Using the cosine law,
>
>
@ d 2 e @ 2 5 32 1 52 2 2(3)(5) cos 50°
>
>
@ d 2 e @ 8 3.84
>
>
>
>
c.> e 2> d is the vector in the opposite direction of
d 2 e , but with the same magnitude. So:
>
>
>
>
@ e 2 d @ 5 @ d 2 >e @ 8> 3.84>
(i 1 j ) ? (i )
>
28. a. Scalar:
51
@i@
>
>
i
Vector: 1a > b 5 i
@i@
b. Scalar:
Vector: 1a
c. Scalar:
Vector:
1
>
>
>
>
@ j@
(i 1 j ) ? (j )
>
j
>
b5j
>
>
@ j@
51
>
>
>
(i 1 j ) ? (k 1 j )
"2
@k 1 j@
>
5
(k 1 j )
>
>
?
>
>
>
>
1
"2
>
1
5 (k 1 j )
2
@k 1 j@
29. a. If its magnitude is 1, it’s a unit vector:
>
0 a 0 5 "14 1 19 1 361 2 1 not a unit vector
@ b @ 5 "13 1 13 1 13 5 1, unit vector
>
>
0 c 0 5 "14 1 12 1 14 5 1, unit vector
@ d @ 5 "1 1 1 1 1 2 1, not a unit vector
>
>
b. a is. When dotted with d, it equals 0.
30. 25 ? sin> (30°) ? 0.6 5 7.50 J
>
31. >a. a ? b 5 6 2 5 2 1 5 0
b. a with the x-axis:
>
>
0 a 0 5 "4 1 25 1 1 5 "30
2
cos (a) 5
"30
>
a with the y-axis:
5
cos (b) 5
"30
>
a with the z-axis:
21
cos (g) 5
"30
>
@ b> @ 5 "9 1 1 1 1 5 "11
b with the x-axis:
3
cos (a) 5
"11
7-29
>
b with the y-axis:
21
cos (b) 5
"11
>
b with the z-axis:
1
cos (g) 5
"11
6
5
1
>
>
c. m 1 ? m 2 5
2
2
50
!330
!330
!330
32. Need to show that the magnitudes of the
diagonals are equal to show that it is a rectangle.
>
>
>
@ 3i 1 3j 1 10k @ 5 "9 1 9 1 100 5 "118
@ 2i 1 9j 2 6k @ 5 "1 1 81 1 36 5 "118
33. a. Direction cosine for x-axis:
>
>
>
"3
2
We know the identity
cos2 a 1 cos2 b 1 cos2 g 5 1.
Since a 5 30g, and b 5 g, we get
3
2 cos2 b 5 1 2
4
1
cos b 5 cos g 5 6
2"2
"3
cos a 5
2
So there are two possibilities, depending upon
whether b 5 g is acute or obtuse.
b. If g is acute, then
1
cos g 5
2"2
g 8 69.3°
If Á is obtuse, then
1
cos g 5
2"2
g 8 110.7°
1
> >
> >
34. a ? b 5 0 a 0 @ b @ cos (u) 5
2
>
>
>
>
(a 2 3b ) ? (ma
)
5
0
1
b
>
>
> >
> >
> >
ma ? a 1 a ? b 2 3ma ? b 2 3b ? b 5 0
1
3
6
m1 2 m2 50
2
2
2
1
5
2 m5
2
2
m 5 25
> >
a ? b> 5 0 2 20 1 12 5 28
35.
>
a 1 b 5 (21, 21, 28)
>
>
@ a 1 b @ 5 "1 1 1 1 64 5 "66
cos (30°) 5
7-30
>
>
a 2 b 5 (1, 9, 24)
>
>
@ a 2 b @ 5 "1 1 81 1 16 5 "98
>
>
1 >
66
98
1 >
@a 1 b@ 2 2 @a 2 b@ 2 5
2
5 28
4
4
4
4
>
>
>
36.> c 5 b >2 a >
0 c 02 5 @b 2 a @ 2
>
>
>
>
5 (b 2 a ) ? (b 2 a )
>
>
>
>
> >
> >
5b?b2a?b1a?a2a?b
>
>
> >
5 0 a 0 2 1 @ b @ 2 2 2a ? b
>
>
> >
5 0 a 0 2 1 @ b @ 2 2 2 0 a 0 @ b @ cos u
>
37. AB> 5 (2, 0, 4)
@ AB @ 5 "4 1 0 1 16 5 2"5
>
@ AC @ 5 (1, 0, 2)
>
@ AC @ 5 "1 1 0 1 4 5 "5
>
BC 5 (21, 0, 22)
>
@ BC @ 5 "1 1 0 1 4 5 "5
>
cos A 5
>
AB ? AC
@ AB @ @ AC @
>
>
10
10
51
But this means that angle A 5 0°, so that this
triangle is degenerate. For completeness, though,
>
>
>
>
notice that BC 5 2AC and AB 5 2 AC . This
means that point C sits at the midpoint of the line
segment joining A and B. So angle
C 5 180° and angle B 5 0°. So
cos B 5 1;
cos C 5 21.
The area of triangle ABC is, of course, 0.
5
Chapter 7 Test, p. 422
>
>
1. a. We use the diagram to calculate a 3 b, noting
a1 5 21, a2 5 1, a3 5 1 and b1 5 2, b2 5 1,
b3 5 23.>
>
b
a
1
1
x
x 5 1(23) 2 1(1) 5 24
23
1
y
y 5 1(2) 2 (21)(23) 5 21
21
2
z
z 5 21(1) 2 1(2) 5 23
1
1>
>
So, a 3 b 5 (24, 21, 23)
Chapter 7: Applications of Vectors
b. We
use the diagram again:
>
>
b
1
x
23
x 5 1(27) 2 (23)(1) 5 24
27
y
2
z
c. The area of the parallelogram is the magnitude of
c
1
y 5 23(5) 2 (2)(27) 5 21
5
z 5 2(1) 2 1(5) 5 23
1 > 1
>
So, b 3 >c 5 (24, 21, 23)
>
>
c. a ? (b 3 c ) 5 (21, 1, 1) ? (24, 21, 23)
5 (21)(24) 1 (1)(21)
1 (1)(23)
50
d. We could use the diagram
method again, or, we
> > >
>
>
>
note that for any vectors x , y , x 3 y 5 2 (y 3 x ),
>
>
>
>
so letting y 5 x , we have
x> 3 x 5 0 from the last
>
>
>
equation. Since a 3 b 5 b 3> c from
> the> first two
>
parts of the problem, (a 3 b ) 3 (b 3 c ) 5 0.
>
2. a.> To find the scalar and vector
projections of a
>
>
>
>
>
@
@
on b, we
need
to
calculate
and
a
b
5
"b
?
b
?
b
> >
a ? b 5 (1, 21, 1) ? (2, 21, 22)
5 (1)(2) 1 (21)(21) 1 (1)(22)
> 51
@ b @ 5 "22 1 (21)2 1 (22)2
> 53
So, @ b @ 5 3
>
>
> >
The scalar projection of a on b is
a?b
1
> 5 3 , and
@b@
>
>
the
vector
projection
of
on
is
a
b
>
a
>
a?b
@b@
>2
bb 5 19 (2, 21, 22).
>
>
b. We find the direction cosines for b:
b1
2
3
@b@
a 8 48.2°.
b
21
cos (b) 5 2> 5
3
@b@
b 8 109.5°.
b
22
cos (g) 5 3> 5
3
@b@
g 8 131.8°.
cos (a) 5
>
5
Calculus and Vectors Solutions Manual
the cross >product.
>
b
a
21
21
x
x 5 (21)(22) 2 1(21) 5 3
1
22
y
y 5 1(2) 2 (1)(22) 5 4
1
2
z
z 5 (1)(21) 2 (21)(2) 5 1
21
21
>
>
So, a 3 b 5 (3, 4, 1) and thus,
>
>
@ a 3 b @ 5 "32 1 42 1 12
5 "26
>
>
So the area of the parallelogram formed by a and b
is "26 or 5.10 square units.
3. We first draw a diagram documenting the
situation:
E
F
120°
40 N
40 N
R
60°
60°
50 N
D
50 N
G
In triangle DEF, we use the cosine law:
>
@ R @ 5 "402 1 502 2 2(40)(50) cos (120°)
>
@ R @ 8 78.10
We now use the sine law to find /EDF:
sin /EDF
sin /DEF
>
>
5
@ EF @
@R@
sin /EDF
sin 120°
8
50
78.10
sin /EDF 8 0.5544
/EDF 8 33.7°
The equilibrant force is equal in magnitude and
opposite in direction to the resultant force, so both
forces have a magnitude of 78.10 N. The resultant
makes an angle 33.7° to the 40 N force and 26.3° to
the 50 N force. The equilibrant makes an angle 146.3°
to the 40 N force and 153.7° to the 50 N force.
7-31
4. We find the resultant velocity of the airplane.
E
F
F
R
1000 km/h
G
D
Vector diagram
Position diagram
Since the airplane’s velocity is perpendicular to the
wind, the resultant’s magnitude is given by the
Pythagorean theorem:
>
@ R @ 5 "10002 1 1002
>
@ R @ 8 1004.99
The angle is determined using the tangent ratio:
100
tan /EDF 5
1000
/EDF 8 5.7°
Thus, the resultant velocity is 1004.99 km> h,
N 5.7° W (or W 84.3° N).
5. a. The canoeist will travel 200 m across the
stream, so the total time he will paddle is:
d
t5
rcanoeist
200 m
t5
2.5 m>s
t 5 80 s
The current is flowing 1.2 m> s downstream, so the
distance that the canoeist travels downstream is:
d 5 rcurrent 3 t
d 5 (1.2 m>s)(80 s)
d 5 96 m
So, the canoeist will drift 96 m south.
b. In order to arrive directly across stream, the
canoeist must take into account the change in his
velocity caused by the current. That is, he must
initially paddle upstream in a direction such that
the resultant velocity is directed straight across
the stream. The resultant velocity:
E
2.5 m/s
D
7-32
R
Since the resultant velocity is perpendicular to the
current, the direction in which the canoeist should
head is determined by the sine ratio.
1.2
sin /EDF 5
2.5
/EDF 8 28.7°
The canoeist should head 28.7° upstream.
6. The area of the triangle is exactly:
>
>
1
ADABC 5 @ AB 3 BC @
2
>
AB 5 (2, 1, 3) 2 (21, 3, 5)
5 (3, 22, 22)
>
BC 5 (21, 1, 4) 2 (2, 1, 3)
5 (23, 0, 1)
>
>
BC
AB
22 x 0
x 5 (22)(1) 2 (22)(0) 5 22
22 y 1
y 5 (22)(23) 2 (3)(1) 5 3
3 z 23
z 5 (3)(0) 2 (22)(23) 5 26
22
0
>
>
So, AB 3 BC 5 (22, 3, 26) and
>
>
@ AB 3 BC @ 5 "(22)2 1 32 1 (26)2
5 "49
57
>
>
1
7
So, ADABC 5 @ AB 3 BC @ 5 .
2
2
The area of the triangle is 3.50 square units.
7.
458
T1
708
T2
25 kg
The system is in equilibrium (i.e. it is not moving),
>
so we> know that the horizontal components of T1
and T2 are equal:
@ T1 @ sin (45°) 5 @ T2 @ sin (70°)
>
>
@ T2 @ 5
>
sin (45°) >
@T @
sin (70°) 1
1.2 m/s
F
Chapter 7: Applications of Vectors
>
>
Also, the vertical component of T1 1 T2 must equal
the> gravitational > force on the block:
@ T1 @ cos 45° 1 @ T2 @ cos 70° 5 (25 kg)(9.8 m>s2 )
>
Substituting in for T2, we find that:
>
@ T1 @ cos 45° 1
> sin 45°
@ T1 @
cos 70° 5 (25 kg) (9.8 m>s 2 )
sin 70°
>
sin 45°
@ T1 @ acos 45° 1
cos 70°b 5 245 N
sin 70°
@ T1 @ (0.9645) 8 245 N
>
@ T1 @ 8 254.0 N
>
So, we can now find
>
sin (45°) >
@ T2 @ 5
@T @
sin (70°) 1
>
sin (45°)
@ T2 @ 8
(254.0 N)
sin (70°)
>
@ T2 @ 8 191.1 N
The direction of the tensions are indicated in the
diagram.
8. a. We explicitly calculate both sides of the
equation. The left side is:
> >
x ? y 5 (3, 3, 1) ? (21, 2, 23)
5 (3)(21) 1 (3)(2) 1 (1)(23)
50
We perform a few computations before computing
the right side:
>
>
x 1 y 5 (3, 3, 1) 1 (21, 2, 23)
5 (2, 5, 22)
Calculus and Vectors Solutions Manual
>
>
>
>
>
>
0 x 1 y 0 2 5 (x 1 y ) ? (x 1 y )
5 22 1 52 1 (22)2
5 33
>
>
x 2 y 5 (3, 3, 1) 2 (21, 2, 23)
5 (4, 1, 4)
>
>2
>
>
>
>
0 x 2 y 0 5 (x 2 y ) ? (x 2 y )
5 42 1 12 1 42
5 33
Thus, the right side is
1 >
1
1
1 >
>
>
0 x 1 y 0 2 2 0 x 2 y 0 2 5 (33) 2 (33)
4
4
4
4
50
So, the equation holds for these vectors.
b. We now verify that the formula holds in general.
We will compute the right side of the equation, but
we first perform some intermediary computations:
>
>
>
>
>
>
0 x 1 y 0 2 5 (x 1 y ) ? (x 1 y )
> >
> >
> >
> >
5 (x ? x ) 1 (x ? y ) 1 (y ? x ) 1 (y ? y )
> >
> >
> >
5 (x ? x ) 1 2(x ? y ) 1 (y ? y )
>
>
>
>
>
>
0 x 2 y 0 2 5 (x 2 y ) ? (x 2 y )
> >
>
>
> >
5 (x ? x ) 1 (x ? 2y ) 1 (2y ? x )
>
>
1 (2y ? 2y )
> >
> >
> >
5 (x ? x ) 2 2(x ? y ) 1 (y ? y )
So, the right side of the equation is:
1 >
1
1 >
>
>
> >
0 x 1 y 0 2 2 0 x 2 y 0 2 5 (4(x ? y ))
4
4
4
> >
5x?y
Thus, the equation holds for arbitrary vectors.
7-33
CHAPTER 7
Applications of Vectors
m 2 5 52 1 122 2 2(5)(12) cos 135
2 !2
5 25 1 144 2 120a
b
2
5 169 1 84.85
5 253.85
Review of Prerequisite Skills, p. 350
1. The velocity relative to the ground has a
magnitude equivalent to the hypotenuse of a
triangle with sides 800 and 100. So, by the
Pythagorean theorem we can find the magnitude of
the velocity.
v 2 5 8002 1 1002
5 640 000 1 10 000
5 650 000
v 5 "650 000
8 806 km>h
100
800
m 5 "253.85
8 15.93 units
122 5 15.932 1 52 2 2(15.93)(5) cos u
144 5 253.76 1 25 2 159.3 cos u
2134.76 5 2159.3 cos u
134.76
cos u 5
159.3
134.76
u 5 cos21 a
b
159.3
8 32.2°
So the displacement is 15.93 units, W 32.2° N.
3.
z
u
C(–2, 0, 1)
B(–3, 2, 0)
y
A(0, 1, 0)
tan u 5
u 5 tan
100
800
21
x
100
a
b
800
u 8 7.1°
The velocity of the airplane relative to the ground is
about 806 km> h N 7.1° E.
2.
displacement
12
u
5
The angle between the two displacements is 135°.
The magnitude, m, and the angle, u, of the
displacement can be found using the cosine law.
Calculus and Vectors Solutions Manual
D(0, 2, –3)
4. a. (3, 22, 7)
l 5 magnitude
5 "32 1 (22)2 1 72
5 "9 1 4 1 49
5 "62
8 7.87
b. (29, 3, 14)
l 5 magnitude
5 "(29)2 1 32 1 142
5 !81 1 9 1 196
5 !286
8 16.91
7-1
c. (1, 1, 0)
l 5 magnitude
5 "12 1 12 1 02
5 "2
8 1.41
d. (2, 0, 29)
l 5 magnitude
5 "22 1 02 1 (29)2
5 "4 1 0 1 81
5 "85
8 9.22
5. a. A(x, y, 0)
In the xy-plane at the point (x, y).
b. B(x, 0, z)
In the xz-plane at the point (x, z).
c. C(0, y, z)
In the yz-plane at the point (y, z).
6. a. (26, 0) 1 7(1, 21)
>
>
>
>
5 (26i 1 0j ) 1 7(i 2 j )
>
>
>
>
5 (26i 1 0j ) 1 (7i 2 7j )
>
>
5 i 2 7j
b. (4, 21, 3) 2 (22, 1, 3)
>
>
>
>
>
>
5 (4i 2 j 1 3k ) 2 (22i 1 j 1 3k )
>
>
5 6i 2 2j
c. 2(21, 1, 3) 1 3(22, 3, 21)
>
>
>
>
>
>
5 2(2i 1 j 1 3k ) 1 3(22i 1 3j 2 k )
>
>
>
>
>
>
5 (22i 1 2j 1 6k ) 1 (26i 1 9j 2 3k )
>
>
>
5 28i 1 11j 1 3k
1
3
d. 2 (4, 26, 8) 1 (4, 26, 8)
2
2
>
>
>
>
3 >
1 >
5 2 (4i 2 6j 1 8k ) 1 (4i 2 6j 1 8k )
2 >
2>
>
>
>
>
5 (22i 1 3j 2 4k ) 1 (6i 2 9j 1 12k )
>
>
>
5 4i 2 6j 1 8k
>
>
7. a. a> 1 b >
>
>
>
5 (3i
1
2j
2
k
)
1
(22i
1
j
)
>
>
>
5 i 1 3j 2 k
>
>
b. a 2> b >
>
>
>
1
j
)
5 (3i> 1 2j> 2 k>) 2 (22i
>
>
5 (3i
1
2j
2
k
)
1
(2i
2
j
)
>
>
>
5 5i 1 j 2 k
>
>
c. 2a 2> 3b >
>
>
>
1 2j
2 k>) 2 3(22i
1> j )
5 2(3i
>
>
>
5 (6i> 1 4j
2 2k
) 1 (6i 2 3j )
>
>
5 12i 1 j 2 2k
7-2
7.1 Vectors as Forces, pp. 362–364
1. a. 10 N is a melon, 50 N is a chair, 100 N is a
computer
b. Answers will vary.
2. a.
10 N
20 N
30 N
b. 180°
3.
10 N
20 N
The forces should be placed in a line along the
same direction.
4. For three forces to be in equilibrium, they must
form a triangle, which is a planar figure.
5.
equilibrant
f2
resultant
f1
a. The resultant is equivalent in magnitude to the
hypotenuse, h, of the triangle with 5 and 12 as sides
12
and is directed northeast at an angle of sin21 h .
Thus, the resultant is "52 1 122 5 13 N at an angle
of sin21 12
13 5 N 22.6° E. The equlibrant is equal in
magnitude and opposite in direction of the resultant.
Thus, the equilibrant is 13 N at an angle of S 22.6° W.
b. The resultant is "92 1 122 5 15 N at an angle of
sin21 12
15 5 S 36.9° W. The equilibrant, then, is 15 N
at N 36.9° E.
6. For three forces to form equilibrium, they must
be able to form a triangle or a balanced line, so
a. Yes, since 3 1 4 . 7 these can form a triangle.
b. Yes, since 9 1 40 . 41 these can form a triangle.
c. No, since "5 1 6 , 9 these cannot form a
triangle.
d. Yes, since 9 1 10 5 19, placing the 9 N and 10 N
force in a line directly opposing the 19 N force
achieves equilibrium.
Chapter 7: Applications of Vectors
c. (1, 1, 0)
l 5 magnitude
5 "12 1 12 1 02
5 "2
8 1.41
d. (2, 0, 29)
l 5 magnitude
5 "22 1 02 1 (29)2
5 "4 1 0 1 81
5 "85
8 9.22
5. a. A(x, y, 0)
In the xy-plane at the point (x, y).
b. B(x, 0, z)
In the xz-plane at the point (x, z).
c. C(0, y, z)
In the yz-plane at the point (y, z).
6. a. (26, 0) 1 7(1, 21)
>
>
>
>
5 (26i 1 0j ) 1 7(i 2 j )
>
>
>
>
5 (26i 1 0j ) 1 (7i 2 7j )
>
>
5 i 2 7j
b. (4, 21, 3) 2 (22, 1, 3)
>
>
>
>
>
>
5 (4i 2 j 1 3k ) 2 (22i 1 j 1 3k )
>
>
5 6i 2 2j
c. 2(21, 1, 3) 1 3(22, 3, 21)
>
>
>
>
>
>
5 2(2i 1 j 1 3k ) 1 3(22i 1 3j 2 k )
>
>
>
>
>
>
5 (22i 1 2j 1 6k ) 1 (26i 1 9j 2 3k )
>
>
>
5 28i 1 11j 1 3k
1
3
d. 2 (4, 26, 8) 1 (4, 26, 8)
2
2
>
>
>
>
3 >
1 >
5 2 (4i 2 6j 1 8k ) 1 (4i 2 6j 1 8k )
2 >
2>
>
>
>
>
5 (22i 1 3j 2 4k ) 1 (6i 2 9j 1 12k )
>
>
>
5 4i 2 6j 1 8k
>
>
7. a. a> 1 b >
>
>
>
5 (3i
1
2j
2
k
)
1
(22i
1
j
)
>
>
>
5 i 1 3j 2 k
>
>
b. a 2> b >
>
>
>
1
j
)
5 (3i> 1 2j> 2 k>) 2 (22i
>
>
5 (3i
1
2j
2
k
)
1
(2i
2
j
)
>
>
>
5 5i 1 j 2 k
>
>
c. 2a 2> 3b >
>
>
>
1 2j
2 k>) 2 3(22i
1> j )
5 2(3i
>
>
>
5 (6i> 1 4j
2 2k
) 1 (6i 2 3j )
>
>
5 12i 1 j 2 2k
7-2
7.1 Vectors as Forces, pp. 362–364
1. a. 10 N is a melon, 50 N is a chair, 100 N is a
computer
b. Answers will vary.
2. a.
10 N
20 N
30 N
b. 180°
3.
10 N
20 N
The forces should be placed in a line along the
same direction.
4. For three forces to be in equilibrium, they must
form a triangle, which is a planar figure.
5.
equilibrant
f2
resultant
f1
a. The resultant is equivalent in magnitude to the
hypotenuse, h, of the triangle with 5 and 12 as sides
12
and is directed northeast at an angle of sin21 h .
Thus, the resultant is "52 1 122 5 13 N at an angle
of sin21 12
13 5 N 22.6° E. The equlibrant is equal in
magnitude and opposite in direction of the resultant.
Thus, the equilibrant is 13 N at an angle of S 22.6° W.
b. The resultant is "92 1 122 5 15 N at an angle of
sin21 12
15 5 S 36.9° W. The equilibrant, then, is 15 N
at N 36.9° E.
6. For three forces to form equilibrium, they must
be able to form a triangle or a balanced line, so
a. Yes, since 3 1 4 . 7 these can form a triangle.
b. Yes, since 9 1 40 . 41 these can form a triangle.
c. No, since "5 1 6 , 9 these cannot form a
triangle.
d. Yes, since 9 1 10 5 19, placing the 9 N and 10 N
force in a line directly opposing the 19 N force
achieves equilibrium.
Chapter 7: Applications of Vectors
7. Arms 90 cm apart will yield a resultant with a
smaller magnitude than at 30 cm apart. A resultant
with a smaller magnitude means less force to
counter your weight, hence a harder chin-up.
8. Using the cosine law, the resultant has a magnitude,
r, of
>
>
> >
r 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos 120°
Now we look at x1 and x2. We know
>
x1 5 @ f1 @ sin 15
r 5 "148
8 12.17 N
Using the sine law, the resultant’s angle, u, can be
found by
sin u
sin 120°
5
8
12.17
@ f2 @ (1.035) 5 10
1
5 62 1 82 2 2(6)(8)a2 b
2
5 36 1 64 1 48
5 148
sin u 5 8
"3
2
12.17
u 5 sin21 8
"3
2
12.17
8 34.7° from the 6 N force toward the 8 N
force. The equilibrant, then, would be 12.17 N at
180° 2 34.7° 5 145.3° from the 6 N force away
from the 8 N force.
9.
10 N
f1
f2
x2 5 @ f2 @ sin 75
x1 1 x2 5 10
>
So @ f1 @ sin 15 1 @ f2 @ sin 75 5 10
>
Substituting then solving for f2 yields
> cos 75
>
@ f2 @
sin 15 1 @ f2 @ sin 75 5 10
cos 15
> cos 75
@ f2 @ a
sin 15 1 sin 75b 5 10
cos 15
>
>
>
@ f2 @ 5 9.66 N
>
Now we solve for f1 :
>
@ f1 @ 5 @ f2 @
cos 75
cos 15
>
cos 75
@ f1 @ 5 (9.66)
cos 15
>
@ f1 @ 5 (9.66)(0.268)
>
>
@ f1 @ 5 2.59 N
So the force 15° from the 10 N force is 9.66 N and
the force perpendicular to it is 2.59 N.
10. The force of the block is
(10 kg)(9.8 N>kg) 5 98 N. The component of this
force parallel to the ramp is
(98) sin 30° 5 (98)A 12 B 5 49 N, directed down the
ramp. So the force preventing this block from
moving would be 49 N directed up the ramp.
11. a.
>
7N
13 N
>
f1 5 force 15° from the 10 N force
>
>
f2 5 force perpendicular to f1
>
x1 5 component of f1 parallel to the 10 N force
>
x2 5 component of f2 parallel to the 10 N force
>
>
We know that the components of f1 and f2
perpendicular to the 10 N force must be equal, so we
can write
>
>
@ f1 @ cos 15 5 @ f2 @ cos 75
>
> cos 75
@ f1 @ 5 @ f2 @
cos 15
Calculus and Vectors Solutions Manual
8N
b. Using the cosine law for the angle, u, we have
132 5 82 1 72 2 2(8)(7) cos u
169 5 64 1 49 2 112 cos u
56 5 2112 cos u
256
cos u 5
112
21
u 5 cos21
2
5 120
7-3
This is the angle between the vectors when placed
head to tail. So the angle between the vectors when
placed tail to tail is 180° 2 120° 5 60°.
12. The 10 N force and the 5 N force result in a 5 N
force east. The 9 N force and the 14 N force result
in a 5 N force south. The resultant of these is now
equivalent to the hypotenuse of the right triangle
with 5 N as both bases and is directed 45° south of
east. So the resultant is "52 1 52 5 "50 8 7.1 N
45° south of east.
13.
of an equilateral triangle are 60°, the resultant will be
at a 60° angle with the other two vectors.
b. Since the equilibrant is directed opposite the
resultant, the angle between the equilibrant and the
other two vectors
>
>is 180° 2 60° 5 120°.
15. Since f1 and f2 act opposite one another,
they
>
>
net a 10 N force directed west. Since f3 and f4 act
opposite one another, they net a 10 N force directed
45° north of east.
So using the cosine law to find
>
the resultant, fr,
>
@ fr @ 2 5 102 1 102 2 2(10)(10) cos 45°
5 200 2 200 cos 45°
5 200 2 200 a
f1 = 24 N
@ fr @ 5
>
resultant = 25 N
equlibrant
f2
a. Using the Pythagorean theorem,
>
>
@ f1 @ 2 1 @ f2 @ 2 5 252
>
>
@ f2 @ 2 5 252 2 @ f1 @ 2
5 252 2 242
5 49
>
@ f2 @ 5 7
>
b. The angle, u, between f1 and the resultant is
given by
>
@ f2 @
sin u 5
25
7
sin u 5
25
7
u 5 sin21
25
8 16.3°
>
So the angle between f1 and the equilibrant is
180° 2 16.3° 5 163.7°.
14. a.
60°
1N
1N
60°
60°
1N
For these three equal forces to be in equilibrium, they
must form an equilateral triangle. Since the resultant
will lie along one of these lines, and since all angles
7-4
Å
"2 b
2
200 2 200a
"2
b
2
8 7.65 N
Since our net forces are equal at 10 N, the angle of
the resultant is directed halfway
between
the two, or
>
>
1
at 2 (135°) 5 67.5° from f2 toward f3.
16.
T2
T1
20 kg
Let T1 be the tension in the 30° rope and T2 be the
tension in the 45° rope.
Since this system is in equilibrium, we know that
the horizontal components of T1 and T2 are equal
and opposite and the vertical components add to be
opposite the action of the mass. Also, the force
produced by the mass is (20 kg)(9.8 N>kg) 5 196 N.
So we have a system of two equations: the first,
(T1 ) cos 30° 5 (T2 ) cos 45° represents the balance
of the horizontal components, and the second,
(T1 ) sin 30° 1 (T2 ) sin 45° 5 196 represents the
balance of the vertical components with the mass.
So solving this system of two equations with two
variable gives the desired tensions.
T1 cos 30° 5 T2 cos 45°
cos 45°
T1 5 T2
cos 30°
T1 sin 30° 1 T2 sin 45° 5 196
cos 45°
b sin 30° 1 T2 sin 45° 5 196
aT2
cos 30°
cos 45°
T2 aa
b sin 30° 1 sin 45°b 5 196
cos 30°
Chapter 7: Applications of Vectors
T2 (1.12) 5 196
T2 8 175.73 N
cos 45°
T1 5 (175.73)
cos 30°
8 143.48 N
Thus the tension in the 45° rope is 175.73 N and the
tension in the 30° rope is 143.48 N.
17.
40 cm
24 cm
Thus the tension in the 24 cm string is 39.2 N and
the tension in the 32 cm string is 29.4 N.
18.
resultant
x
35°
u
2x
32 cm
5 kg
First, use the Cosine Law to find the angles the
strings make at the point of suspension. Let u1 be the
angle made by the 32 cm string and u2 be the angle
made by the 24 cm string.
242 5 322 1 402 2 2(32)(40) cos u1
22048 5 22560 cos u1
2048
u1 5 cos21
2560
8 36.9°
322 5 242 1 402 2 2(24)(40) cos u2
21152 5 21920 cos u2
1152
u2 5 cos21
1920
8 53.1°
A keen eye could have recognized this triangle as a
3-4-5 right triangle and simply used the Pythagorean
theorem as well. Now we set up the same system of
equations as in problem 16, with T1 being the tension
in the 32 cm string and T2 being the tension in the
24 cm string, and the force of the mass being
(5 kg)(9.8 N>kg) 5 49 N.
T1 cos 36.9° 5 T2 cos 53.1°
cos 53.1°
T1 5 T2
cos 36.9°
T1 sin 36.9° 1 T2 sin 53.1° 5 49
cos 53.1°
aT2
b sin 36.9° 1 T2 sin 53.1° 5 49
cos 36.9°
cos 53.1°
T2 aa
b sin 36.9° 1 sin 53.1°b 5 49
cos 36.9°
T2 (1.25) 5 49
T2 8 39.2 N
cos 53.1°
T1 5 (39.2)
cos 36.9°
8 29.4 N
Calculus and Vectors Solutions Manual
N
W
E
S
(Port means left and starboard means right.) We are
looking for the resultant of these two force vectors
that are 35° apart. We don’t know the exact value
of the force, so we will call it x. So the small tug
is pulling with a force of x and the large tug is
pulling with a force of 2x. To find the magnitude
of the resultant, r, in terms of x, we use the cosine
law.
r 2 5 x 2 1 (2x)2 2 2(x)(2x) cos 145°
5 x 2 1 4x 2 2 4x 2 cos 145°
8 5x 2 2 4x 2 (20.8192)
8 5x 2 1 3.2768x 2
8 8.2768x 2
r 8 "8.2768x 2
8 2.8769x
Now we use the cosine law again to find the angle,
u, made by the resultant.
x 2 5 r 2 1 (2x)2 2 2(2.8769x)(2x) cos u
x 2 5 8.2768x 2 1 4x 2 2 11.5076x 2 cos u
x 2 5 12.2768x 2 2 11.5076x 2 cos u
211.2768x 2 5 211.5076x 2 cos u
11.2768
cos u 5
11.5076
11.2768
u 5 cos21 a
b
11.5076
8 11.5° from the large tug toward the
small tug, for a net of 8.5° to the starboard side.
7-5
19.
u10N 5 cos21 a
5N
8N
10N
a. First we will find the resultant of the 5 N and
8 N forces. Use the Pythagorean theorem to find the
magnitude, m.
m 2 5 52 1 82
5 25 1 64
5 89
m 5 "89 8 9.4
Next we use the Pythagorean theorem again to find
the magnitude, M, of the resultant of this net force
and the 10 N force.
M 2 5 m 2 1 102
5 89 1 100
5 189
M 5 "189 8 13.75
Since the equilibrant is equal in magnitude to the
resultant, we have the magnitude of the equilibrant
equal to approximately 13.75 N.
b. To find each angle, use the definition of cosine
with respect each force as a leg and the resultant as
the hypotenuse. Let u5N be the angle from the
5 N force to the resultant, u8N be the angle from the
8 N force to the resultant, and u10N be the angle
from the 10 N force to the resultant.
Let the sign of the resultant be negative, since it is
in a direction away from the head of each of the
given forces.
5
cos u5N 5
213.75
5
u5N 5 cos21 a
b
213.75
8 111.3°
8
cos u8N 5
213.75
8
u8N 5 cos21 a
b
213.75
8 125.6°
10
cos u10N 5
213.75
7-6
10
b
213.75
8 136.7°
20. We know that the resultant of these two forces is
equal in magnitude and angle to the> diagonal
line
>
of the parallelogram formed with f1 and f2 as legs
>
>
and has diagonal length @ f1 1 f2 @ . We also know
from the cosine law that
>
>
>
>
>
>
@ f1 1 f2 @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos f
where f is the supplement to u in our parallelogram.
Since we know f 5 180 2 u, then
cos f 5 cos (180 2 u) 5 2cos u.
Thus
we> have >
>
>
>
>
@ f1 1 f2 @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos f
>
>
>
>
5 @ f1 @ 2 1 @ f2 @ 2 1 2 @ f1 @ @ f2 @ cos u
@ f1 1 f2 @ 5 " @ f1 @ 2 1 @ f2 @ 2 1 2 @ f1 @ @ f2 @ cos u
>
>
>
>
>
>
7.2 Velocity, pp. 367–370
1. a. Both the woman and the train’s velocities are
in the same direction, so we add them.
80 km>h 1 4 km>h 5 84 km>h
b. The woman’s velocity is directed opposite that of
train, so we subtract her velocity from the train’s.
80 km>h 2 4 km>h 5 76 km>h. The resultant is in
the same direction as the train’s movement.
2. a. The velocity of the wind is directed opposite that
of the airplane, so we subtract the wind’s velocity
from the airplane’s.
600 km>h 2 100 km>h 5 500 km>h north.
b. Both the wind and the airplane’s velocities are in
the same direction, so we add them.
600 km>h 1 100 km>h 5 700 km>h north.
3. We use the Pythagorean theorem to find the
magnitude, m, of the resultant velocity and we use
the definition of sine to find the angle, u, made.
m 2 5 3002 1 502
5 90 000 1 2500
5 92 500
m 5 "92 500
8 304.14 km>h
50
tan u 5
300
50
u 5 tan21
300
8 9.5°. The resultant is 304.14 km> h, W 9.5° S.
Chapter 7: Applications of Vectors
19.
u10N 5 cos21 a
5N
8N
10N
a. First we will find the resultant of the 5 N and
8 N forces. Use the Pythagorean theorem to find the
magnitude, m.
m 2 5 52 1 82
5 25 1 64
5 89
m 5 "89 8 9.4
Next we use the Pythagorean theorem again to find
the magnitude, M, of the resultant of this net force
and the 10 N force.
M 2 5 m 2 1 102
5 89 1 100
5 189
M 5 "189 8 13.75
Since the equilibrant is equal in magnitude to the
resultant, we have the magnitude of the equilibrant
equal to approximately 13.75 N.
b. To find each angle, use the definition of cosine
with respect each force as a leg and the resultant as
the hypotenuse. Let u5N be the angle from the
5 N force to the resultant, u8N be the angle from the
8 N force to the resultant, and u10N be the angle
from the 10 N force to the resultant.
Let the sign of the resultant be negative, since it is
in a direction away from the head of each of the
given forces.
5
cos u5N 5
213.75
5
u5N 5 cos21 a
b
213.75
8 111.3°
8
cos u8N 5
213.75
8
u8N 5 cos21 a
b
213.75
8 125.6°
10
cos u10N 5
213.75
7-6
10
b
213.75
8 136.7°
20. We know that the resultant of these two forces is
equal in magnitude and angle to the> diagonal
line
>
of the parallelogram formed with f1 and f2 as legs
>
>
and has diagonal length @ f1 1 f2 @ . We also know
from the cosine law that
>
>
>
>
>
>
@ f1 1 f2 @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos f
where f is the supplement to u in our parallelogram.
Since we know f 5 180 2 u, then
cos f 5 cos (180 2 u) 5 2cos u.
Thus
we> have >
>
>
>
>
@ f1 1 f2 @ 2 5 @ f1 @ 2 1 @ f2 @ 2 2 2 @ f1 @ @ f2 @ cos f
>
>
>
>
5 @ f1 @ 2 1 @ f2 @ 2 1 2 @ f1 @ @ f2 @ cos u
@ f1 1 f2 @ 5 " @ f1 @ 2 1 @ f2 @ 2 1 2 @ f1 @ @ f2 @ cos u
>
>
>
>
>
>
7.2 Velocity, pp. 367–370
1. a. Both the woman and the train’s velocities are
in the same direction, so we add them.
80 km>h 1 4 km>h 5 84 km>h
b. The woman’s velocity is directed opposite that of
train, so we subtract her velocity from the train’s.
80 km>h 2 4 km>h 5 76 km>h. The resultant is in
the same direction as the train’s movement.
2. a. The velocity of the wind is directed opposite that
of the airplane, so we subtract the wind’s velocity
from the airplane’s.
600 km>h 2 100 km>h 5 500 km>h north.
b. Both the wind and the airplane’s velocities are in
the same direction, so we add them.
600 km>h 1 100 km>h 5 700 km>h north.
3. We use the Pythagorean theorem to find the
magnitude, m, of the resultant velocity and we use
the definition of sine to find the angle, u, made.
m 2 5 3002 1 502
5 90 000 1 2500
5 92 500
m 5 "92 500
8 304.14 km>h
50
tan u 5
300
50
u 5 tan21
300
8 9.5°. The resultant is 304.14 km> h, W 9.5° S.
Chapter 7: Applications of Vectors
4. Adam must swim at an angle, u, upstream so as
to counter the 1 km> h velocity of the stream. This is
equivalent to Adam swimming along the hypotenuse
of a right traingle with 1 km> h leg and a 2 km> h
hypotenuse. So the angle is found using the definition
of cosine.
1
cos u 5
2
1
u 5 cos21
2
5 60° upstream
5. a. 2 m> s forward
b. 20 m> s 1 2 m> s 5 22 m> s in the direction of the car
6. Since the two velocities are at right angles we
can use the Pythagorean theorem to find the
magnitude, m, of the resultant velocity and we use
the definition of sine to find the angle, u, made.
m 2 5 122 1 52
5 144 1 25
5 169
m 5 "169
5 13 m>s
5
sin u 5
13
5
u 5 sin21
13
8 22.6° from the direction of the boat toward
the direction of the current. This results in a net of
22.6° 1 15° 5 37.6°, or N 37.6° W.
7. a. First we find the components of the resultant
directed north and directed west. The component
directed north is the velocity of the airplane, 800,
minus 100 sin 45°, since the wind forms a 45°
angle south of west. The western component of
the resultant is simply 100 cos 45°. So we use the
Pythagorean theorem to find the magnitude, m, of
the resultant and the definition of sine to find the
angle, u, of the resultant.
m 2 5 (800 2 100 sin 45°)2 1 (100 cos 45°)2
8 (729.29)2 1 (71.71)2
8 536 863.8082
m 8 732.71 km>h
Use the sine law to determine the direction.
sin u
sin 45°
5
100
732.71
u 8 5.5°
The direction is N 5.5° W.
Calculus and Vectors Solutions Manual
b. The airplane is travelling at approximately
732.71 km> h, so in 1 hour the airplane will travel
about 732.71 km.
8. a. First we find the velocity of the airplane. We
use the Pythagorean theorem to find the magnitude,
m, of the resultant.
m 2 5 4502 1 1002
5 202 500 1 10 000
5 212 500
m 5 "212 500
8 461 km>h
So in 3 hours, the airplane will travel about
(461 km>h)(3 h) 5 1383 km.
b. To find the angle, u, the airplane travels, we use
the definition of sine.
100
sin u 5
461
100
u 5 sin21
461
8 12.5° east of north.
9. a. To find the angle, u, at which to fly is the
equivalent of the angle of a right triangle with 44 as
the opposite leg and 244 as the hypotenuse. So we
use the definition of sine to find this angle.
44
sin u 5
244
44
u 5 sin21
244
8 10.4° south of west.
b. By the Pythagorean Theorem, the resultant ground
speed of the airplane is "(2442 2 442 ) 5 240 km>h.
Since time 5 distance>rate, the duration of the
flight is simply (480 km)> (240 km>h) 5 2 h.
10. a. Since Judy is swimming perpendicular to
the flow of the river, her resultant velocity is simply
the hypotenuse of a right triangle with 3 and 4 as
bases, which is a 3-4-5 right triangle. Thus, Judy’s
resultant velocity is 5 km> h. The direction is
determined by tan u 5 43. u 8 53.1° downstream
b. Judy’s distance traveled down the river would be the
“4” leg of the 3-4-5 triangle formed by the vectors, but
scaled down so that 1m (the width of the river) is
equivalent to the “3” leg. So her distance traveled is
4
3 8 1.33 km. This makes her about 0.67 km from
Helen’s cottage.
c. While in the river, Judy is swimming at
5 km> h for a distance of 53 km. Since
time 5 distance>rate, her time taken is
5
3 km
5 1 hours 5 20 minutes.
5 km>h 3
7-7
11.
h
m/
v
5k
20 30°
u
212 km/h
a. and b. Here, 205 km> h directed 30° north of east
is the resultant of 212 km> h directed east, and the
wind speed, v, directed at some angle. This problem
is more easily approached finding the wind speed,
v, first. So we will do that using the cosine law.
v 2 5 2052 1 2122 2 2(205)(212) cos 30°
5 42 025 1 44 944 2 86 920 cos 30°
5 86 969 2 75 275
5 11 694
v 5 "11 694
8 108 km>h
Now to find the wind’s direction, we simply find the
angle supplementary to the lesser angle, u, formed
by the parallelogram of these three velocities. We
can use the sine law for this.
sin u
sin 30°
5
205
108
sin 30°
sin u 5 205a
b
108
sin 30°
u 5 sin21 a205a
bb
108
8 71.6°
Thus, the direction of v is the angle supplementary
to u in the parallelogram:
180° 2 71.6° 5 108.4° 5 18.4° west of north.
12.
4
5
Since her swimming speed is a maximum of 4 km> h,
this is her maximum resultant magnitude, which is
also the hypotenuse of the triangle formed by her and
the river’s velocity vector. Since one of these legs is
5 km> h, we have a triangle with a leg larger than its
hypotenuse, which is impossible.
13. a. First we need to find Mary’s resultant
velocity, v. Since this resultant is the diagonal of the
parallelogram formed by hers and the river’s
velocity, we can use the cosine law with the angle, u,
of the parallelogram adjacent 30°.
7-8
v 2 5 32 1 42 2 2(3)(4) cos 150°
5 9 1 16 2 24 cos 150°
5 25 1 20.8
5 45.8
v 5 "45.8
8 6.8 m>s
So in 10 seconds, Mary travels about
(6.8 m>s)(10 s) 5 68 m.
b. Since Mary is travelling at 3 m>s at an angle of 30°,
to find the component of her velocity, v, perpendicular
to the current, we use the definition of sine.
v 5 3 sin 30
1
5 3a b
2
5 1.5 m>s perpendicular to the current.
So since time 5 distance>rate, the time taken is
(150 m)> (1.5 m>s) 5 100 s.
14. a. So we have a 5.5 m> s vector and a 4 m> s
vector with a resultant vector that is directed 45°
south of west. Letting u be the angle between the
4 km> h vector and the resultant, we can construct
a parallelogram using these three vectors and a
subsequent triangle with u opposite the 5.5 m> s
vector and 45° opposite the 4 m> s vector. We now
use the sine law to find u.
sin u
sin 45°
5
5.5
4
sin 45°
sin u 5 5.5a
b
4
sin 45°
u 5 sin21 a5.5a
bb
4
8 76.5° from the resultant.
Since the resultant is 45° west of south, Dave’s
direction is 76.5° 1 45° 5 121.5° west of south,
which is equivalent to about 180° 2 121.5° 5 58.5°
upstream.
b. First, we find the magnitude, m, of Dave’s 4 m> s
velocity in the direction perpendicular to the river.
This is done using the definition of sine.
m 5 4 sin 58.5°
8 3.41 m>s perpendicular to the river.
Since time is distance>rate, we have
(200 m)> (3.41 m>s) 8 58.6 s.
15. Let b represent the speed of the steamboat and c
represent the speed of the current. On the way
downstream, the effective speed is b 1 c, and
upstream is b 2 c. The distance upstream and
downstream is the same, so 5(b 1 c) 5 7(b 2 c).
So, b 5 6c. This means that the speed of the boat
is 6 times the speed of the current. So, (6c 1 c) ? 5
Chapter 7: Applications of Vectors
or 35c is the distance. This means that it would take
a raft 35 hours moving with the speed of the current
to get from A to B.
7.3 The Dot Product of Two Geometric
Vectors, pp. 377–378
> >
> >
>
1. a ? b 5 0 a 0 @ b @ cos u 5 0. This means 0 a 0 5 0,
>
or @ b @ 5 0, or cos u 5 0. To be guaranteed that the
two vectors are perpendicular, the vectors must be
nonzero.
> >
2. a ? b is a scalar, and
a dot product is only defined
> >
>
for vectors, so (a ? b ) ? c is meaningless.
> >
> >
>
>
3. Answers
may vary. Let a 5 i, b 5 j, c 5 2i.
>
>
>
>
>
>
a ? b 5 0, b ? c 5 0, but a 5 2c .
>
>
>
>
>
>
>
>
4. a ? b 5 b ? a 5 b ? c because c 5 a
>
>
>
>
5. Since a and b are unit vectors, 0 a 0 5 @ b @ 5 1 and
since they are pointing in opposite directions
then
> >
u 5 180° so cos u 5 21. Therefore a ? b 5 21.
> >
> >
6. a. p ? q 5 0 p 0 0 q 0cos u
5 (4)(8) cos (60°)
5 (32)(.5)
5 16
> >
> >
b. x ? y 5 0 x 0 0 y 0cos u
5 (2)(4) cos (150°)
"3
5 (8)a2
b
2
8 26.93
> >
> >
c. a ? b 5 0 a 0 @ b @ cos u
5 (0)(8) cos (100°)
50
> >
> >
d. p ? q 5 0 p 0 0 q 0cos u
5 (1)(1) cos (180°)
5 (1)(21)
5 21
> >
> >
e. m ? n 5 0 m 0 0 n 0cos u
5 (2)(5) cos (90°)
5 (10)(0)
50
> >
> >
f. u ? v 5 0 u 0 0 v 0cos u
5 (4)(8) cos 145°
8 226.2
> >
> >
7. a. x ? y 5 0 x 0 0 y 0cos u
12"3 5 (8)(3) cos u
"3
5 cos u
2
u 5 30°
Calculus and Vectors Solutions Manual
> >
> >
b. m ? n 5 0 m 0 0 n 0cos u
(6) 5 (6)(6) cos u
1
5 cos u
6
u 8 80°
> >
> >
c. p ? q 5 0 p 0 0 q 0cos u
3 5 (5)(1) cos u
3
5 cos u
5
u 8 53°
> >
> >
d. p ? q 5 0 p 0 0 q 0cos u
23 5 (5)(1) cos u
3
2 5 cos u
5
u 8 127°>
> >
>
e. a ? b 5 0 a 0 @ b @ cos u
10.5 5 (7)(3) cos u
1
5 cos u
2
u 5 60°
> >
> >
f. u ? v 5 0 u 0 0 v 0cos u
250 5 (10)(10) cos u
1
2 5 cos u
2
u 5 120°>
> >
>
8. a ? b 5 0 a 0 @ b @ cos u
5 (7.5)(6) cos (180° 2 120°)
1
5 (45)a b
2
5 22.5
Note: u is the angle between the two vectors when
they are tail to> tail, so u 2> 120°.
>
>
>
>
>
>
9. a. (a 1 5b ) ? (2a 2 3b ) 5 a ? 2a 2 a ? 3b
>
>
>
>
1 5b ? 2a 2> 5b ? 3b
>2
5 2 0 a 0 2 15 @ b @ 2
> >
> >
2 3a ? b 1 10a
?b
>
>
2
@
@
5 2 0 a 0 2 2 15
b
> >
1 7a ? b
>
>
>
>
>
>
>
b. 3x ? (x 2 3y ) 2 (x 2 3y ) ? (23x 1 y )
>
>
>
>
> >
>
>
5 3 0 x 0 2 2 3x ? 3y 1 3 0 x 0 2 2 x ? y 2 (23y ? 23x )
>2
1 30 y 0
>
> >
> >
> >
>
5 6 0 x 0 2 2 9x ? y 2 x ? y 2 9x ? y 1 3 0 y 0 2
>2
> >
>2
5 6 0 x 0 2 19x ? y 1 30 y 0
>
10. @ 0 @ 5 0 so the dot product of any vector with
>
0 is 0.
7-9
or 35c is the distance. This means that it would take
a raft 35 hours moving with the speed of the current
to get from A to B.
7.3 The Dot Product of Two Geometric
Vectors, pp. 377–378
> >
> >
>
1. a ? b 5 0 a 0 @ b @ cos u 5 0. This means 0 a 0 5 0,
>
or @ b @ 5 0, or cos u 5 0. To be guaranteed that the
two vectors are perpendicular, the vectors must be
nonzero.
> >
2. a ? b is a scalar, and
a dot product is only defined
> >
>
for vectors, so (a ? b ) ? c is meaningless.
> >
> >
>
>
3. Answers
may vary. Let a 5 i, b 5 j, c 5 2i.
>
>
>
>
>
>
a ? b 5 0, b ? c 5 0, but a 5 2c .
>
>
>
>
>
>
>
>
4. a ? b 5 b ? a 5 b ? c because c 5 a
>
>
>
>
5. Since a and b are unit vectors, 0 a 0 5 @ b @ 5 1 and
since they are pointing in opposite directions
then
> >
u 5 180° so cos u 5 21. Therefore a ? b 5 21.
> >
> >
6. a. p ? q 5 0 p 0 0 q 0cos u
5 (4)(8) cos (60°)
5 (32)(.5)
5 16
> >
> >
b. x ? y 5 0 x 0 0 y 0cos u
5 (2)(4) cos (150°)
"3
5 (8)a2
b
2
8 26.93
> >
> >
c. a ? b 5 0 a 0 @ b @ cos u
5 (0)(8) cos (100°)
50
> >
> >
d. p ? q 5 0 p 0 0 q 0cos u
5 (1)(1) cos (180°)
5 (1)(21)
5 21
> >
> >
e. m ? n 5 0 m 0 0 n 0cos u
5 (2)(5) cos (90°)
5 (10)(0)
50
> >
> >
f. u ? v 5 0 u 0 0 v 0cos u
5 (4)(8) cos 145°
8 226.2
> >
> >
7. a. x ? y 5 0 x 0 0 y 0cos u
12"3 5 (8)(3) cos u
"3
5 cos u
2
u 5 30°
Calculus and Vectors Solutions Manual
> >
> >
b. m ? n 5 0 m 0 0 n 0cos u
(6) 5 (6)(6) cos u
1
5 cos u
6
u 8 80°
> >
> >
c. p ? q 5 0 p 0 0 q 0cos u
3 5 (5)(1) cos u
3
5 cos u
5
u 8 53°
> >
> >
d. p ? q 5 0 p 0 0 q 0cos u
23 5 (5)(1) cos u
3
2 5 cos u
5
u 8 127°>
> >
>
e. a ? b 5 0 a 0 @ b @ cos u
10.5 5 (7)(3) cos u
1
5 cos u
2
u 5 60°
> >
> >
f. u ? v 5 0 u 0 0 v 0cos u
250 5 (10)(10) cos u
1
2 5 cos u
2
u 5 120°>
> >
>
8. a ? b 5 0 a 0 @ b @ cos u
5 (7.5)(6) cos (180° 2 120°)
1
5 (45)a b
2
5 22.5
Note: u is the angle between the two vectors when
they are tail to> tail, so u 2> 120°.
>
>
>
>
>
>
9. a. (a 1 5b ) ? (2a 2 3b ) 5 a ? 2a 2 a ? 3b
>
>
>
>
1 5b ? 2a 2> 5b ? 3b
>2
5 2 0 a 0 2 15 @ b @ 2
> >
> >
2 3a ? b 1 10a
?b
>
>
2
@
@
5 2 0 a 0 2 2 15
b
> >
1 7a ? b
>
>
>
>
>
>
>
b. 3x ? (x 2 3y ) 2 (x 2 3y ) ? (23x 1 y )
>
>
>
>
> >
>
>
5 3 0 x 0 2 2 3x ? 3y 1 3 0 x 0 2 2 x ? y 2 (23y ? 23x )
>2
1 30 y 0
>
> >
> >
> >
>
5 6 0 x 0 2 2 9x ? y 2 x ? y 2 9x ? y 1 3 0 y 0 2
>2
> >
>2
5 6 0 x 0 2 19x ? y 1 30 y 0
>
10. @ 0 @ 5 0 so the dot product of any vector with
>
0 is 0.
7-9
>
>
> >
>
>
>
>
11. (a 2 5b ) ? (a 2 b ) 5 @ a 2 5b @ @ a 2 b @ cos (90°)
>
>
>
>
>
>
0 a 0 2 2 a ? b 2 5b ? a 1 5 @ b @ 2 5 0
>
>
> >
0 a 0 2 1 5 @ b @ 2 5 6a ? b
>
> > 1 >
a ? b 5 ( 0 a 0 2 1 5 @ b @ 2)
6
5
1
>
>
>
>
> >
> >
12. a. (a 1 b ) ? (a 1 b ) 5 a ? >a 1 a ? b
>
>
>
1 b ? a 1 b> ? b
>
>
> >
5 0 a 02 1 a ? b 1 a ? b
>
1 @b@ 2
>
>2
> >
@
@2
1
2a
1
b
5
0
a
0
?
b
>
>
>
>
>
> >
> >
>
b. (a 1 b ) ? (a 2 b ) 5 a ? a 2 a ? b 1 b ? a
>
>
2b?b
>
>
> >
> >
5 0 a 02 2 a ? b 1 a ? b 2 @b@ 2
>
>
5 0 a 02 2 @b@ 2
>
> >
13. a. 0 a 0 2 5 a ?> a
>
>
>
5 (b 1 c ) ? (b 1 c )
>
>
>
>
5 @ b @ 2 1 2b ? c 1 0 c 0 2
>
>
>
>
b. b ? c 5 @ b @ 0 c 0cos (90°) 5 0
>
>
>
Therefore 0 a 0 2 5 @ b @ 2 1 0 c 0 2.
This is just what the Pythagorean theorem says,
>
>
where b and c are the legs of the right triangle.
>
>
>
>
>
>
14. (u 1 v 1 w ) ? (u 1 v 1 w )
> >
> >
>
>
> >
> >
5u?u1u?v1u?w1v?u1v?v
>
>
> >
> >
>
>
1v?w1w?u1w?v1w?w
>2
>2
>2
> >
5 0 u 0 1 0 v 0 1 0 w 0 1 2 0 u 0 0 v 0cos (90°)
> >
> >
1 2 0 u 0 0 w 0cos (90°) 1 2 0 v 0 0 w 0cos (90°)
5 (1)2 1 (2)2 1 (3)2
5 14
>
>
>
>
15. 0 u 1 v 0 2 1 0 u 2 v 0 2
>
>
>
>
>
>
>
>
5 (u 1 v ) ? (u 1 v ) 1 (u 2 v ) ? (u 2 v )
>
> >
>
>
> >
>
5 0 u 0 2 1 2u ? v 1 0 v 0 2 1 0 u 0 2 2 2u ? v 1 0 v 0 2
>
>
5 20 u 0 2 1 2 0 v 0 2
>
>
>
>
>
16. (a 1 b ) ? (a 1 b 1 c )
>
>
>
>
> >
> >
>
>
5 0 a 02 1 a ? b 1 a ? c 1 b ? a 1 @b@ 2 1 b ? c
> >
> >
5 1 1 20 a 0 @ b @ cos (60°) 1 0 a 0 0 c 0cos (60°) 1 1
> >
1 @ b @ 0 c 0cos (120°)
1
1
1
5 2 1 2a b 1 2
2
2
2
53
>
>
>
>
17. a 1 b >1 c 5 0 >
>
>
>
>
>
>
a ? (a 1 b 1 c ) 1 b ? (a 1 b 1 c )
>
>
>
>
1 c ? (a 1 b 1 c ) 5 0
>
>
>
>
>
>
> >
>
>
0 a 02 1 a ? b 1 a ? c 1 b ? a 1 @b@ 2 1 b ? c
>
> >
>
>
1 c ? a 1 c ? b 1 0 c 02 5 0
7-10
>
> >
> >
>
1 1 4 1 9 1 2(a ? b 1 a ? c 1 b ? c ) 5 0
>
>
>
> >
>
2(a ? b 1 a ? c 1 b ? c ) 5 214
>
>
>
> >
>
a ? b 1 a ? c 1 b ? c 5 27
>
>
>
18. d> 5 b> 2 c
>
b 5 d 1> c
> >
> >
>
c ? a 5 ((b ? a ) a ) ? a
>
>
> >
> > >
>
c ? a 5 (b ? a )(a ? a ) because b ? a is a scalar
>
> >
> >
c ? a 5 (b ? a ) 0 a 0 2
>
> >
>
>
>
c ? a 5 (d 1 c ) ? a because 0 a 0 5 1
>
> >
>
> >
c> ? a 5 d ? a 1 c ? a
>
d?a50
7.4 The Dot Product for Algebraic
Vectors, pp. 385–387
> >
a?b50
1.
(21)b1 1 b2 5 0
b2 5 b1
Any vector of the form (c, c) is perpendicular
>
to a . Therefore there are infinitely many vectors
>
perpendicular to a . Answers may vary. For example:
(1, 1), (2, > 2), (3, 3).
>
2. a. a ? b 5 (22)(1) 1 (1)(2)
50
u
5 90°
> >
b. a ? b 5 (2)(4) 1 (3)(3) 1 (21)(217)
5 8 1 9 1 17
5 34 . 0
cos u . 0
u is acute
> >
c. a ? b 5 (1)(3) 1 (22)(22) 1 (5)(22)
5 3 1 4 2 10
5 23 , 0
cos u , 0
u is obtuse
3. Any vector in the xy-plane
is of the form
>
>
a 5 (a1, a2, 0). Let b 5 (0, 0, 1).
> >
a ? b 5 (0)(a1 ) 1 (0)(a2 ) 1 (0)(1)
50
Therefore (0, 0, 1) is perpendicular to every vector
in the xy-plane.
Any vector in the xz-plane
is of the form
>
>
c 5 (c1, 0, c3 ). Let d 5 (0, 1, 0).
> >
c ? d 5 (0)(c1 ) 1 (0)(1) 1 (0)(c3 )
50
Therefore (0, 1, 0) is perpendicular to every vector
in the xz-plane.
Chapter 7: Applications of Vectors
>
>
> >
>
>
>
>
11. (a 2 5b ) ? (a 2 b ) 5 @ a 2 5b @ @ a 2 b @ cos (90°)
>
>
>
>
>
>
0 a 0 2 2 a ? b 2 5b ? a 1 5 @ b @ 2 5 0
>
>
> >
0 a 0 2 1 5 @ b @ 2 5 6a ? b
>
> > 1 >
a ? b 5 ( 0 a 0 2 1 5 @ b @ 2)
6
5
1
>
>
>
>
> >
> >
12. a. (a 1 b ) ? (a 1 b ) 5 a ? >a 1 a ? b
>
>
>
1 b ? a 1 b> ? b
>
>
> >
5 0 a 02 1 a ? b 1 a ? b
>
1 @b@ 2
>
>2
> >
@
@2
1
2a
1
b
5
0
a
0
?
b
>
>
>
>
>
> >
> >
>
b. (a 1 b ) ? (a 2 b ) 5 a ? a 2 a ? b 1 b ? a
>
>
2b?b
>
>
> >
> >
5 0 a 02 2 a ? b 1 a ? b 2 @b@ 2
>
>
5 0 a 02 2 @b@ 2
>
> >
13. a. 0 a 0 2 5 a ?> a
>
>
>
5 (b 1 c ) ? (b 1 c )
>
>
>
>
5 @ b @ 2 1 2b ? c 1 0 c 0 2
>
>
>
>
b. b ? c 5 @ b @ 0 c 0cos (90°) 5 0
>
>
>
Therefore 0 a 0 2 5 @ b @ 2 1 0 c 0 2.
This is just what the Pythagorean theorem says,
>
>
where b and c are the legs of the right triangle.
>
>
>
>
>
>
14. (u 1 v 1 w ) ? (u 1 v 1 w )
> >
> >
>
>
> >
> >
5u?u1u?v1u?w1v?u1v?v
>
>
> >
> >
>
>
1v?w1w?u1w?v1w?w
>2
>2
>2
> >
5 0 u 0 1 0 v 0 1 0 w 0 1 2 0 u 0 0 v 0cos (90°)
> >
> >
1 2 0 u 0 0 w 0cos (90°) 1 2 0 v 0 0 w 0cos (90°)
5 (1)2 1 (2)2 1 (3)2
5 14
>
>
>
>
15. 0 u 1 v 0 2 1 0 u 2 v 0 2
>
>
>
>
>
>
>
>
5 (u 1 v ) ? (u 1 v ) 1 (u 2 v ) ? (u 2 v )
>
> >
>
>
> >
>
5 0 u 0 2 1 2u ? v 1 0 v 0 2 1 0 u 0 2 2 2u ? v 1 0 v 0 2
>
>
5 20 u 0 2 1 2 0 v 0 2
>
>
>
>
>
16. (a 1 b ) ? (a 1 b 1 c )
>
>
>
>
> >
> >
>
>
5 0 a 02 1 a ? b 1 a ? c 1 b ? a 1 @b@ 2 1 b ? c
> >
> >
5 1 1 20 a 0 @ b @ cos (60°) 1 0 a 0 0 c 0cos (60°) 1 1
> >
1 @ b @ 0 c 0cos (120°)
1
1
1
5 2 1 2a b 1 2
2
2
2
53
>
>
>
>
17. a 1 b >1 c 5 0 >
>
>
>
>
>
>
a ? (a 1 b 1 c ) 1 b ? (a 1 b 1 c )
>
>
>
>
1 c ? (a 1 b 1 c ) 5 0
>
>
>
>
>
>
> >
>
>
0 a 02 1 a ? b 1 a ? c 1 b ? a 1 @b@ 2 1 b ? c
>
> >
>
>
1 c ? a 1 c ? b 1 0 c 02 5 0
7-10
>
> >
> >
>
1 1 4 1 9 1 2(a ? b 1 a ? c 1 b ? c ) 5 0
>
>
>
> >
>
2(a ? b 1 a ? c 1 b ? c ) 5 214
>
>
>
> >
>
a ? b 1 a ? c 1 b ? c 5 27
>
>
>
18. d> 5 b> 2 c
>
b 5 d 1> c
> >
> >
>
c ? a 5 ((b ? a ) a ) ? a
>
>
> >
> > >
>
c ? a 5 (b ? a )(a ? a ) because b ? a is a scalar
>
> >
> >
c ? a 5 (b ? a ) 0 a 0 2
>
> >
>
>
>
c ? a 5 (d 1 c ) ? a because 0 a 0 5 1
>
> >
>
> >
c> ? a 5 d ? a 1 c ? a
>
d?a50
7.4 The Dot Product for Algebraic
Vectors, pp. 385–387
> >
a?b50
1.
(21)b1 1 b2 5 0
b2 5 b1
Any vector of the form (c, c) is perpendicular
>
to a . Therefore there are infinitely many vectors
>
perpendicular to a . Answers may vary. For example:
(1, 1), (2, > 2), (3, 3).
>
2. a. a ? b 5 (22)(1) 1 (1)(2)
50
u
5 90°
> >
b. a ? b 5 (2)(4) 1 (3)(3) 1 (21)(217)
5 8 1 9 1 17
5 34 . 0
cos u . 0
u is acute
> >
c. a ? b 5 (1)(3) 1 (22)(22) 1 (5)(22)
5 3 1 4 2 10
5 23 , 0
cos u , 0
u is obtuse
3. Any vector in the xy-plane
is of the form
>
>
a 5 (a1, a2, 0). Let b 5 (0, 0, 1).
> >
a ? b 5 (0)(a1 ) 1 (0)(a2 ) 1 (0)(1)
50
Therefore (0, 0, 1) is perpendicular to every vector
in the xy-plane.
Any vector in the xz-plane
is of the form
>
>
c 5 (c1, 0, c3 ). Let d 5 (0, 1, 0).
> >
c ? d 5 (0)(c1 ) 1 (0)(1) 1 (0)(c3 )
50
Therefore (0, 1, 0) is perpendicular to every vector
in the xz-plane.
Chapter 7: Applications of Vectors
Any vector in the yz-plane
is of the form
>
>
e 5 (0, e2, e3 ). Let f 5 (1, 0, 0).
> >
e ? f 5 (1)(0) 1 (0)(e2 ) 1 (0)(e3 )
50
Therefore (1, 0, 0) is perpendicular to every vector
in the yz-plane.
4. a.
(1, 2, 21) ? (4, 3, 10) 5 4 1 6 2 10
50
5
(24, 25, 26) ? a5, 23, 2 b 5 220 1 15 1 5
6
50
b. If any of the vectors were collinear then one
would be a scalar multiple of the other. Comparing
the signs of the individual components of each
vector eliminates (1, 2, 21) and ( 5, 23, 2 56) . All of
the components of (24, 25, 26) have the same
sign and the same is true for (4, 3, 10), but (4, 3, 10)
is not a scalar multiple of (24, 25, 26). Therefore
none of the vectors are collinear.
5. a. Using the strategy of Example 5 yields
(x, y) ? (1, 22) 5 0 and (x, y) ? (1, 1) 5 0
x 2 2y 5 0 and x 1 y 5 0
3y 5 0
Therefore the only result is x 5 y 5 0, or (0, 0).
This is because (1, 22) and (1, 1) both lie on the
xy-plane and are not collinear, so any vector that is
perpendicular to both vectors must be in R 3 which
does not exist in R 2.
b. If we select any two vectors that are not collinear
in R 2, then any vector that is perpendicular to both
cannot be in R 2 and must be in R 3. This is not
possible since R 3 does
not exist in R 2.
> >
a?b
6. a. cos u 5 > >
0 a 0 @b@
(5)(21) 1 (3)(22)
5
"25 1 9"1 1 4
211
5
"(34)(5)
211
5
"170
u 8 148°
> >
a?b
b. cos u 5 > >
0 a 0 @b@
(21)(6) 1 (4)(22)
5
"1 1 16"36 1 4
214
5
"680
u 8 123°
Calculus and Vectors Solutions Manual
> >
a?b
c. cos u 5 > >
0 a 0 @b@
(2)(2) 1 (2)(1) 1 (1)(22)
5
"4 1 4 1 1"4 1 1 1 4
4
5
(3)(3)
4
5
9
u 8 64° >
>
a?b
d. cos u 5 > >
0 a 0 @b@
(2)(25) 1 (3)(0) 1 (26)(12)
5
"4 1 9 1 36"25 1 144
282
5
(7)(13)
282
5
91
u 8 154°
> >
> >
7. a.
a ? b 5 0 a 0 @ b @ cos u
> >
(21)(26k) 1 (2)(21) 1 (23)(k) 5 0 a 0 @ b @
cos (90°)
6k 2 2 2 3k 5 0
3k 5 2
2
k5
3
> >
> >
a ? b 5 0 a 0 @ b @ cos u
b.
(1)(0) 1 (1)(k) 5 "1 1 1"k 2 cos (45°)
1
k 5 "2 0 k 0
"2
k 5 0k0
k$0
8. a.
y
2
(0, 1)
1
(1, 0) x
–2 –1 0 1 2
–1
–2
b.
y
2
(0, 1)
1
(1, 0) x
0
–2 –1
1 2
–1
–2
7-11
The diagonals are (1, 0) 1 (0, 1) 5 (1, 1) and
(1, 0) 2 (0, 1) 5 (1, 21) or
(1, 0) 1 (0, 1) 5 (1, 1) and
(0, 1) 2 (1, 0) 5 (21, 0).
c. (1, 1) ? (1, 21)
5121
50
or (1, 1) ? (21, 1)
5 21 1 1
50
> >
a?b
9. a. cos u 5 > >
0 a 0 @b@
(1 2 "2)(1) 1 ("2 2 1)(1)
5
> >
0 a 0 @b@
50
u 5> 90°>
a?b
b. cos u 5 > >
0 a 0 @b@
5
5
"2 2 1 1 "2 1 1 1 "2
" (2 2 2"2 1 1) 1 (2 1 2"2 1 1) 1 2 "1 1 1 1 1
3"2
"8"3
"3
5
2
u 5 30°
>
>
10. a. i. a 5 kb
8 5 12k
2
k5
3
2
p 5 4a b
3
8
p5
3
2
25 q
3
q53
ii. Answers may
vary. For example:
> >
a?b50
2q 1 4p 1 96 5 0
q 5 22p 2 48
Let p 5 1
q 5 250
b. In part a., the values are unique because both
vectors have their third component specified, and >
the ratios must be the same for each component b.
In part b. the values are not unique; any value of
p could have been chosen, each resulting in a
different value of q.
7-12
>
>
>
11. AB 5 (2, 6), BC 5 (25, 25), CA 5 (3, 21)
>
>
AB ? CA
cos (180° 2 uA ) 5
>
>
@ AB @ @ CA @
626
5
>
>
@ AB @ @ CA @
50
180° 2 uA 5 90°
uA 5 90° >
>
AB ? BC
cos (180° 2 uB ) 5
>
>
@ AB @ @ BC @
210 2 30
5
"4 1 36"25 1 25
240
5
"(40)(50)
4
52
Å5
180° 2 uB 8 153.4°
uB 8 26.6°
uC 5 180° 2 uA 2 uB
uC 8 63.4°
12. a. O 5 (0, 0, 0), A 5 (7, 0, 0), B 5 (7, 4, 0),
C 5 (0, 4, 0), D 5 (7, 0, 5), E 5 (0, 4, 5),
F 5 (0, 0, 5)
>
>
>
>
b.
AE ? BF 5 @ AE @ @ BF @ cos u
(27, 4, 5) ? (27, 24, 5) 5 "49 1 16 1 25
3 "49 1 16 1 25 cos u
49 2 16 1 25 5 90 cos u
58
5 cos u
90
u 8 50°
13. a. Answers may vary. For example:
(x, y, z) ? (21, 3, 0) 5 0
2x 1 3y 5 0
x 5 3y
(x, y, z) ? (1, 25, 2) 5 0
x 2 5y 1 2z 5 0
22y 1 2z 5 0
y5z
Let y 5 1.
(3, 1, 1) is perpendicular to (21, 3, 0) and
(1, 25, 2).
b. Answers may vary. For example:
(x, y, z) ? (1, 3, 24) 5 0
x 1 3y 2 4z 5 0
x 5 4z 2 3y
(x, y, z) ? (21, 22, 3) 5 0
Chapter 7: Applications of Vectors
2x 2 2y 1 3z 5 0
3y 2 4z 2 2y 1 3z 5 0
y5z
Let y 5 1.
(1, 1, 1) is perpendicular to (1, 3, 24) and
(21, 22, 3).
14. (p, p, 1) ? (p, 22, 23) 5 0
p 2 2 2p 2 3 5 0
2 6 "22 2 4(23)
2
p5162
p 5 3 or 21
15. a. (23, p, 21) ? (1, 24, q) 5 0
23 2 4p 2 q 5 0
3 1 4p 1 q 5 0
b. 3 1 4p 2 3 5 0
p50
16. Answers may vary. For example: Note that
>
>
s 5 22r , so they are collinear. Therefore any
>
vector that is perpendicular to s is also
>
perpendicular to r .
(x, y, z) ? (1, 2, 21) 5 0
x 1 2y 2 z 5 0
Let x 5 z 5 1.
(1, 0, 1) is perpendicular to (1, 2, 21) and
(22, 24, 2).
Let x 5 y 5 1.
(1, 1, 3) is perpendicular to (1, 2 2 1) and
(22, 24, 2).
> >
> >
17. x ? y 5 0 x 0 0 y 0cos u
(24, p, 22) ? (22, 3, 6)
p5
5 "16 1 p 2 1 4"4 1 9 1 36 cos u
8 1 3p 2 12 5 "20 1 p 2 (7) cos u
(3p 2 4)2 5 a7"20 1 p 2 cos ub
9p 2 2 24p 1 16 5 49(20 1 p 2 )a
2
4 2
b
21
320
16
1 p2
9
9
65p 2 2 216p 2 176 5 0
9p 2 2 24p 1 16 5
p5
216 6 "(2216)2 2 4(65)(2176)
2(65)
p 5 4 or 2
44
65
> >
18. a. a ? b 5 23 1 3
50
Therefore, since the two diagonals are perpendicular,
all the sides must be the same length.
Calculus and Vectors Solutions Manual
>
>
1 >
b. AB 5 (a 1 b )
2
5 (1, 2, 21)
>
>
1 >
BC 5 (a 2 b )
2
5 (2, 1, 1)
>
>
@ AB @ 5 @ BC @ 5 "6
>
>
>
>
c. AB ? BC 5 @ AB @ @ BC @ cos u1
2 1 2 2 1 5 6 cos u1
1
5 cos u1
2
u1 5 60°
2u1 1 2u2 5 360°
u2 5 120°
>
>
19. a. AB 5 (3, 4, 212), DA 5 (24, 2 2 q, 25)
>
>
AB ? DA 5 0
212 1 8 2 4q 1 60 5 0
21 2 q 1 15 5 0
q 5 14
>
>
DA 5 CB
(24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)
x 5 6, y 5 18, z 5 24
The coordinates of vertex C are (6, 18, 24).
>
>
>
>
b.
AC ? BD 5 @ AC @ @ BD @ cos u
(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 49
3 "1 1 64 1 289 cos u
7 1 128 2 119 5 354 cos u
16
5 cos u
354
u 8 87.4°
20. The two vectors representing the body diagonals
are (0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1) and
(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)
(21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u
1 2 1 1 1 5 3 cos u
1
5 cos u
3
u 8 70.5°
a 5 180° 2 u
a 8 109.5°
Mid-Chapter Review, pp. 388–389
> >
1. a. a ? b 5 (3)(2) cos (60°)
1
5 (6)
2
53
7-13
2x 2 2y 1 3z 5 0
3y 2 4z 2 2y 1 3z 5 0
y5z
Let y 5 1.
(1, 1, 1) is perpendicular to (1, 3, 24) and
(21, 22, 3).
14. (p, p, 1) ? (p, 22, 23) 5 0
p 2 2 2p 2 3 5 0
2 6 "22 2 4(23)
2
p5162
p 5 3 or 21
15. a. (23, p, 21) ? (1, 24, q) 5 0
23 2 4p 2 q 5 0
3 1 4p 1 q 5 0
b. 3 1 4p 2 3 5 0
p50
16. Answers may vary. For example: Note that
>
>
s 5 22r , so they are collinear. Therefore any
>
vector that is perpendicular to s is also
>
perpendicular to r .
(x, y, z) ? (1, 2, 21) 5 0
x 1 2y 2 z 5 0
Let x 5 z 5 1.
(1, 0, 1) is perpendicular to (1, 2, 21) and
(22, 24, 2).
Let x 5 y 5 1.
(1, 1, 3) is perpendicular to (1, 2 2 1) and
(22, 24, 2).
> >
> >
17. x ? y 5 0 x 0 0 y 0cos u
(24, p, 22) ? (22, 3, 6)
p5
5 "16 1 p 2 1 4"4 1 9 1 36 cos u
8 1 3p 2 12 5 "20 1 p 2 (7) cos u
(3p 2 4)2 5 a7"20 1 p 2 cos ub
9p 2 2 24p 1 16 5 49(20 1 p 2 )a
2
4 2
b
21
320
16
1 p2
9
9
65p 2 2 216p 2 176 5 0
9p 2 2 24p 1 16 5
p5
216 6 "(2216)2 2 4(65)(2176)
2(65)
p 5 4 or 2
44
65
> >
18. a. a ? b 5 23 1 3
50
Therefore, since the two diagonals are perpendicular,
all the sides must be the same length.
Calculus and Vectors Solutions Manual
>
>
1 >
b. AB 5 (a 1 b )
2
5 (1, 2, 21)
>
>
1 >
BC 5 (a 2 b )
2
5 (2, 1, 1)
>
>
@ AB @ 5 @ BC @ 5 "6
>
>
>
>
c. AB ? BC 5 @ AB @ @ BC @ cos u1
2 1 2 2 1 5 6 cos u1
1
5 cos u1
2
u1 5 60°
2u1 1 2u2 5 360°
u2 5 120°
>
>
19. a. AB 5 (3, 4, 212), DA 5 (24, 2 2 q, 25)
>
>
AB ? DA 5 0
212 1 8 2 4q 1 60 5 0
21 2 q 1 15 5 0
q 5 14
>
>
DA 5 CB
(24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)
x 5 6, y 5 18, z 5 24
The coordinates of vertex C are (6, 18, 24).
>
>
>
>
b.
AC ? BD 5 @ AC @ @ BD @ cos u
(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 49
3 "1 1 64 1 289 cos u
7 1 128 2 119 5 354 cos u
16
5 cos u
354
u 8 87.4°
20. The two vectors representing the body diagonals
are (0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1) and
(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)
(21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u
1 2 1 1 1 5 3 cos u
1
5 cos u
3
u 8 70.5°
a 5 180° 2 u
a 8 109.5°
Mid-Chapter Review, pp. 388–389
> >
1. a. a ? b 5 (3)(2) cos (60°)
1
5 (6)
2
53
7-13
>
>
>
>
>
> >
b. (3a 1 2b ) ? (4a 2 3b ) 5 12 0 a 0 2> 2 9a ? b >
>
1 8b ? a 2 6 @ b @ 2
5 12(3) 2 2 3 2 6(2) 2
5 81
2.
20 cm
u2
15 cm
25 cm
u1
Let T1 be the tension in the 15 cm cord and T2 be
the tension in the 20 cm cord. Let u1 be the angle
the 15 cm cord makes with the ceiling and u2 be the
angle the 20 cm cord makes with the ceiling. By the
cosine law:
(15)2 5 (20)2 1 (25)2 2 2(20)(25) cos (u2 )
cos (u2 ) 5 0.8
sin (u2 ) 5 "1 2 cos2 (u2 )
sin (u2 ) 5 0.6
(20)2 5 (15)2 1 (25)2 2 (2)(15)(25) cos (u1 )
cos (u1 ) 5 0.6
sin (u1 ) 5 0.8
Horizontal Components:
2T1 cos (u1 ) 1 T2 cos (u2 ) 5 0
(0.8)T2 5 (0.6)T1
T2 5 (0.75)T1
Vertical Components:
T1 sin (u1 ) 1 T2 sin (u2 ) 2 (15)(9.8) 5 0
(0.8)T1 1 (0.6)(0.75)T1 5 147
(1.25)T1 5 147
T1 5 117.6 N
T2 5 (0.75)T1
T2 5 88.2 N
Therefore the tension in the 15 cm cord is 117.60 N
and the tension in the 20 cm cord is 88.20 N.
3. The diagonals of a square are perpendicular, so
the dot product is 0.
4. a.
a
v
v +w
135°
w>
>
0 v 0 5 500, 0 w 0 5 100
By the cosine law:
>
>
0 v 1 w 0 2 5 (500)2 1 (100)2
2 2(500)(100) cos (135°)
>
>
0 v 1 w 0 8 575.1
7-14
By the cosine law:
sin (a)
sin (135°)
5
100
575.1
sin (a) 8 0.123
a 8 7.06°
The resultant velocity of the airplane is 575.1 km>h
at S7.06°E
b. (distance) 5 (rate)(time)
1000
km
t8
?
575.1 (km/h)
t 8 1.74 hours
5. a.
E
F
@ E ' @ 5 @ E @ cos (40°)
>
@ E ' @ 5 (9.8)(15)cos (40°)
>
@ E ' @ 8 112.61 N
>
>
b. @ F @ 5 @ E @ sin (40°)
>
@ F @ 8 94.49 N
6. 6u 5 360°
>
>
u 5 60° >
> >
>
a ? b 5 0 a 0 @ b @ cos (60°)
5 (3)(3)(0.5)
5 4.5
> >
7. a. a ? b 5 (4)(1) 1 (25)(2) 1 (20)(2)
5 34>
> >
>
b. a ? b 5 0 a 0 @ b @ cos (u)
34 5 "16 1 25 1 400 "1 1 4 1 4 cos (u)
34
cos (u) 5
63 >
>
>
>
>
>
> >
8. a. a ? b 5 (i 1 2j 1 k ) ? (2i 2 3j 1 4k )
522614
5 0>
>
>
>
>
>
>
>
b. b ? c 5 (2i 2 3j 1 4k ) ? (3i 2 j 2 k )
561324
55 >
>
>
>
>
>
>
>
c. b 1 c 5 (2i 2 3j 1 4k ) 1 (3i 2 j 2 k )
>
>
>
5 5i 2 4j >1 3k>
>
>
>
>
>
>
>
d. a ? (b 1 c ) 5 (i 1 2j 1 k ) ? (5i 2 4j 1 3k )
552813
5
>
> 0>
>
>
>
>
e. (a 1 b ) ? (b 1 c ) 5 (3i 2 j 1 5k )
>
>
>
? (5j 2 4j 1 3k )
5 15 1 4 1 15
5 34
Chapter 7: Applications of Vectors
>
>
>
>
>
>
>
f. (2a 2 3b ) ? (2a 1 c ) 5 ((2i 1 4j 1 2k )
>
>
>
2 (6i 2 9j 1 12k ))
>
>
>
? ((2i 1 4j 1 2k )
>
>
>
1 (3i 2 j 1 k ))
>
>
>
5 (24i 1 13j 2 10k )
>
>
>
? (5i 1 3j 1 k )
5 220 1 39 2 10
59
> >
9. a.
?
q
5
0
p
>
>
>
>
>
>
(xi 1 j 1 3k ) ? (3xi 1 10xj 1 k ) 5 0
3x 2 1 10x 1 3 5 0
210 6 "(10)2 2 4(3)(3)
x5
2(3)
210 6 8
x5
6
1
x 5 23 or x 5 2
3
>
>
b. If p and q are parallel then one is a scalar
multiple of the other.
>
>
p 5 nq where n is a constant
>
>
>
>
>
>
xi 1 j 1 3k 5 n(3xi 1 10xj 1 k )
>
n 5 3 by the k > component
x 5 9x by the i component
x50
>
1 5 30(0) by the j component
120
Therefore there is no value of x that will make these
two vectors parallel.
>
>
>
>
>
>
>
>
10. a. 3x 2 2y 5 (3i 2 6j 2 3k ) 2 (2i 2 2j 2 2k )
>
>
>
5 i 2 4j 2 k
>
>
>
>
>
>
>
>
b. 3x ? 2y 5 (3i 2 6j 2 3k ) ? (2i 2 2j 2 2k )
5 6 1 12 1 6
5 24 >
>
>
>
>
>
>
>
c. 0 x 2 2y 0 5 @ (i 2 2j 2 k ) 2 (2i 2 2j 2 2k ) @
>
>
5 @ 2i 1 k @
5 " (2i 1 k ) ? (2i 1 k )
>
>
>
>
5 "2 or 1.41
>
>
>
>
>
>
>
d. (2x 2 3y ) ? (x 1 4y ) 5 ((2i 2 4j 2 2k )
>
>
>
2 (3i 2 3j 2 3k )) ?
>
>
>
1 (( i 2 2j 2 k )
>
>
>
1 (4i 2 4j 2 4k )
>
>
>
5 (2i 2 j 1 k )
>
>
>
? (5i 2 6j 2 5k )
5 25 1 6 2 5
5 24
Calculus and Vectors Solutions Manual
> >
> >
> >
> >
e. 2x ? y 2 5y ? x 5 2x ? y 2 5x ? y
> >
5 23x ? y
>
>
>
>
>
5 23(i 2 2j 2 k) ? (i 2 j 2 k )
5 23(1 1 2 1 1)
5 212
11.
5N
4N
180° - u
3N
2
2
(4) 5 (5) 1 (3)2 2 2(3)(5) cos (180° 2 u)
0.6 5 cos (180° 2 u)
180° 2 u 8 53.1
u 8 126.9°
12. (F)2 5 (3)2 1 (4)2 2 2(3)(4) cos (180° 2 60°)
(F)2 5 25 2 24 cos (120°)
(F)2 5 37
F 8 6.08 N
(3)2 5 (4)2 1 "37
(
cos u 5
44
)2 2 2(4)("37) cos u
8"37
u 8 25.3°
>
F 8 6.08 N, 25.3° from the 4 N force towards the
3 >N force.
E 8 6.08 N, 180° 2 25.3° 5 154.7° from the
4 N force away from the 3 N force.
>
>
>
>
13. a. The diagonals are m 1 n and m 2 n
>
>
m 1 n 5 (1, 4, 10)
>
>
m 2 n 5 (3, 210, 0)
>
>
>
>
>
> >
>
(m 1 n ) ? (m 2 n ) 5 0 m 1 n 0 0 m 2 n 0cos u
3 2 40 5 "1 1 16 1 100 "9 1 100 cos u
cos u 8 20.3276
u 8 109.1°
>
>
>
>
> >
b. 0 m 2 n 0 2 5 0 m 0 2 1 0 n 0 2 2 2 0 m 0 0 n 0cos u
(9 1 100) 5 (4 1 9 1 25) 1 (1 1 49 1 25)
2 2"38 "75 cos u
cos u 8 0.0374
u 8 87.9°
7-15
14. a. 45 sin (150°) 5 500 sin u
u 8 N 2.6° E
b. v 5 500 cos (2.6°) 2 45 cos (30°)
8 460.5 km> h
1000
t8
460.5
t 8 2.17 hours
> >
a?x50
15.
2x1 1 2x2 1 5x3 5 0
x 5 2x2 1 5x3
> 1>
b?x50
x1 1 3x2 1 5x3 5 0
2x2 1 5x3 1 3x2 1 5x3 5 0
x2 1 2x3 5 0
choose x3 5 1
x2 5 22
x1 5 1
1
>
x5
(1, 22, 1)
!6
1 2
1
2 1
1
>
x5a
,2
,
b or a2
,
,2
b
!6 !6 !6
!6 !6
!6
16. a. v 5 4 1 3 cos (45°)
8 6.12 m> s
d 8 (6.12)(10)
8 61.2 m
b. w 5 3 sin (45°)
8 2.12 m> s
180
t8
2.12
t 8 84.9 seconds
>
>
>
>
17. a. (x 1 y ) ? (x 2 y ) 5 0
>2
> >
> >
>2
0x0 2 x ? y 1 y ? x 2 0y0 5 0
>
>
0 x 02 5 0 y 02
>
>
>
>
>
>
(x 1 y ) ? (x 2 y ) 5 0 when x and y have
the same length. >
>
b. Vectors >a and b determine a parallelogram. Their
>
sum a 1 b is one diagonal of the parallelogram
formed, with
its tail in the same location
as the tails
>
>
>
>
of a and b. Their difference a 2 b is the other
diagonal
of the parallelogram.
>
18. @ F @ 5 350 cos (40°)
8 268.12 N
7-16
7.5 Scalar and Vector Projections,
pp. 398–400
>
>
>
>
a ?b
1. a. Scalar projection of a on b is > where
@b@
>
>
a 5 (2, 3) and b is the positive x-axis (X, 0).
> >
a ? b 5 (2X) 1 (3 3 0)
5 2X 1 0
5 2X
@ b @ 5 "X 2 1 02
>
5X
2X
> 5
X
@b@
5 2;
The vector projection
is the
scalar projection
>
>
> >
a?b
multiplied by
b
@ b@
>
where
b
@ b@
>
is the x-axis divided by
>
the magnitude of the x-axis which is equal
> to i.
>
The scalar projection of 2 multiplied> by i equals 2i.
>
>
>
a ?b
b. Scalar projection of a on b is > where
@b@
>
>
a 5 (2, 3) and b is now the positive y-axis (0, Y).
> >
a ? b 5 (2 3 0) 1 (3Y)
5 0 1 3Y
> 5 3Y
@ b @ 5 "02 1 Y 2
5Y
> >
a @b@
3Y
> 5
Y
@b@
5 3;
The vector projection is the scalar projection
multiplied by
>
>
b
b
@ b@
>
where
@ b@
>
is the y-axis divided
>
by the magnitude of the y-axis which is equal
to j. >
>
The scalar projection of 3 multiplied by j equals 3j.
2. Using the formula
would cause a division by 0.
>
Generally the 0 has any direction and 0 magnitude.
You can not project onto nothing.
>
>
3. You are projecting a onto the tail of b which
>
is a point with magnitude
0. Therefore it is 0; the
>
>
projections
of b onto the tail of a are also 0
>
and 0.
Chapter 7: Applications of Vectors
14. a. 45 sin (150°) 5 500 sin u
u 8 N 2.6° E
b. v 5 500 cos (2.6°) 2 45 cos (30°)
8 460.5 km> h
1000
t8
460.5
t 8 2.17 hours
> >
a?x50
15.
2x1 1 2x2 1 5x3 5 0
x 5 2x2 1 5x3
> 1>
b?x50
x1 1 3x2 1 5x3 5 0
2x2 1 5x3 1 3x2 1 5x3 5 0
x2 1 2x3 5 0
choose x3 5 1
x2 5 22
x1 5 1
1
>
x5
(1, 22, 1)
!6
1 2
1
2 1
1
>
x5a
,2
,
b or a2
,
,2
b
!6 !6 !6
!6 !6
!6
16. a. v 5 4 1 3 cos (45°)
8 6.12 m> s
d 8 (6.12)(10)
8 61.2 m
b. w 5 3 sin (45°)
8 2.12 m> s
180
t8
2.12
t 8 84.9 seconds
>
>
>
>
17. a. (x 1 y ) ? (x 2 y ) 5 0
>2
> >
> >
>2
0x0 2 x ? y 1 y ? x 2 0y0 5 0
>
>
0 x 02 5 0 y 02
>
>
>
>
>
>
(x 1 y ) ? (x 2 y ) 5 0 when x and y have
the same length. >
>
b. Vectors >a and b determine a parallelogram. Their
>
sum a 1 b is one diagonal of the parallelogram
formed, with
its tail in the same location
as the tails
>
>
>
>
of a and b. Their difference a 2 b is the other
diagonal
of the parallelogram.
>
18. @ F @ 5 350 cos (40°)
8 268.12 N
7-16
7.5 Scalar and Vector Projections,
pp. 398–400
>
>
>
>
a ?b
1. a. Scalar projection of a on b is > where
@b@
>
>
a 5 (2, 3) and b is the positive x-axis (X, 0).
> >
a ? b 5 (2X) 1 (3 3 0)
5 2X 1 0
5 2X
@ b @ 5 "X 2 1 02
>
5X
2X
> 5
X
@b@
5 2;
The vector projection
is the
scalar projection
>
>
> >
a?b
multiplied by
b
@ b@
>
where
b
@ b@
>
is the x-axis divided by
>
the magnitude of the x-axis which is equal
> to i.
>
The scalar projection of 2 multiplied> by i equals 2i.
>
>
>
a ?b
b. Scalar projection of a on b is > where
@b@
>
>
a 5 (2, 3) and b is now the positive y-axis (0, Y).
> >
a ? b 5 (2 3 0) 1 (3Y)
5 0 1 3Y
> 5 3Y
@ b @ 5 "02 1 Y 2
5Y
> >
a @b@
3Y
> 5
Y
@b@
5 3;
The vector projection is the scalar projection
multiplied by
>
>
b
b
@ b@
>
where
@ b@
>
is the y-axis divided
>
by the magnitude of the y-axis which is equal
to j. >
>
The scalar projection of 3 multiplied by j equals 3j.
2. Using the formula
would cause a division by 0.
>
Generally the 0 has any direction and 0 magnitude.
You can not project onto nothing.
>
>
3. You are projecting a onto the tail of b which
>
is a point with magnitude
0. Therefore it is 0; the
>
>
projections
of b onto the tail of a are also 0
>
and 0.
Chapter 7: Applications of Vectors
>
>
4. Answers
may
vary.
For
example:
p
,
5
AE
>
>
q 5 AB
D
p
A
E
C
q
B
>
>
>
Scalar projection p on q 5 @ AC @ ;
>
>
>
Vector projection p on q 5 AC;
>
>
>
Scalar projection q on p 5 @ AD @ ;
>
>
>
Vector projection q on p 5 AD
>
>
5. When a 5 (21, 2, 5) and b 5 (1, 0, 0) then
> >
a ? b 5 (21 3 1 1 2 3 0 1 5 3 0)
5 21
>
@ b @ 5 "12 1 02 1 02
51
> >
a ?b
21
Therefore the scalar projection is > 5
1
@b@
5 21;
>
b
(1, 0, 0)
The vector equation is 21 3 > 5 21 3
1
@b@
5 21;
>
Under> the same approach, when a 5 (21, 2, 5)
and b 5 (0, 1, 0), then
> >
a ? b 5 (21 3 0 1 2 3 1 1 5 3 0)
52
>
@ b @ 5 "02 1 1 1 02
51
> >
2
a ?b
Therefore the scalar projection is > 5
1
@b@
5 2,
>
b
(0, 1, 0)
The vector equation is 2 3 > 5 2 3
1
@b@
5 2;
>
The
same is also true when a 5 (21, 2, 5) and
>
b 5 (0, 0, 1) then
> >
a ? b 5 (21 3 0 1 2 3 0 1 5 3 1)
55
>
@ b @ 5 "02 1 02 1 12
51
> >
a ?b
5
Therefore the scalar projection is > 5
1
@b@
5 5,
>
b
(0, 0, 1)
The vector equation is 5 3 > 5 5 3
1
@b@
5 5;
Calculus and Vectors Solutions Manual
Without having to use formulae, a projection of
> >
>
(21, 2, 5) on i, j, or k is the same as a projection
>
>
>
of (21, 0, 0) on i, (0, 2, 0) on j, and (0, 0, 5) on k
which intuitively yields the same result.
> >
6. a. p ? q 5 (3 3 24) 1 (6 3 5)
1 (222 3 220)
5 212 1 30 1 440
5 458
>
0 q 0 5 "(24)2 1 52 1 (220)2
5 "16 1 25 1 400
5 "441
5 21
Therefore the scalar projection is
> >
p?q
458
,
> 5
0q0
21
>
458
q
The vector equation 5
3 >
21
0q0
458 (24, 5, 220)
5
.
21
21
458
(24, 5, 20).
5
441
>
>
b. Direction angles for p where p 5 (a, b, c)
a
include a, b, and g. cos a 5
"a 2 1 b 2 1 c 2
3
5
"32 1 62 1 (222)2
3
5
"9 1 36 1 484
3
5
"529
3
5 ,
23
3
Therefore a 5 cos21 a b
23
8 82.5°;
b
cos b 5
"a 2 1 b 2 1 c 2
6
5
2
2
"3 1 6 1 (222)2
6
5
"9 1 36 1 484
6
5
"529
6
5 ,
23
6
Therefore b 5 cos21 a b
23
8 74.9°;
7-17
cos g 5
5
5
5
>
8. a. The scalar projection of a on the x-axis
c
"a 2 1 b 2 1 c 2
222
(X, 0, 0) is
"3 1 6 1 (222)
222
2
2
2
"9 1 36 1 484
222
"529
222
5
,
23
Therefore g 5 cos21 a
222
b
23
8 163.0°
> >
7. a. x ? y 5 (1 3 1) 1 (1 3 21)
5 1 1 (21)
50
>
0 y 0 5 "12 1 (21)2
5 "2
> >
x ?y
0
> 5
0y0
"2
5 0;
>
>
y
The vector projection is 0 3 > 5 0
0y0
> >
b. x ? y 5 (2 3 1) 1 (2"3 3 0)
52
>
0 y 0 5 "12 1 02
51
> >
x ?y
2
The scalar projection is > 5
0y0
1
5 2;
>
y
(1, 0)
The vector projection is 2 3 > 5 2 3
0y0
1
>
5 2i
> >
c. x ? y 5 (2 3 25) 1 (5 3 12)
5 210 1 60
5 50
>
0 y 0 5 "(25)2 1 122
The scalar projection is
5 "25 1 144
5 "169
5 13
> >
x ?y
50
The scalar projection is > 5 .
0y0
13
>
50
y
(25, 12)
50
The vector projection is
3 > 5
3
13
0y0
13
13
50
(25, 12)
5
169
7-18
>
a ? (X, 0, 0)
0 (X, 0, 0) 0
>
a ? (X, 0, 0)
(21 3 X) 1 (2 3 0) 1 (4 3 0)
5
0 (X, 0, 0) 0
"X 2 1 02 1 02
2X
5
X
5 21;
>
The vector projection of a on the x-axis is
(X, 0, 0)
(X, 0, 0)
5 21 3
21 3
2
2
2
X
"X 1 0 1 0
>
5 2i;
>
The scalar projection of a on the y-axis (0, Y, 0) is
>
(21 3 0) 1 (2 3 Y) 1 (4 3 0)
a ? (0, Y, 0)
5
0 (0, Y, 0) 0
"02 1 Y 2 1 02
2Y
5
Y
52
>
The vector projection of a on the y-axis is
(0, Y, 0)
(0, Y, 0)
23
523
2
2
2
Y
"0 1 Y 1 0
>
5 2j;
>
The scalar projection of a on the z-axis (0, 0, Z) is
>
(21 3 0) 1 (2 3 0) 1 (4 3 Z)
a ? (0, 0, Z)
5
0 (0, 0, Z) 0
"02 1 02 1 Z 2
4Z
5
Z
5 4;
>
The vector projection of a on the z-axis is
(0, 0, Z)
(0, 0, Z)
43
543
2
2
2
Z
"0 1 0 1 Z
>
5 4k.
>
b. The scalar projection of m a on the x-axis
(X, 0, 0) is
>
ma ? (X, 0, 0)
(2m 3 X) 1 (2m 3 0)
5
0 (X, 0, 0) 0
"X 2 1 02 1 02
(4m 3 0)
1
"X 2 1 02 1 02
2mX
5
X
5 2m
>
The vector projection of ma on the x-axis is
(X, 0, 0)
(X, 0, 0)
5 2m 3
2m 3
2
2
2
X
"X 1 0 1 0
>
5 2mi;
Chapter 7: Applications of Vectors
>
The scalar projection of ma on the y-axis (0, Y, 0) is
>
ma ? (0, Y, 0)
(2m 3 0) 1 (2m 3 Y)
5
0 (0, Y, 0) 0
"0 2 1 Y 2 1 02
(4m 3 0)
1
"0 2 1 y2 1 02
2mY
5
Y
5 2m;
>
The vector projection of ma on the y-axis is
(0, Y, 0)
(0, Y, 0)
5 2m 3
2m 3
2
2
2
Y
"0 1 Y 1 0
>
5 2mj;
>
The scalar projection of ma on the z-axis (0, 0, Z) is
>
ma ? (0, 0, Z)
(2m 3 0) 1 (2m 3 0)
5
0 (0, 0, Z) 0
"0 2 1 02 1 Z2
(4m 3 Z)
1
"0 2 1 02 1 Z2
4mZ
5
Z
5 4m;
>
The vector projection of ma on the z-axis is
(0, 0, Z)
(0, 0, Z)
4m 3
5 4m 3
2
2
2
Z
"0 1 0 1 Z
>
5 4mk.
9. a.
a
>
a projected onto itself will yield itself. The scalar
projection will be the magnitude of itself.
b. Using the formula for the scalar projection
>
>
0 a 0cos u 5 0 a 0cos 0
>
5 0 a 0 (1)
>
5 0 a 0.
The vector projection is the scalar projection
>
>
>
>
a
a
multiplied by 0 > 0 , 0 a 0 3 0 > 0 5 a .
a
a
10. a. B –a
O
a
>
>
>
(2a ) ? a
2 0 a 02
b.
5
>
>
0a0
0a0
>
5 20a 0
A
> 0a0
>
So the vector projection is 2 0 a 0 a > b 5 2 a .
0a0
>
Calculus and Vectors Solutions Manual
>
11. a. AB 5 Point B 2 Point A
5 (21, 3, 4) 2 (1, 2, 2)
5 (22, 1, 2)
>
The scalar projection of AB on the x-axis (X, 0, 0) is
>
(22 3 X) 1 (1 3 0) 1 (2 3 0)
a ? (X, 0, 0)
5
0 (X, 0, 0) 0
"X 2 1 02 1 02
22X
5
X
5 22;
>
The vector projection of AB on the x-axis is
(X, 0, 0)
(X, 0, 0)
22 3
5 22 3
X
"X 2 1 02 1 02
>
5 22i;
>
The scalar projection of AB on the y-axis (0, Y, 0) is
>
(22 3 0) 1 (1 3 Y) 1 (2 3 0)
a ? (0, Y, 0)
5
0 (0, Y, 0) 0
"02 1 Y 2 1 02
Y
5
Y
5 1;
>
The vector projection of AB on the y-axis is
(0, Y, 0)
(0, Y, 0)
13
513
2
2
2
Y
"0 1 Y 1 0
>
5 j;
>
The scalar projection of AB on the z-axis (0, 0, Z) is
>
a ? (0, 0, Z)
(22 3 0) 1 (1 3 0) 1 (2 3 Z)
5
0 (0, 0, Z) 0
"02 1 02 1 Z 2
2Z
5
Z
5 2;
>
The vector projection of AB on the z-axis is
(0, 0, Z)
(0, 0, Z)
23
523
2
2
2
Z
"0 1 0 1 Z
>
5 2k
b. The angle made with the y-axis is b
b
cos b 5
2
"a 1 b 2 1 c 2
1
5
2
"(22) 1 12 1 22
1
5
"4 1 1 1 4
1
5
"9
1
5 ,
3
7-19
1
Therefore b 5 cos21 a b
3
8 70.5°
>
12. a. @ BD @
C
a
u
B
b. @ BD @
D
>
b
B
c
b
u
A
C
a
u
5
D
c
b
u
A
c. In an isosceles triangle, CD is a median
and a
>
>
right bisector of BA. Therefore a and b have the
>
same magnitude projected on c .
d. Yes, not only do they have the same magnitude,
but they are in the same direction as well which
makes them have equivalent vector projections.
>
13.
the formula for the scalar projection of a on
> a. Use
>
b 5 0 a 0cos u
5 10 cos 135°
5 27.07
>
And the> formula for the scalar projection of b on
>
a 5 @ b @ cos u
5 12 cos 135°
5 28.49
b.
b
12
135° 10
Q
O
a
P
>
>
>
OQ> is the vector projection of b on a>
>
OP is the vector projection of a on b
>
14. a. AB 5 Point B 2 Point A
5 (1, 3, 3) 2 (22, 1, 4)
5 (3, 2, 21)
>
>
The scalar projection of AB on OD is
>
>
AB ? OD
(3 3 21) 1 (2 3 2) 1 (21 3 2)
>
5
@ OD @
"(21)2 1 22 1 22
(23) 1 4 1 (22)
5
"1 1 4 1 4
7-20
21
"9
1
52
3
>
b. BC 5 Point C 2 Point B
5 (26, 7, 5) 2 (1, 3, 3)
5 (27, 4, 2)
>
>
The scalar projection of BC on OD is
>
>
BC ? OD
(27 3 21) 1 (4 3 2) 1 (2 3 2)
>
5
@ OD @
"(21)2 1 22 1 22
71814
5
"1 1 4 1 4
19
5
"9
19
5
3
>
>
>
>
AB ? OD
BC ? OD
1
19
>
>
1
52 1
3
3
@ OD @
@ OD @
18
5
3
5
6
>
AC 5 Point C 2 Point A
5 (26, 7, 5) 2 (22, 1, 4)
5 (24, 6, 1)
>
>
The> scalar
> projection of AC on OD is
AC ? OD
(24 3 21) 1 (6 3 2) 1 (1 3 2)
>
5
@ OD @
"(21)2 1 22 1 22
4 1 12 1 2
5
"1 1 4 1 4
18
5
"9
18
5
3
56
>
c. Same lengths and both are in the direction of OD.
Add to get one vector.
15. a. 1 5 cos2 a 1 cos2 b 1 cos2 g
2
2
a
b
5a
b 1a
b
"a 2 1 b 2 1 c 2
"a 2 1 b 2 1 c 2
2
c
1a
b
2
"a 1 b 2 1 c 2
a2
b2
5 2
1
a 1 b2 1 c2
a2 1 b2 1 c2
2
c
1 2
a 1 b2 1 c2
Chapter 7: Applications of Vectors
a2 1 b2 1 c2
a2 1 b2 1 c2
51
b. a 5 90°, b 5 30°, g 5 60°
cos a 5 cos 90°
5 0,
x50
cos b 5 cos 30°
"3
5
,
2
y is a multiple of "3
2 .
cos g 5 cos 60°
1
5 ,
2
1
z is a multiple of .
2
1
Answers include Q 0, "3
2 , 2 R , Q 0, "3, 1 R , etc.
c. If two angles add to 90°, then all three will add to
180°.
16. a. a 5 b 5 g
cos a 5 cos b 5 cos g
cos2 a 5 cos2 b 5 cos2 g
1 5 cos2 a 1 cos2 b 1 cos2 g
1 5 3 cos2 x
1
5 cos2 x
3
1
5 cos x
Å3
1
x 5 cos21
Å3
x 8 54.7°
1
.
b. For obtuse, use cos x 5 2
Å3
1
x 5 cos21 a2
b
Å3
x 8 125.3°
17. cos2 x 1 sin2 x 5 1
cos2 x 5 1 2 sin2 x
1 5 cos2 a 1 cos2 b 1 cos2 g
1 5 (1 2 sin2 a) 1 (1 2 sin2 b) 1 (1 2 sin2 g)
1 5 3 2 (sin2 a 1 sin2 b 1 sin2 g)
sin2 a 1 sin2 b 1 sin2 g 5 2
5
Calculus and Vectors Solutions Manual
18. Answers may vary. For example:
z
B (0, c, d)
y
x
A (a, b, 0)
7.6 The Cross Product of Two
Vectors, pp. 407–408
z
1. a.
b
axb
y
a
x
>
>
>
a 3 b is perpendicular to a . Thus, their dot product
must equal 0. The same applies to the second case.
z
a3b
b
a1b
y
a
x >
>
>
b.> a 1 b is still
> in the same plane formed
> by a and
>
>
b, thus a 1 b is perpendicular to a 3 b making the
dot product 0 again. >
>
c. Once again, a 2> b is still in> the same plane
>
>
formed
> by a and b, thus a 2 b is perpendicular to
>
the dot product 0 again.
a 3 b making
>
>
2. a 3 b produces a vector, not a scalar. Thus, the
equality is meaningless.
3. a. It’s possible because there is a vector crossed
with a vector, then dotted with another vector,
producing a scalar.
> >
b. This is meaningless because a ? b produces a
scalar. This results in a scalar crossed with a vector,
which is meaningless.
7-21
a2 1 b2 1 c2
a2 1 b2 1 c2
51
b. a 5 90°, b 5 30°, g 5 60°
cos a 5 cos 90°
5 0,
x50
cos b 5 cos 30°
"3
5
,
2
y is a multiple of "3
2 .
cos g 5 cos 60°
1
5 ,
2
1
z is a multiple of .
2
1
Answers include Q 0, "3
2 , 2 R , Q 0, "3, 1 R , etc.
c. If two angles add to 90°, then all three will add to
180°.
16. a. a 5 b 5 g
cos a 5 cos b 5 cos g
cos2 a 5 cos2 b 5 cos2 g
1 5 cos2 a 1 cos2 b 1 cos2 g
1 5 3 cos2 x
1
5 cos2 x
3
1
5 cos x
Å3
1
x 5 cos21
Å3
x 8 54.7°
1
.
b. For obtuse, use cos x 5 2
Å3
1
x 5 cos21 a2
b
Å3
x 8 125.3°
17. cos2 x 1 sin2 x 5 1
cos2 x 5 1 2 sin2 x
1 5 cos2 a 1 cos2 b 1 cos2 g
1 5 (1 2 sin2 a) 1 (1 2 sin2 b) 1 (1 2 sin2 g)
1 5 3 2 (sin2 a 1 sin2 b 1 sin2 g)
sin2 a 1 sin2 b 1 sin2 g 5 2
5
Calculus and Vectors Solutions Manual
18. Answers may vary. For example:
z
B (0, c, d)
y
x
A (a, b, 0)
7.6 The Cross Product of Two
Vectors, pp. 407–408
z
1. a.
b
axb
y
a
x
>
>
>
a 3 b is perpendicular to a . Thus, their dot product
must equal 0. The same applies to the second case.
z
a3b
b
a1b
y
a
x >
>
>
b.> a 1 b is still
> in the same plane formed
> by a and
>
>
b, thus a 1 b is perpendicular to a 3 b making the
dot product 0 again. >
>
c. Once again, a 2> b is still in> the same plane
>
>
formed
> by a and b, thus a 2 b is perpendicular to
>
the dot product 0 again.
a 3 b making
>
>
2. a 3 b produces a vector, not a scalar. Thus, the
equality is meaningless.
3. a. It’s possible because there is a vector crossed
with a vector, then dotted with another vector,
producing a scalar.
> >
b. This is meaningless because a ? b produces a
scalar. This results in a scalar crossed with a vector,
which is meaningless.
7-21
>
>
c. This> is possible. a 3 b produces a vector, and
>
c 1 d also produces a vector. The result is a vector
dotted with a vector producing
a scalar.
> >
d. This
is
possible.
produces
a scalar, and
a
?
b
>
>
c 3 d produces a vector. The product of a scalar
and vector produces a vector.
>
>
e. This> is possible. a 3 b produces a vector, and
>
c 3 d produces a vector. The cross product of a
vector and vector produces
> a vector.
>
f. This is possible. a 3 b produces a vector. When
added to another vector, it produces another vector.
4. a. (2, 23, 5) 3 (0, 21, 4)
5 (23(4) 2 5(21), 5(0) 2 2(4),
2(21) 2 (23)(0))
5 (27, 28, 22)
(2, 23, 5) ? (27, 28, 22) 5 0
(0, 21, 4) ? (27, 28, 22) 5 0
b. (2, 21, 3) 3 (3, 21, 2)
5 (21(2) 2 3(21), 3(3) 2 2(2),
2(21) 2 (21)(3))
5 (1, 5, 1)
(2, 21, 3) ? (1, 5, 1) 5 0
(3, 21, 2) ? (1, 5, 1) 5 0
c. (5, 21, 1) 3 (2, 4, 7)
5 (21(7) 2 1(4), 1(2) 2 5(7),
5(4) 2 (21)(2))
5 (211, 233, 22)
(5, 21, 1) ? (211, 233, 22) 5 0
(2, 4, 7) ? (211, 233, 22) 5 0
d. (1, 2, 9) 3 (22, 3, 4)
5 (2(4) 2 9(3), 9(22) 2 1(4),
1(3) 2 2(22))
5 (219, 222, 7)
(1, 2, 9) ? (219, 222, 7) 5 0
(22, 3, 4) ? (219, 222, 7) 5 0
e. (22, 3, 3) 3 (1, 21, 0)
5 (3(0) 2 3(21), 3(1) 2 (22)(0),
22(21) 2 3(1))
5 (3, 3, 21)
(22, 3, 3) ? (3, 3, 21) 5 0
(1, 21, 0) ? (3, 3, 21) 5 0
f. (5, 1, 6) 3 (21, 2, 4)
5 (1(4) 2 6(2), 6(21) 2 5(4),
5(2) 2 1(21))
5 (28, 226, 11)
(5, 1, 6) ? (28, 226, 11) 5 0
(21, 2, 4) ? (28, 226, 11) 5 0
5. (21, 3, 5) 3 (0, a, 1)
5 (3(1) 2 5(a), 5(0) 2 (21)(1),
21(a) 2 3(0))
7-22
If we look at the x component, we know that:
3(1) 2 5(a) 5 22
25(a) 5 25
>a 5 1
>
6. a. a 3 b 5 (1(1) 2 1(5), 1(0) 2 0(1),
0(5) 2 0(1))
5 (24, 0, 0)
b. Vectors of the form (0, b, c) are in the
yz-plane. Thus, the only vectors perpendicular to the
yz-plane are those of the form (a, 0, 0) because they
are parallel to the x-axis.
7. a. (1, 2, 1) 3 (2, 4, 2)
5 (2(2) 2 1(4), 1(2) 2 1(2), 1(4) 2 2(2))
5 (0, 0, 0)
b. (a, b, c) 3 (ka, kb, kc)
5 (b(kc) 2 c(kb), c(ka) 2 a(kc),
a(kb) 2 b(ka))
Using the commutative law of multiplication we
can rearrange this:
5 (bck 2 bck, ack 2 ack, abk 2 abk)
5 (0, 0, 0)
>
>
>
8. a. p 3 (q 1 r ) 5 (1, 22, 4) 3 3(1, 2, 7)
1 (21, 1, 0)4
5 (1, 22, 4) 3 (1 2 1, 2 1 1, 7 1 0)
5 (1, 22, 4) 3 (0, 3, 7)
5 (22(7) 2 4(3), 4(0) 2 1(7),
1(3) 1 2(0))
5 (226, 27, 3)
>
>
>
>
p 3 q 1 p 3 r 5 (22(7) 2 4(2),
4(1) 2 1(7), 1(2) 1 2(1))
1 (22(0) 2 4(1),
4(21) 2 1(0), 1(1) 1 2(21))
5 (222, 23, 4) 1 (24, 24, 21)
5 (226, 27, 3)
>
>
>
b. p 3 (q 1 r ) 5 (4, 1, 2) 3 3(3, 1, 21)
1 (0, 1, 2)4
5 (4, 1, 2) 3 (3, 1 1 1, 21 1 2)
5 (4, 1, 2) 3 (3, 2, 1)
5 (1(1) 2 2(2), 3(2) 2 4(1),
4(2) 2 1(3))
5 (23, 2, 5)
>
>
>
>
p 3 q 1 p 3 r 5 (1(21) 2 2(1), 2(3) 2 4(21),
4(1) 2 1(3)) 1 (1(2) 2 2(1),
2(0) 2 4(2), 4(1) 2 1(0))
5 (23, 10, 1) 1 (0, 28, 4)
5 (23, 2, 5)
>
>
9. a. i 3 j 5 (1, 0, 0) 3 (0, 1, 0)
5 (0 2 0, 0 2 0, 1 2 0)
5 (0, 0, 1)
>
5k
Chapter 7: Applications of Vectors
>
>
2j 3 i 5 (0, 21, 0) 3 (1, 0, 0)
5 (0 2 0, 0 2 0, 0 2 (21))
5 (0, 0, 1)
>
>
> 5k
b. j 3 k 5 (0, 1, 0) 3 (0, 0, 1)
5 (1 2 0, 0 2 0, 0 2 0)
5 (1, 0, 0)
>
>
>5i
2k 3 j 5 (0, 0, 21) 3 (0, 1, 0)
5 (0 2 (21), 0 2 0, 0 2 0)
5 (1, 0, 0)
>
>
>5i
c. k 3 i 5 (0, 0, 1) 3 (1, 0, 0)
5 (0 2 0, 1 2 0, 0 2 0)
5 (0, 1, 0)
>
>
>5j
2i 3 k 5 (21, 0, 0) 3 (0, 0, 1)
5 (0 2 0, 0 2 (21), 0 2 0)
5 (0, 1, 0)
>
5j
10. k(a2b3 2 a3b2, a3b1 2 a1b3, a1b2 2 a2b1 )
? (a1, a2, a3 )
5 k(a1a2b3 2 a1a3b2 1 a2a3b1 2 a2a1b3
1 a3a1b2 2 a3a2b1 )
5 k(0)
50
>
>
>
k(a
a is perpendicular
to
).
3
b
>
>
11. a. a 3 b 5 (2, 0, 0) 3 (0, 3, 0)
5 (0 2 0, 0 2 0, 6 2 0)
5 (0, 0, 6)
>
>
c 3 d 5 (2, 3, 0) 3 (4, 3, 0)
5 (0 2 0, 0 2 0, 6 2 12)
5
(0, 0,> 26)
>
>
>
b. (a 3 b ) 3 (c 3 d ) 5 (0, 0, 6) 3 (0, 0, 26)
(by part a.)
5 (0 2 0, 0 2 0, 0 2 0)
5 (0, 0, 0)
c. All the vectors are in the xy-plane. Thus, the cross
product in part b. is between vectors parallel to the
z-axis and so parallel to each other.
The cross
>
product of parallel vectors is 0.
>
12. Let x 5 (1, 0, 1)
>
y 5 (1, 1, 1)
>
z 5 (1, 2, 3)
>
>
Then x 3 y 5 (0 2 1, 1 2 1, 1 2 0)
5 (21, 0, 1)
>
>
>
(x 3 y ) 3 z 5 (0 2 2, 1 2 (23), 23 2 0)
5 (22, 4, 23)
Calculus and Vectors Solutions Manual
>
>
y 3 z 5 (3 2 2, 1 2 3, 2 2 1)
5 (1, 22, 1)
>
>
>
x 3 (y 3 z ) 5 (0 1 2, 1 2 1, 22 2 0)
5 (2, 0, 22)
>
>
>
>
>
>
Thus (x 3> y ) 3 z 2 x> 3 (y 3 z ).
>
>
13. (a 2 b ) 3 (a 1 b )
By the distributive property of cross product:
>
>
>
>
>
>
5 (a 2 b ) 3 a 1 (a 2 b ) 3 b
By the distributive
property >again:
>
>
>
>
>
>
>
5 a 3 a 2 b 3 a 1 a 3 b 2 b 3 >b
A vector
with> itself equals 0, thus:
> crossed
>
>
5 2b 3> a 1> a 3 b
>
>
5 a 3 b> 2 b 3 a >
>
>
5 a 3 b 2 (2a 3 b )
>
>
5 2a 3 b
7.7 Applications of the Dot Product
and Cross Product, pp. 414–415
1. By pushing as far away from the hinge as
>
possible, 0 r 0 is increased making the cross product
bigger. By pushing at right angles, sine is its largest
value, 1, making
the cross product larger.
>
>
2. a. a 3 b 5 (1, 2, 1) 3 (2, 4, 2)
5 (2(2) 2 1(4), 1(2)
2 1(2), 1(4) 2 2(2))
5 (0, 0, 0)
>
>
@a 3 b @ 5 0
b. This makes sense because the vectors lie on the
same line. Thus, the parallelogram would just be a
line making
its area 0.
> >
3. a. f ? s 5 3 ? 150 5 450 J
b.
x
y
50°
392 N
40 m
The axes are tilted to illustrate the force of gravity
can be split up into components to find the part in
the direction of the motion. Let x be the component
of force going in the motion’s direction.
x
cos (50°) 5
392
x 5 (392) cos (50°)
Now we have our force, so:
(392) cos 50° N ? 40 m 8 10 078.91 J
7-23
>
>
2j 3 i 5 (0, 21, 0) 3 (1, 0, 0)
5 (0 2 0, 0 2 0, 0 2 (21))
5 (0, 0, 1)
>
>
> 5k
b. j 3 k 5 (0, 1, 0) 3 (0, 0, 1)
5 (1 2 0, 0 2 0, 0 2 0)
5 (1, 0, 0)
>
>
>5i
2k 3 j 5 (0, 0, 21) 3 (0, 1, 0)
5 (0 2 (21), 0 2 0, 0 2 0)
5 (1, 0, 0)
>
>
>5i
c. k 3 i 5 (0, 0, 1) 3 (1, 0, 0)
5 (0 2 0, 1 2 0, 0 2 0)
5 (0, 1, 0)
>
>
>5j
2i 3 k 5 (21, 0, 0) 3 (0, 0, 1)
5 (0 2 0, 0 2 (21), 0 2 0)
5 (0, 1, 0)
>
5j
10. k(a2b3 2 a3b2, a3b1 2 a1b3, a1b2 2 a2b1 )
? (a1, a2, a3 )
5 k(a1a2b3 2 a1a3b2 1 a2a3b1 2 a2a1b3
1 a3a1b2 2 a3a2b1 )
5 k(0)
50
>
>
>
k(a
a is perpendicular
to
).
3
b
>
>
11. a. a 3 b 5 (2, 0, 0) 3 (0, 3, 0)
5 (0 2 0, 0 2 0, 6 2 0)
5 (0, 0, 6)
>
>
c 3 d 5 (2, 3, 0) 3 (4, 3, 0)
5 (0 2 0, 0 2 0, 6 2 12)
5
(0, 0,> 26)
>
>
>
b. (a 3 b ) 3 (c 3 d ) 5 (0, 0, 6) 3 (0, 0, 26)
(by part a.)
5 (0 2 0, 0 2 0, 0 2 0)
5 (0, 0, 0)
c. All the vectors are in the xy-plane. Thus, the cross
product in part b. is between vectors parallel to the
z-axis and so parallel to each other.
The cross
>
product of parallel vectors is 0.
>
12. Let x 5 (1, 0, 1)
>
y 5 (1, 1, 1)
>
z 5 (1, 2, 3)
>
>
Then x 3 y 5 (0 2 1, 1 2 1, 1 2 0)
5 (21, 0, 1)
>
>
>
(x 3 y ) 3 z 5 (0 2 2, 1 2 (23), 23 2 0)
5 (22, 4, 23)
Calculus and Vectors Solutions Manual
>
>
y 3 z 5 (3 2 2, 1 2 3, 2 2 1)
5 (1, 22, 1)
>
>
>
x 3 (y 3 z ) 5 (0 1 2, 1 2 1, 22 2 0)
5 (2, 0, 22)
>
>
>
>
>
>
Thus (x 3> y ) 3 z 2 x> 3 (y 3 z ).
>
>
13. (a 2 b ) 3 (a 1 b )
By the distributive property of cross product:
>
>
>
>
>
>
5 (a 2 b ) 3 a 1 (a 2 b ) 3 b
By the distributive
property >again:
>
>
>
>
>
>
>
5 a 3 a 2 b 3 a 1 a 3 b 2 b 3 >b
A vector
with> itself equals 0, thus:
> crossed
>
>
5 2b 3> a 1> a 3 b
>
>
5 a 3 b> 2 b 3 a >
>
>
5 a 3 b 2 (2a 3 b )
>
>
5 2a 3 b
7.7 Applications of the Dot Product
and Cross Product, pp. 414–415
1. By pushing as far away from the hinge as
>
possible, 0 r 0 is increased making the cross product
bigger. By pushing at right angles, sine is its largest
value, 1, making
the cross product larger.
>
>
2. a. a 3 b 5 (1, 2, 1) 3 (2, 4, 2)
5 (2(2) 2 1(4), 1(2)
2 1(2), 1(4) 2 2(2))
5 (0, 0, 0)
>
>
@a 3 b @ 5 0
b. This makes sense because the vectors lie on the
same line. Thus, the parallelogram would just be a
line making
its area 0.
> >
3. a. f ? s 5 3 ? 150 5 450 J
b.
x
y
50°
392 N
40 m
The axes are tilted to illustrate the force of gravity
can be split up into components to find the part in
the direction of the motion. Let x be the component
of force going in the motion’s direction.
x
cos (50°) 5
392
x 5 (392) cos (50°)
Now we have our force, so:
(392) cos 50° N ? 40 m 8 10 078.91 J
7-23
c.
140 N
20°
250 m
First find the x component of the force:
(140) cos (20°) 5 x
Calculate work:
140 cos 20° N ? 250 m 8 32 889.24 J
d.
100 N
45°
500 m
First calculate the x component of the force:
x 5 (100) cos (45°)
Calculate work:
100 cos
m 5 35 355.34 J
> 45°
> ? 500
>
4. a. i 3 j 5 k
The square formed by the 2 vectors has an area of 1.
The 2 vectors are >on the xy-plane, thus, the cross
product must be k by the right hand rule.
>
>
>
b. 2i 3 j 5 2k
Once again, the area is 1, making the possible vector
have a magnitude of 1. Also, the 2 vectors are on the
xy-plane again so the cross product must lie on the
z axis. However, because
of the right hand rule, the
>
product
must
be
this
time.
2k
>
>
>
c. i 3 k 5 2j
The square has an area of 1, so the magnitude of the
vector produced must be 1. The 2 vectors are on the
xz-plane. The> new vector must be on the y axis
making it – j because of the right hand rule.
>
>
>
d. 2i 3 k 5 2j
The square has an area of 1. The 2 vectors
> are on
the xz-plane. So the new vector must be j because
of the right> hand rule.
>
5. a. a 3 b 5 (1, 1, 0) 3 (1, 0, 1)
5 (1 2 0, 0 2 1, 0 2 1)
5 (1, 21, 21)
>
>
@ a 3 b @ 5 "1 1 1 1 1 5 "3
So the area
> of the parallelogram is "3 square units.
>
b. a 3 b 5 (1, 22, 3) 3 (1, 2, 4)
5 (28 2 6, 3 2 4, 2 1 2)
5 (214, 21, 4)
>
>
@ a 3 b @ 5 "196 1 1 1 16 5 "213
So the area of the parallelogram is "213 square units.
7-24
>
>
6. p 3 q 5 (a, 1, 21) 3 (1, 1, 2)
5 (2 1 1, 22a 2 1, a 2 1)
5 (3, 2a 1 1, a 2 1)
>
>
0 p 3 q 0 5 "9 1 (2a 1 1)2 1 (a 2 1)2 5 "35
9 1 (2a 1 1)2 1 (a 2 1)2 5 35
2
9 1 4a 1 4a 1 1 1 a 2 2 2a 1 1 5 35
5a 2 1 2a 2 24 5 0
22 6 "22 2 4(5)(224)
a5
2(5)
22 6 22
5
10
212
5 2,
5
7. a.
B
AB
A
AC
C
As we see from the picture, the area of the triangle
ABC is just half the area of> the parallelogram
>
determined by vectors AB and AC. Thus, we use the
magnitude of the cross product to calculate the area.
>
AB> 5 (1 1 2, 0 2 1, 1 2 3) 5 (3, 21, 22)
AC> 5 (2 1
> 2, 3 2 1, 2 2 3) 5 (4, 2, 21)
AB 3 AC 5 (1 1 4, 23 1 8, 6 1 4) 5 (5, 5, 10)
>
>
@ AB 3 AC @ 5 "25 1 25 1 100 5 5"6
Since triangle ABC is half the area of the
parallelogram, its area is 5"6
2 square units.
b. This is just a different way of describing the first
triangle, thus the area is 5"6
2 square units.
c. Any two sides of a triangle can be used to
calculate> its area.
>
>
>
8. @ r 3 f @ 5 ( 0 r 0sin (u)) @ f @
5 (0.14) sin (45°) ? 10
8 0.99 J
9.
A
BN
OA
C
O
OB
B
We know that the area of a parallelogram is equal to
its height multiplied with its base. Its height is BN
>
>
and its base is AC 5 OB as can be seen from the
picture. We can calculate the area using the given
vectors,
then
>
> use the area to find BN.
OA 3 OB 5 (8 2 4, 12 2 16, 4 2 6)
5 (4, 24, 22)
>
>
@ OA 3 OB @ 5 "16 1 16 1 4 5 "36 5 6
Chapter 7: Applications of Vectors
Now we need to calculate @ OB @ to know the length
of the base.
>
>
AC 5 @ OB @ 5 "9 1 1 1 16 5 "26
Substituting these results into the equation for area:
>
@ OB @ ? BN 5 6
>
"26 BN 5 6
BN 5
6
or about 1.18
"26
>
>
10. a.
p 3 q 5 (26 2 3, 6 2 3, 1 1 4)
5 (29, 3, 5)
>
>
>
(p 3 q ) 3 r 5 (0 2 5, 5 1 0, 29 2 3)
5 (25, 5, 212)
a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)
Looking at x components:
a 1 2b 5 25; a 5 25 2 2b
y components:
22a 1 b 5 5
Substitute in a:
10 1 4b 1 b 5 5
5b 5 25
b 5 21
Substitute b back into the x components:
a 5 25 1 2; a 5 23
Check in z components:
3a 1 3b 5 212
29 2 3 5 212
> >
b. p ? r 5 1 2 2 1 0 5 21
> >
q?r52111053
> > >
> > >
(p ? r )q 2 (q ? r )p 5 21(2, 1, 3) 2 3(1, 22, 3)
5 (2, 21, 23) 2 (3, 26, 9)
5 (22 2 3, 21 1 6, 23 2 9)
5 (25, 5, 212)
Review Exercise, pp. 418–421
>
>
1. a. a 3 b 5 (2 2 0, 21 1 1, 0 1 2)
5 (2, 0, 2)
>
b. b 3 c 5 (0 2 4, 25 1 5, 24 2 0)
5 (24, 0, 24)
c. 16
d. The cross products are parallel, so the original
vectors are in the same plane.
>
2. a.> 0 a 0 5 "22 1 (21)2 1 22 5 3
b. @ b @ 5> "62 1 32 1 (22)2 5 7
>
c. a 2 b 5 (2 2 6, 21 2 3, 2 1 2)
5 (24, 24, 4)
>
>
@ a 2 b @ 5 "(24)2 1 (24)2 1 42 5 4"3
>
Calculus and Vectors Solutions Manual
>
>
d. a 1 b 5 (2 1 6, 21 1 3, 2 2 2)
5 (8, 2, 0)
>
>
@ a 1 b> @ 5 "82 1 22 1 02 5 2"17
>
e. a ? b 5 2(6)
2 1(3) 1 2(22) 5 5
>
>
a 2 2b 5 (2 2 12, 21 2 6, 2 1 4)
f.
> 5 (210, 27, 6)
>
>
a ? (a 2 2b ) 5 2(210) 2 1(27) 1 2(6) 5 21
>
>
3. a. If a 5> 6, then
y will be twice x , thus collinear.
>
x 3 y 5 (3, a, 9) ? (a, 12, 18) 5 0
b.
3a 1 12a 1 162 5 0
15a 5 2162
254
a5
5
> >
a?b
4. cos (u) 5 > >
0 a 0 @b@
> >
a ? b 5 4(23) 1 5(6) 1 20(22) 5 458
>
0 a 0 5 "42 1 52 1 202 5 21
>
@ b @ 5 "(23)2 1 62 1 222 5 23
458
u 5 cos21 a
b
483
u 8 18.52°
5. a.
y
4
2 OA
OB
x
0
–4 –2
2 4
–2
–4
b. We can use the dot product of the 2 diagonals to
calculate
the
angle. >The diagonals
are the vectors
>
>
>
OA> 1 OB> and OA 2 OB.
OA 1 OB 5 (5 2 1, 1 1 4) 5 (4, 5)
>
>
OA 2 OB 5 (5 1 1, 1 2 4) 5 (6, 23)
>
>
>
>
(OA 1 OB ) ? (OA 2 OB )
>
>
>
>
cos (u) 5
@ OA 1 OB @ @ OA 2 OB @
>
>
>
>
(OA 1 OB ) ? (OA 2 OB ) 5 4(6) 1 5(23) 5 9
>
>
@ OA 1 OB @ 5 "42 1 52 5 "41
@ OA 2 OB @ 5 "62 1 (23)2 5 3"5
>
>
u 5 cos21 a
9
3"205
b
u 8 77.9°
7-25
Now we need to calculate @ OB @ to know the length
of the base.
>
>
AC 5 @ OB @ 5 "9 1 1 1 16 5 "26
Substituting these results into the equation for area:
>
@ OB @ ? BN 5 6
>
"26 BN 5 6
BN 5
6
or about 1.18
"26
>
>
10. a.
p 3 q 5 (26 2 3, 6 2 3, 1 1 4)
5 (29, 3, 5)
>
>
>
(p 3 q ) 3 r 5 (0 2 5, 5 1 0, 29 2 3)
5 (25, 5, 212)
a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)
Looking at x components:
a 1 2b 5 25; a 5 25 2 2b
y components:
22a 1 b 5 5
Substitute in a:
10 1 4b 1 b 5 5
5b 5 25
b 5 21
Substitute b back into the x components:
a 5 25 1 2; a 5 23
Check in z components:
3a 1 3b 5 212
29 2 3 5 212
> >
b. p ? r 5 1 2 2 1 0 5 21
> >
q?r52111053
> > >
> > >
(p ? r )q 2 (q ? r )p 5 21(2, 1, 3) 2 3(1, 22, 3)
5 (2, 21, 23) 2 (3, 26, 9)
5 (22 2 3, 21 1 6, 23 2 9)
5 (25, 5, 212)
Review Exercise, pp. 418–421
>
>
1. a. a 3 b 5 (2 2 0, 21 1 1, 0 1 2)
5 (2, 0, 2)
>
b. b 3 c 5 (0 2 4, 25 1 5, 24 2 0)
5 (24, 0, 24)
c. 16
d. The cross products are parallel, so the original
vectors are in the same plane.
>
2. a.> 0 a 0 5 "22 1 (21)2 1 22 5 3
b. @ b @ 5> "62 1 32 1 (22)2 5 7
>
c. a 2 b 5 (2 2 6, 21 2 3, 2 1 2)
5 (24, 24, 4)
>
>
@ a 2 b @ 5 "(24)2 1 (24)2 1 42 5 4"3
>
Calculus and Vectors Solutions Manual
>
>
d. a 1 b 5 (2 1 6, 21 1 3, 2 2 2)
5 (8, 2, 0)
>
>
@ a 1 b> @ 5 "82 1 22 1 02 5 2"17
>
e. a ? b 5 2(6)
2 1(3) 1 2(22) 5 5
>
>
a 2 2b 5 (2 2 12, 21 2 6, 2 1 4)
f.
> 5 (210, 27, 6)
>
>
a ? (a 2 2b ) 5 2(210) 2 1(27) 1 2(6) 5 21
>
>
3. a. If a 5> 6, then
y will be twice x , thus collinear.
>
x 3 y 5 (3, a, 9) ? (a, 12, 18) 5 0
b.
3a 1 12a 1 162 5 0
15a 5 2162
254
a5
5
> >
a?b
4. cos (u) 5 > >
0 a 0 @b@
> >
a ? b 5 4(23) 1 5(6) 1 20(22) 5 458
>
0 a 0 5 "42 1 52 1 202 5 21
>
@ b @ 5 "(23)2 1 62 1 222 5 23
458
u 5 cos21 a
b
483
u 8 18.52°
5. a.
y
4
2 OA
OB
x
0
–4 –2
2 4
–2
–4
b. We can use the dot product of the 2 diagonals to
calculate
the
angle. >The diagonals
are the vectors
>
>
>
OA> 1 OB> and OA 2 OB.
OA 1 OB 5 (5 2 1, 1 1 4) 5 (4, 5)
>
>
OA 2 OB 5 (5 1 1, 1 2 4) 5 (6, 23)
>
>
>
>
(OA 1 OB ) ? (OA 2 OB )
>
>
>
>
cos (u) 5
@ OA 1 OB @ @ OA 2 OB @
>
>
>
>
(OA 1 OB ) ? (OA 2 OB ) 5 4(6) 1 5(23) 5 9
>
>
@ OA 1 OB @ 5 "42 1 52 5 "41
@ OA 2 OB @ 5 "62 1 (23)2 5 3"5
>
>
u 5 cos21 a
9
3"205
b
u 8 77.9°
7-25
6.
>
T1
30°
5 (250, 220, 18)
>
>
0 x 3 y 0 5 "502 1 202 1 182 5 "3224 8 56.78
9. (0, 3, 25) 3 (2, 3, 1)
5 (3 1 15, 210 2 0, 0 2 6) 5 (18, 210, 26)
The cross product is perpendicular to the given
vectors, but its magnitude is
T2
45°
x
98 N
The vertical components of the tensions must equal
the downward force:
T1 sin (30°) 1 T2 sin (45°) 5 98 N
1
1
T 1
T 5 98
2 1 "2 2
T1 5 196 2 "2T2
The horizontal components:
T1 cos (30°) 1 T2 cos (45°) 5 0 N
"3
1
T 2
T 50
2 1 "2 2
Substitute in T1:
"6
98"3 2
T 5 298"3
2 2
2"6 2 "2
T2 5 298"3
2
T2 8 87.86N
Substitute this back in to get T1:
T1 8 71.74N
7.
x
50 km/h
300 km/h
x 5 "50 1 3002 8 304.14
50
tan21 a
b 8 9.46°
300
The resultant velocity is 304.14 km> h, W 9.46° N.
8. a.
z
2
y
x
x
7-26
>
b. x 3 y 5 (215 2 35, 25 2 15, 21 2 3)
y
y
"18 2 1 (210)2 1 (26)2, or 2"115. A unit vector
perpendicular to the given vectors is
a !115, 2 !115, 2 !115 b.
9
5
3
>
10. a.cos (a) 5
>
>
AB ? AC
@ AB @ @ AC @
>
>
AB> 5 (0, 23, 4) 2 (2, 3, 7) 5 (22, 26, 23)
> AC> 5 (5, 2, 24) 2 (2, 3, 7) 5 (3, 21, 211)
AB ? AC 5 22(3) 2 6(21) 2 3(211) 5 33
>
@ AB @ 5 "(22)2 1 (26)2 1 (23)2 5 7
>
2
@ AC @ 5 "32 1 (21)
1
(211)2 5 "131
>
>
AB ? AC
>
>
a 5 cos21
@ AB @ @ AC @
33
5 cos21
7"131
8 65.68° >
>
21 BA ? BC
>
>
b 5 cos
@
@
@
@
BA
BC
>
>
BA> 5 2AB 5 (2, 6, 3)
> BC> 5 (5 2 0, 2 1 3, 24 2 4, ) 5 (5, 5, 28)
BA ? BC 5 2(5) 1 6(5) 1 3(28) 5 16
>
@ BA @ 5 "22 1 62 1 32 5 7
>
@ BC @ 5 "52 1 52 1 (28)2 5 "144
16
b 5 cos21
7"114
8 77.64°
g 5 180 2 a 2 b 8 36.68°
So b 8 77.64° is the largest angle.
b. The area is >half the magnitude
of the cross
>
product of AB and AC.
>
>
1
1
AB 3 AC 5 0 (63, 231, 20) 0 8 36.50
2
2
11. The triangle formed by the two strings and the
ceiling is similar to a 3-4-5 right triangle, with the
30 cm and 40 cm strings as legs. So the angle
adjacent to the 30 cm leg satisfies
3
cos u 5
5
Chapter 7: Applications of Vectors
The angle adjacent to the 40 cm leg satisfies
4
cos f 5
5
Also,
4
3
sin u 5 and sin f 5 .
5
5
Let T1 be the tension in the 30 cm string, and T2 be
the tension in the 40 cm string. Then
T1 cos u 2 T2 cos f 5 0
3
4
T1 2 T2 5 0
5
5
4
T1 5 T2
3
Also,
T1 sin u 1 T2 sin f 5 (10)(9.8) 5 98
4
3
T1 2 T2 5 98
5
5
4
3
4
a T2 b 1 T2 5 98
3
5
5
5
T 5 98
3 2
T2 5 58.8 N
4
T1 5 (58.8)
3
5 78.4 N
So the tension in the 30 cm string is 78.4 N and the
tension in the 40 cm string is 58.8 N.
12. a.
54 N
30 N
25 N
42 N
b. The east- and west-pulling forces result in a force
of 5 N west. The north- and south-pulling forces
result in a force of 12 N north. The 5 N west and
12 N north forces result in a force pulling in the
north-westerly direction with a force of
"52 1 122 5 13 N,
by using the Pythagorean theorem. To find the exact
direction of this force, use the definition of sine.
Calculus and Vectors Solutions Manual
If u is the angle west of north, then
5
sin u 5
13
u 8 22.6°
So the resultant is 13 N in a direction
N22.6°W. The equilibrant is 13 N in a direction
S22.6°E.
13. a. Let D be the origin, then:
A 5 (2, 0, 0), B 5 (2, 4, 0), C 5 (0, 4, 0),
D 5 (0, 0, 0), E 5 (2, 0, 3), F 5 (2, 4, 3),
G 5 (0, 4, 3) H 5 (0, 0, 3)
>
b. AF> 5 (0, 4, 3)
AC 5 (22, 4, 0)
>
>
AF ? AC> 5 0 1 16 1 0 5 16
@ AF @ 5 "02 1 42 1 32 5 5
>
@ AC @ 5 "(22)2 1 42 1 02 5 2"5
>
>
AF ? AC
>
>
cos (u) 5
@ AF @ @ AC @
16
b
u 5 cos21 a
10"5
u 8 44.31°
>
c. Scalar projection 5 @ AF @ cos (u)
By part b.:
5 (5) cos (44.31°)
8 3.58 >
>
> >
14. a ? b 5 0 a 0 @ b @ cos (u) 5 cos (u)
1
cos (u) 5 2 (cosine law)
2 >
>
>
>
(2a 2 5b ) ? (b 1 3a )
>
>
> >
> >
5 213a ? b 1 6a ? a 2 5b ? b
>
>
5 213a ? b 1 1
5 213 cos (u) 1 1
5 7.5
15. a. The angle to the bank, u, will satisfy
2
sin (90° 2 u) 5 3
90° 2 u 8 41.8°
u 8 48.2°
b. By the Pythagorean theorem, Kayla’s net
swimming speed will be
"32 2 22 5 "5 km> h.
So since distance 5 rate 3 time, it will take her
0.3
t5
"5
8 0.13 h
8 8 min 3 sec
to swim across.
7-27
c. Such a situation would have resulted in a right
triangle where one of the legs is longer than the
hypotenuse, which is impossible.
>
>
16. > a. The > diagonals are OA 1 OB and
OA> 2 OB>.
OA 1 OB 5 (3 2 6, 2 1 6, 26 2 2)
5 (23, 8, 28)
>
>
OA 2 OB 5 (3 1 6, 2 2 6, 26 1 2)
5 (9, 24, 24)
>
>
b. OA ? OB> 5 3(26) 1 2(6) 2 6(22) 5 6
@ OA @ 5 "32 1 22 1 (26)2 5 7
>
2
@ OB @ 5 "(26)
1
62 1 (22)2 5 2"19
>
>
OA ? OB
>
>
cos (u) 5
@ OA @ @ OB @
6
b
u 5 cos21 5 a
14"19
8 84.36°
17. a. The z value is >double, so if a 5 4 and
b 5 24, the vector q will be collinear.
>
>
b. If p and q are perpendicular, then their dot
product will equal 0.
> >
p ? q 5 2a 2 2b 2 18 5 0
c. Let a 5 9, and b> 5 0, then we have a vector
perpendicular to p . Now it must be divided by its
magnitude to make it a unit vector:
>
0 p 0 5 "81 1 0 1 324 5 9"5
So the unit vector is:
1
2
a
, 0,
b
!5 !5
> >
18. a. m ? n 5 2"3 2 2"3 1 3 5 3
>
0 m 0 5 "3 1 4 1 9 5 4
>
0 n 0 5 "4 1 3 1 1 5 2"2
> >
m?n
cos (u) 5 > >
0m00n0
3
u 5 cos21 a
b
8"2
8 74.62°
>
b. Scalar projection 5 0 n 0cos (u)
5 2"2 cos (74.62°)
8 0.75
with the unit vector
c. Scalar projection multiplied
>
in the direction of m :
>
m
5 (0.75) >
0m0
("3, 22, 23)
5 (0.75)
4
5 (0.1875)("3, 22, 23)
7-28
>
>
d. m ? k 5 23
u 5 cos21 a
23
b
4
8 138.59°
19. a. If the dot product is 0, then the vectors are
perpendicular:
(1, 0, 0) ? (0, 0, 21) 5 0 1 0 1 0 5 0
(1, 0, 0) ? (0, 1, 0) 5 0 1 0 1 0 5 0
(0, 0, 21) ? (0, 1, 0) 5 0 1 0 1 0 5 0 special
1
1
21 1
1
b. a ,
, 0b ? a
,
,
b
!2 !2
!3 !3 !3
1
1
52
1
10
!6
!6
50
1
1
a
,
, 0b ? (0, 0, 21) 5 0 1 0 1 0 5 0
!2 !2
21 1
1
a
,
,
b ? (0, 0, 21)
!3 !3 !3
1
1
52
not special
501012
!3
>
> !3
20. a. p 3 q
5 (22(1) 2 1(21), 1(2) 2 1(1), 1(21) 1 2(2))
5 (21, 1, 3)
>
>
b. p> 2 q> 5 (21, 21, 0)
p 1 q 5 (3, 23, 2)
>
>
>
>
(p 2 q ) 3 (p 1 q ) 5 (22 2 0, 0 1 2, 3 2 (23))
5 (22, 2, 6)
>
>
c. p 3 r 5 (4 2 1, 0 1 2, 1 2 0)
5 (3, 2, 1)
>
>
>
(p 3 r ) ? r 5 0 1 2 2 2 5 0
>
>
d. p 3 q 5 (22 1 1, 2 2 1, 21 1 4)
5 (21, 1, 3)
21. Since the angle between the two vectors is 60°,
the angle formed when they are placed head-to-tail
is 120°. So the resultant, along with these two
vectors, forms an isosceles triangle with top angle
120° and two equal angles 30°. By the cosine law,
the two equal forces satisfy
202 5 2F 2 2 2F 2 cos 120°
400
F2 5
3
20
F5
"3
8 11.55
> N
>
22. a 3 b 5 (2 2 0, 25 2 3, 0 2 10)
5 (2, 28, 210)
Chapter 7: Applications of Vectors
23.
First we need to determine the dot product of
>
>
x and y :
> >
> >
x ? y 5 0 x 0 0 y 0cos u
5 (10) cos (60°)
55
>
>
>
>
(x 2 2y ) ? (x 1 3y )
By the distributive property:
> >
> >
> >
> >
5 x ? x 1 3x ? y 2 2x ? y 2 6y ? y
5 4 1 15 2 10 2 150
5 2141
24. 0 (2, 2, 1) 0 5 "22 1 22 1 12 5 3
Since the magnitude of the scalar projection is 4,
the scalar projection itself has value 4 or 24.
If it is 4, we get
(1, m, 0) ?(2, 2, 1)
54
3
2 1 2m 5 12
m 5 5
If it is 24, we get
(1, m, 0) ?(2, 2, 1)
5 24
3
2 1 2m 5 212
m 5 27
So the two possible values for m are 5 and 27.
> >
25. a ? j 5 23
>
0 a 0 5 "144 1 9 1 16 5 13
23
u 5 cos21 a
b
13
8 103.34°
26. a.> C 5 (3, 0, 5), F 5 (0, 4, 0)
b. CF 5 (0, 4, 0) 2 (3, 0, 5) 5 (23, 4, 25)
c. @ CF @ > 5 "9 1 16 1 25 5 5"2
>
OP 5 (3, 4, 5)
>
@
@ > 5 "9 1 16 1 25 5 5"2
OP
>
CF ? OP 5 29 1 16 2 25 5 218
218
u 5 cos21 a
b
50
8 111.1°
27.
d
50°
130°
e
a. Using properties of parallelograms, we know that
the other angle is 130° (Angles must add up to
360°, opposite angles are congruent).
Using the cosine law,
Calculus and Vectors Solutions Manual
@ d 1 e @ 2 5 32 1 52 2 2(3)(5) cos 130°
>
>
@ d 1 e @ 8 7.30
b. Using the cosine law,
>
>
@ d 2 e @ 2 5 32 1 52 2 2(3)(5) cos 50°
>
>
@ d 2 e @ 8 3.84
>
>
>
>
c.> e 2> d is the vector in the opposite direction of
d 2 e , but with the same magnitude. So:
>
>
>
>
@ e 2 d @ 5 @ d 2 >e @ 8> 3.84>
(i 1 j ) ? (i )
>
28. a. Scalar:
51
@i@
>
>
i
Vector: 1a > b 5 i
@i@
b. Scalar:
Vector: 1a
c. Scalar:
Vector:
1
>
>
>
>
@ j@
(i 1 j ) ? (j )
>
j
>
b5j
>
>
@ j@
51
>
>
>
(i 1 j ) ? (k 1 j )
"2
@k 1 j@
>
5
(k 1 j )
>
>
?
>
>
>
>
1
"2
>
1
5 (k 1 j )
2
@k 1 j@
29. a. If its magnitude is 1, it’s a unit vector:
>
0 a 0 5 "14 1 19 1 361 2 1 not a unit vector
@ b @ 5 "13 1 13 1 13 5 1, unit vector
>
>
0 c 0 5 "14 1 12 1 14 5 1, unit vector
@ d @ 5 "1 1 1 1 1 2 1, not a unit vector
>
>
b. a is. When dotted with d, it equals 0.
30. 25 ? sin> (30°) ? 0.6 5 7.50 J
>
31. >a. a ? b 5 6 2 5 2 1 5 0
b. a with the x-axis:
>
>
0 a 0 5 "4 1 25 1 1 5 "30
2
cos (a) 5
"30
>
a with the y-axis:
5
cos (b) 5
"30
>
a with the z-axis:
21
cos (g) 5
"30
>
@ b> @ 5 "9 1 1 1 1 5 "11
b with the x-axis:
3
cos (a) 5
"11
7-29
>
b with the y-axis:
21
cos (b) 5
"11
>
b with the z-axis:
1
cos (g) 5
"11
6
5
1
>
>
c. m 1 ? m 2 5
2
2
50
!330
!330
!330
32. Need to show that the magnitudes of the
diagonals are equal to show that it is a rectangle.
>
>
>
@ 3i 1 3j 1 10k @ 5 "9 1 9 1 100 5 "118
@ 2i 1 9j 2 6k @ 5 "1 1 81 1 36 5 "118
33. a. Direction cosine for x-axis:
>
>
>
"3
2
We know the identity
cos2 a 1 cos2 b 1 cos2 g 5 1.
Since a 5 30g, and b 5 g, we get
3
2 cos2 b 5 1 2
4
1
cos b 5 cos g 5 6
2"2
"3
cos a 5
2
So there are two possibilities, depending upon
whether b 5 g is acute or obtuse.
b. If g is acute, then
1
cos g 5
2"2
g 8 69.3°
If Á is obtuse, then
1
cos g 5
2"2
g 8 110.7°
1
> >
> >
34. a ? b 5 0 a 0 @ b @ cos (u) 5
2
>
>
>
>
(a 2 3b ) ? (ma
)
5
0
1
b
>
>
> >
> >
> >
ma ? a 1 a ? b 2 3ma ? b 2 3b ? b 5 0
1
3
6
m1 2 m2 50
2
2
2
1
5
2 m5
2
2
m 5 25
> >
a ? b> 5 0 2 20 1 12 5 28
35.
>
a 1 b 5 (21, 21, 28)
>
>
@ a 1 b @ 5 "1 1 1 1 64 5 "66
cos (30°) 5
7-30
>
>
a 2 b 5 (1, 9, 24)
>
>
@ a 2 b @ 5 "1 1 81 1 16 5 "98
>
>
1 >
66
98
1 >
@a 1 b@ 2 2 @a 2 b@ 2 5
2
5 28
4
4
4
4
>
>
>
36.> c 5 b >2 a >
0 c 02 5 @b 2 a @ 2
>
>
>
>
5 (b 2 a ) ? (b 2 a )
>
>
>
>
> >
> >
5b?b2a?b1a?a2a?b
>
>
> >
5 0 a 0 2 1 @ b @ 2 2 2a ? b
>
>
> >
5 0 a 0 2 1 @ b @ 2 2 2 0 a 0 @ b @ cos u
>
37. AB> 5 (2, 0, 4)
@ AB @ 5 "4 1 0 1 16 5 2"5
>
@ AC @ 5 (1, 0, 2)
>
@ AC @ 5 "1 1 0 1 4 5 "5
>
BC 5 (21, 0, 22)
>
@ BC @ 5 "1 1 0 1 4 5 "5
>
cos A 5
>
AB ? AC
@ AB @ @ AC @
>
>
10
10
51
But this means that angle A 5 0°, so that this
triangle is degenerate. For completeness, though,
>
>
>
>
notice that BC 5 2AC and AB 5 2 AC . This
means that point C sits at the midpoint of the line
segment joining A and B. So angle
C 5 180° and angle B 5 0°. So
cos B 5 1;
cos C 5 21.
The area of triangle ABC is, of course, 0.
5
Chapter 7 Test, p. 422
>
>
1. a. We use the diagram to calculate a 3 b, noting
a1 5 21, a2 5 1, a3 5 1 and b1 5 2, b2 5 1,
b3 5 23.>
>
b
a
1
1
x
x 5 1(23) 2 1(1) 5 24
23
1
y
y 5 1(2) 2 (21)(23) 5 21
21
2
z
z 5 21(1) 2 1(2) 5 23
1
1>
>
So, a 3 b 5 (24, 21, 23)
Chapter 7: Applications of Vectors
>
b with the y-axis:
21
cos (b) 5
"11
>
b with the z-axis:
1
cos (g) 5
"11
6
5
1
>
>
c. m 1 ? m 2 5
2
2
50
!330
!330
!330
32. Need to show that the magnitudes of the
diagonals are equal to show that it is a rectangle.
>
>
>
@ 3i 1 3j 1 10k @ 5 "9 1 9 1 100 5 "118
@ 2i 1 9j 2 6k @ 5 "1 1 81 1 36 5 "118
33. a. Direction cosine for x-axis:
>
>
>
"3
2
We know the identity
cos2 a 1 cos2 b 1 cos2 g 5 1.
Since a 5 30g, and b 5 g, we get
3
2 cos2 b 5 1 2
4
1
cos b 5 cos g 5 6
2"2
"3
cos a 5
2
So there are two possibilities, depending upon
whether b 5 g is acute or obtuse.
b. If g is acute, then
1
cos g 5
2"2
g 8 69.3°
If Á is obtuse, then
1
cos g 5
2"2
g 8 110.7°
1
> >
> >
34. a ? b 5 0 a 0 @ b @ cos (u) 5
2
>
>
>
>
(a 2 3b ) ? (ma
)
5
0
1
b
>
>
> >
> >
> >
ma ? a 1 a ? b 2 3ma ? b 2 3b ? b 5 0
1
3
6
m1 2 m2 50
2
2
2
1
5
2 m5
2
2
m 5 25
> >
a ? b> 5 0 2 20 1 12 5 28
35.
>
a 1 b 5 (21, 21, 28)
>
>
@ a 1 b @ 5 "1 1 1 1 64 5 "66
cos (30°) 5
7-30
>
>
a 2 b 5 (1, 9, 24)
>
>
@ a 2 b @ 5 "1 1 81 1 16 5 "98
>
>
1 >
66
98
1 >
@a 1 b@ 2 2 @a 2 b@ 2 5
2
5 28
4
4
4
4
>
>
>
36.> c 5 b >2 a >
0 c 02 5 @b 2 a @ 2
>
>
>
>
5 (b 2 a ) ? (b 2 a )
>
>
>
>
> >
> >
5b?b2a?b1a?a2a?b
>
>
> >
5 0 a 0 2 1 @ b @ 2 2 2a ? b
>
>
> >
5 0 a 0 2 1 @ b @ 2 2 2 0 a 0 @ b @ cos u
>
37. AB> 5 (2, 0, 4)
@ AB @ 5 "4 1 0 1 16 5 2"5
>
@ AC @ 5 (1, 0, 2)
>
@ AC @ 5 "1 1 0 1 4 5 "5
>
BC 5 (21, 0, 22)
>
@ BC @ 5 "1 1 0 1 4 5 "5
>
cos A 5
>
AB ? AC
@ AB @ @ AC @
>
>
10
10
51
But this means that angle A 5 0°, so that this
triangle is degenerate. For completeness, though,
>
>
>
>
notice that BC 5 2AC and AB 5 2 AC . This
means that point C sits at the midpoint of the line
segment joining A and B. So angle
C 5 180° and angle B 5 0°. So
cos B 5 1;
cos C 5 21.
The area of triangle ABC is, of course, 0.
5
Chapter 7 Test, p. 422
>
>
1. a. We use the diagram to calculate a 3 b, noting
a1 5 21, a2 5 1, a3 5 1 and b1 5 2, b2 5 1,
b3 5 23.>
>
b
a
1
1
x
x 5 1(23) 2 1(1) 5 24
23
1
y
y 5 1(2) 2 (21)(23) 5 21
21
2
z
z 5 21(1) 2 1(2) 5 23
1
1>
>
So, a 3 b 5 (24, 21, 23)
Chapter 7: Applications of Vectors
b. We
use the diagram again:
>
>
b
1
x
23
x 5 1(27) 2 (23)(1) 5 24
27
y
2
z
c. The area of the parallelogram is the magnitude of
c
1
y 5 23(5) 2 (2)(27) 5 21
5
z 5 2(1) 2 1(5) 5 23
1 > 1
>
So, b 3 >c 5 (24, 21, 23)
>
>
c. a ? (b 3 c ) 5 (21, 1, 1) ? (24, 21, 23)
5 (21)(24) 1 (1)(21)
1 (1)(23)
50
d. We could use the diagram
method again, or, we
> > >
>
>
>
note that for any vectors x , y , x 3 y 5 2 (y 3 x ),
>
>
>
>
so letting y 5 x , we have
x> 3 x 5 0 from the last
>
>
>
equation. Since a 3 b 5 b 3> c from
> the> first two
>
parts of the problem, (a 3 b ) 3 (b 3 c ) 5 0.
>
2. a.> To find the scalar and vector
projections of a
>
>
>
>
>
@
@
on b, we
need
to
calculate
and
a
b
5
"b
?
b
?
b
> >
a ? b 5 (1, 21, 1) ? (2, 21, 22)
5 (1)(2) 1 (21)(21) 1 (1)(22)
> 51
@ b @ 5 "22 1 (21)2 1 (22)2
> 53
So, @ b @ 5 3
>
>
> >
The scalar projection of a on b is
a?b
1
> 5 3 , and
@b@
>
>
the
vector
projection
of
on
is
a
b
>
a
>
a?b
@b@
>2
bb 5 19 (2, 21, 22).
>
>
b. We find the direction cosines for b:
b1
2
3
@b@
a 8 48.2°.
b
21
cos (b) 5 2> 5
3
@b@
b 8 109.5°.
b
22
cos (g) 5 3> 5
3
@b@
g 8 131.8°.
cos (a) 5
>
5
Calculus and Vectors Solutions Manual
the cross >product.
>
b
a
21
21
x
x 5 (21)(22) 2 1(21) 5 3
1
22
y
y 5 1(2) 2 (1)(22) 5 4
1
2
z
z 5 (1)(21) 2 (21)(2) 5 1
21
21
>
>
So, a 3 b 5 (3, 4, 1) and thus,
>
>
@ a 3 b @ 5 "32 1 42 1 12
5 "26
>
>
So the area of the parallelogram formed by a and b
is "26 or 5.10 square units.
3. We first draw a diagram documenting the
situation:
E
F
120°
40 N
40 N
R
60°
60°
50 N
D
50 N
G
In triangle DEF, we use the cosine law:
>
@ R @ 5 "402 1 502 2 2(40)(50) cos (120°)
>
@ R @ 8 78.10
We now use the sine law to find /EDF:
sin /EDF
sin /DEF
>
>
5
@ EF @
@R@
sin /EDF
sin 120°
8
50
78.10
sin /EDF 8 0.5544
/EDF 8 33.7°
The equilibrant force is equal in magnitude and
opposite in direction to the resultant force, so both
forces have a magnitude of 78.10 N. The resultant
makes an angle 33.7° to the 40 N force and 26.3° to
the 50 N force. The equilibrant makes an angle 146.3°
to the 40 N force and 153.7° to the 50 N force.
7-31
4. We find the resultant velocity of the airplane.
E
F
F
R
1000 km/h
G
D
Vector diagram
Position diagram
Since the airplane’s velocity is perpendicular to the
wind, the resultant’s magnitude is given by the
Pythagorean theorem:
>
@ R @ 5 "10002 1 1002
>
@ R @ 8 1004.99
The angle is determined using the tangent ratio:
100
tan /EDF 5
1000
/EDF 8 5.7°
Thus, the resultant velocity is 1004.99 km> h,
N 5.7° W (or W 84.3° N).
5. a. The canoeist will travel 200 m across the
stream, so the total time he will paddle is:
d
t5
rcanoeist
200 m
t5
2.5 m>s
t 5 80 s
The current is flowing 1.2 m> s downstream, so the
distance that the canoeist travels downstream is:
d 5 rcurrent 3 t
d 5 (1.2 m>s)(80 s)
d 5 96 m
So, the canoeist will drift 96 m south.
b. In order to arrive directly across stream, the
canoeist must take into account the change in his
velocity caused by the current. That is, he must
initially paddle upstream in a direction such that
the resultant velocity is directed straight across
the stream. The resultant velocity:
E
2.5 m/s
D
7-32
R
Since the resultant velocity is perpendicular to the
current, the direction in which the canoeist should
head is determined by the sine ratio.
1.2
sin /EDF 5
2.5
/EDF 8 28.7°
The canoeist should head 28.7° upstream.
6. The area of the triangle is exactly:
>
>
1
ADABC 5 @ AB 3 BC @
2
>
AB 5 (2, 1, 3) 2 (21, 3, 5)
5 (3, 22, 22)
>
BC 5 (21, 1, 4) 2 (2, 1, 3)
5 (23, 0, 1)
>
>
BC
AB
22 x 0
x 5 (22)(1) 2 (22)(0) 5 22
22 y 1
y 5 (22)(23) 2 (3)(1) 5 3
3 z 23
z 5 (3)(0) 2 (22)(23) 5 26
22
0
>
>
So, AB 3 BC 5 (22, 3, 26) and
>
>
@ AB 3 BC @ 5 "(22)2 1 32 1 (26)2
5 "49
57
>
>
1
7
So, ADABC 5 @ AB 3 BC @ 5 .
2
2
The area of the triangle is 3.50 square units.
7.
458
T1
708
T2
25 kg
The system is in equilibrium (i.e. it is not moving),
>
so we> know that the horizontal components of T1
and T2 are equal:
@ T1 @ sin (45°) 5 @ T2 @ sin (70°)
>
>
@ T2 @ 5
>
sin (45°) >
@T @
sin (70°) 1
1.2 m/s
F
Chapter 7: Applications of Vectors
>
>
Also, the vertical component of T1 1 T2 must equal
the> gravitational > force on the block:
@ T1 @ cos 45° 1 @ T2 @ cos 70° 5 (25 kg)(9.8 m>s2 )
>
Substituting in for T2, we find that:
>
@ T1 @ cos 45° 1
> sin 45°
@ T1 @
cos 70° 5 (25 kg) (9.8 m>s 2 )
sin 70°
>
sin 45°
@ T1 @ acos 45° 1
cos 70°b 5 245 N
sin 70°
@ T1 @ (0.9645) 8 245 N
>
@ T1 @ 8 254.0 N
>
So, we can now find
>
sin (45°) >
@ T2 @ 5
@T @
sin (70°) 1
>
sin (45°)
@ T2 @ 8
(254.0 N)
sin (70°)
>
@ T2 @ 8 191.1 N
The direction of the tensions are indicated in the
diagram.
8. a. We explicitly calculate both sides of the
equation. The left side is:
> >
x ? y 5 (3, 3, 1) ? (21, 2, 23)
5 (3)(21) 1 (3)(2) 1 (1)(23)
50
We perform a few computations before computing
the right side:
>
>
x 1 y 5 (3, 3, 1) 1 (21, 2, 23)
5 (2, 5, 22)
Calculus and Vectors Solutions Manual
>
>
>
>
>
>
0 x 1 y 0 2 5 (x 1 y ) ? (x 1 y )
5 22 1 52 1 (22)2
5 33
>
>
x 2 y 5 (3, 3, 1) 2 (21, 2, 23)
5 (4, 1, 4)
>
>2
>
>
>
>
0 x 2 y 0 5 (x 2 y ) ? (x 2 y )
5 42 1 12 1 42
5 33
Thus, the right side is
1 >
1
1
1 >
>
>
0 x 1 y 0 2 2 0 x 2 y 0 2 5 (33) 2 (33)
4
4
4
4
50
So, the equation holds for these vectors.
b. We now verify that the formula holds in general.
We will compute the right side of the equation, but
we first perform some intermediary computations:
>
>
>
>
>
>
0 x 1 y 0 2 5 (x 1 y ) ? (x 1 y )
> >
> >
> >
> >
5 (x ? x ) 1 (x ? y ) 1 (y ? x ) 1 (y ? y )
> >
> >
> >
5 (x ? x ) 1 2(x ? y ) 1 (y ? y )
>
>
>
>
>
>
0 x 2 y 0 2 5 (x 2 y ) ? (x 2 y )
> >
>
>
> >
5 (x ? x ) 1 (x ? 2y ) 1 (2y ? x )
>
>
1 (2y ? 2y )
> >
> >
> >
5 (x ? x ) 2 2(x ? y ) 1 (y ? y )
So, the right side of the equation is:
1 >
1
1 >
>
>
> >
0 x 1 y 0 2 2 0 x 2 y 0 2 5 (4(x ? y ))
4
4
4
> >
5x?y
Thus, the equation holds for arbitrary vectors.
7-33