Sm chapter3

3
Vectors
CHAPTER OUTLINE
3.1 Coordinate Systems
3.2 Vector and Scalar Quantities
3.3 Some Properties of Vectors
3.4 Components of a Vector and
Unit Vectors
ANSWERS TO QUESTIONS
Q3.1 Only force and velocity are vectors. None of the other
quantities requires a direction to be described. The
answers are (a) yes (b) no (c) no (d) no (e) no
(f) yes (g) no.
Q3.2 The book’s displacement is zero, as it ends up at the
point from which it started. The distance traveled is
6.0 meters.
*Q3.3 The vector −2D1
will be twice as long as D1 and in the
opposite direction, namely northeast. Adding D2, which
is about equally long and southwest, we get a sum that is
still longer and due east, choice (a).
*Q3.4 The magnitudes of the vectors being added are constant,
and we are considering the magnitude only—not the
direction—of the resultant. So we need look only at the
angle between the vectors being added in each case. The
smaller this angle, the larger the resultant magnitude.
Thus the ranking is c = e > a > d > b.
*Q3.5 (a) leftward: negative. (b) upward: positive (c) rightward: positive (d) downward: negative
(e) Depending on the signs and angles of A and B, the sum could be in any quadrant. (f) Now
−A will be in the fourth quadrant, so − +A B will be in the fourth quadrant.
*Q3.6 (i) The magnitude is 10 102 2
+ m sր , answer (f). (ii) Having no y component means answer (a).
*Q3.7 The vertical component is opposite the 30° angle, so sin 30° = (vertical component)/50 m and the
answer is (h).
*Q3.8 Take the difference of the coordinates of the ends of the vector. Final first means head end first.
(i) −4 − 2 = −6 cm, answer (j) (ii) 1 − (−2) = 3 cm, answer (c)
Q3.9 (i) If the direction-angle of A is between 180 degrees and 270 degrees, its components are both
negative: answer (c). If a vector is in the second quadrant or the fourth quadrant, its components
have opposite signs: answer (b) or (d).
Q3.10 Vectors A and B are perpendicular to each other.
Q3.11 No, the magnitude of a vector is always positive. A minus sign in a vector only indicates
direction, not magnitude.
Q3.12 Addition of a vector to a scalar is not defined. Think of numbers of apples and of clouds.
45
13794_03_ch03_p045-064.indd 4513794_03_ch03_p045-064.indd 45 11/28/06 4:40:06 PM11/28/06 4:40:06 PM

SOLUTIONS TO PROBLEMS
Section 3.1 Coordinate Systems
P3.1 x r= = ( ) = ( ) −( ) = −cos cos .θ 5 50 240 5 50 0 5 2 7. m . m .° 55 m
y r= = ( ) = ( ) −( ) = −sin sinθ 5 50 240 5 50 0 8 4. m . m . 66° ..76 m
P3.2 (a) x r= cosθ and y r= sinθ, therefore
x1 2 50 30 0= ( ). m .cos °, y1 2 50 30 0= ( ). m .sin °, and
x y1 1 2 17 1 25, . , . m( ) = ( )
x2 3 80 120= ( ) °. cosm , y2 3 80 120= ( ) °. sinm , and
x y2 2 1 90 3 29, . , . m( ) = −( )
(b) d x y= + = + =( ) ( ) . . .∆ ∆2 2 2 2
4 07 2 04 4 55m m
P3.3 The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m.
(a) We can use the Pythagorean theorem to find the distance from the origin to the fly.
distance m m m2
= + = ( ) + ( ) = =x y2 2 2 2
2 00 1 00 5 00. . . 22 24. m
(b) θ =
⎛
⎝⎜
⎞
⎠⎟ = °−
tan .1 1
2
26 6 ; r = °2 24 26 6. , .m
P3.4 We have 2 00 30 0. .= °r cos
r =
°
=
2 00
30 0
2 31
.
cos .
.
and y r= = =sin sin .30 0 2 31 30 0 1 15. . .° ° .
P3.5 We have r x y= +2 2
and θ =
⎛
⎝⎜
⎞
⎠⎟
−
tan 1 y
x
.
(a) The radius for this new point is
−( ) + = + =x y x y r
2 2 2 2
and its angle is
tan−
−
⎛
⎝
⎞
⎠
=1
180
y
x
° −θ .
(b) ( ) ( )− + − =2 2 22 2
x y r This point is in the third quadrant if x y,( ) is in the first quadrant
or in the fourth quadrant if x y,( ) is in the second quadrant. It is at an angle of 180° +θ .
(c) ( ) ( )3 3 32 2
x y r+ − = This point is in the fourth quadrant if x y,( ) is in the first quadrant
or in the third quadrant if x y,( ) is in the second quadrant. It is at an angle of −θ .
46 Chapter 3

Vectors 47
Section 3.2 Vector and Scalar Quantities
Section 3.3 Some Properties of Vectors
P3.6 − =R 310 km at 57 S of W°
(Scale: 1 unit = 20 km)
P3.7 tan .
tan . .
35 0
100
100 35 0 70 0
° =
= ( ) ° =
x
x
m
m m
P3.8 Find the resultant F F1 2+ graphically by placing the tail of F2 at the head of F1. The resultant
force vector F F1 2+ is of magnitude 9 5. N and at an angle of 57° abovethe axisx .
FIG. P3.6
FIG. P3.7
0 1 2 3 N
x
y
2
1
F1 2+F
F
F
FIG. P3.8

48 Chapter 3
P3.9 (a) d i= − =10 0 10 0. ˆ . m since the displacement is in a
straight line from point A to point B.
(b) The actual distance skated is not equal to the straight-line
displacement. The distance follows the curved path of the
semi-circle (ACB).
s r= ( ) = =
1
2
2 5 15 7π π . m
(c) If the circle is complete, d begins and ends at point A. Hence, d = 0 .
P3.10 (a) The large majority of people are standing or sitting at this hour. Their instantaneous foot-
to-head vectors have upward vertical components on the order of 1 m and randomly
oriented horizontal components. The citywide sum will be ~ 105
m upward .
(b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented
north, south, east, and west quite at random. Then the horizontal component of their total
vector height is very nearly zero. If their compressed pillows give their height vectors
vertical components averaging 3 cm, and if one-tenth of one percent of the population are
on-duty nurses or police ofﬁcers, we estimate the total vector height as
~ . m m10 0 03 10 15 2
( )+ ( ) ~ 103
m upward .
P3.11 To ﬁnd these vector expressions graphically,
we draw each set of vectors. Measurements of
the results are taken using a ruler and
protractor. (Scale: 1 0 5unit m= . )
(a) A + B = 5.2 m at 60°
(b) A − B = 3.0 m at 330°
(c) B − A = 3.0 m at 150°
(d) A − 2B = 5.2 m at 300°
C
B A
5.00 m
d
FIG. P3.9
FIG. P3.11

Vectors 49
P3.12 The three diagrams shown below represent the graphical solutions for the three vector sums:
R A B C1 = + + , R B C A2 = + + , and R C B A3 = + + . We observe that R R R1 2 3= = ,
illustrating that
the sum of a set of vectors is not affected by the order in which the vectors are added .
2
B
A
R1
C
C C
R 3R
A
A
B
B
100 m
FIG. P3.12
P3.13 The scale drawing for the graphical
solution should be similar to the
figure to the right. The magnitude and
direction of the final displacement
from the starting point are obtained
by measuring d and θ on the drawing
and applying the scale factor used in
making the drawing. The results should be
d = = −420 3ft and θ ° .
Section 3.4 Components of a Vector and Unit Vectors
*P3.14 We assume the floor is level. Take the x axis in the direction of the first displacement.
If both of the 90° turns are to the right or both to the left , the displacements add like
40 0 15 0 20 0 20 0 15 0. ˆ . ˆ . ˆ . ˆ .m m mi i i+ − = +j ˆˆj( )m
to give (a) displacement magnitude (202
+ 152
)1ր2
m = 25.0 m
at (b) tan−1
(15ր20) = 36.9° .
If one turn is right and the other is left , the displacements add like
40 0 15 0 20 0 60 0 15 0. ˆ . ˆ . ˆ . ˆ .m m mi i+ + = +j i ˆˆj( )m
to give (a) displacement magnitude (602
+ 152
)1ր2
m = 61.8 m
at (b) tan−1
(15ր60) = 14.0˚. Just two answers are possible.
FIG. P3.13

50 Chapter 3
P3.15 A
A
A A A
x
y
x y
= −
=
= + = −( ) + ( ) =
25 0
40 0
25 0 40 0 42 2 2 2
.
.
. . 77 2. units
We observe that
tanφ =
A
A
y
x
.
So
φ =
⎛
⎝
⎜
⎞
⎠
⎟ = = ( ) =− −
tan tan
.
.
tan .1 140 0
25 0
1 60 5
A
A
y
x
88 0. °.
The diagram shows that the angle from the +x axis can be found by subtracting from 180°:
θ = − =180 58 122° ° ° .
P3.16 The person would have to walk 3 10 1 31. 25.0 km northsin .°( ) = , and
3 10 25 0 2 81. . km eastcos .°( ) = .
*P3.17 Let v represent the speed of the camper. The northward component of its velocity is v cos 8.5°.
To avoid crowding the minivan we require v cos 8.5° ≥ 28 mրs.
We can satisfy this requirement simply by taking v ≥ (28 mրs)րcos 8.5° = 28.3 mրs.
P3.18 (a) Her net x (east-west) displacement is − + + = +3 00 0 6 00 3 00. . . blocks, while her net
y (north-south) displacement is 0 4 00 0 4 00+ + = +. . blocks. The magnitude of the
resultant displacement is
R x y= ( ) + ( ) = ( ) + ( ) =net net
2 2 2 2
3 00 4 00 5 00. . . blocks
and the angle the resultant makes with the x axis (eastward direction) is
θ =
⎛
⎝⎜
⎞
⎠⎟ = ( ) = °− −
tan
.
.
tan . .1 14 00
3 00
1 33 53 1 .
The resultant displacement is then 5 00 53 1. .blocks at N of E° .
(b) The total distance traveled is 3 00 4 00 6 00 13 0. . . .+ + = blocks .
P3.19 x r= cosθ and y r= sinθ, therefore:
(a) x = °12 8 150. cos , y = °12 8 150. sin , and x y, . . m( ) = − +( )111 6 40ˆ î j
(b) x = °3 30 60 0. cos . , y = °3 30 60 0. sin . , and x y, cm( ) = +( )1 65 2 86. ˆ . î j
(c) x = °22 0 215. cos , y = °22 0 215. sin , and x y, in( ) = − −( )18 0 12 6. ˆ . î j
P3.20 x d= = ( ) ( ) = −cos cosθ 50 0 120 25 0. m . m
y d= = ( ) ( ) =
= −( )
sin sin .
.
θ 50 0 120 43 3
25 0
. m m
md ˆˆ . î j+ ( )43 3 m
FIG. P3.15

Vectors 51
P3.21 Let +x be East and +y be North.
x
y
∑
∑
= =
= − = −
250 125 30 358
75 125 30 150
+
+
cos
sin
°
°
m
112 5
358 12 5 358
2 2 2 2
. m
md x y= ( ) + ( ) = ( ) + −( ) =∑ ∑ .
taan
.
θ
θ
=
( )
( )
= − = −
= −
=
∑
∑
y
x
12 5
358
0 0349
2 00
35
.
. °
d 88 2 00m at S of E. °
P3.22 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums
of the east and north components of the displacements from Dallas to Atlanta (A) and from
Atlanta to Chicago. In equation form:
d d dDC east DA east AC east .= =+ −730 5 00 560cos s° iin 21 0 527. miles.
DC north DA north AC
° =
=d d d+ nnorth . . miles= =730 5 00 560 21 0 586sin cos° °+
By the Pythagorean theorem, d d d= + =( ) ( )DC east DC north mi2 2
788 .
Then tanθ = =
d
d
DC north
DC east
.111 and θ = °48 0. .
Thus, Chicago is 788 miles at 48.0° northeast of Dallas .
P3.23 We have B R A= − :
A
A
x
y
= = −
= =
150 75 0
150 120 130
cos120 cm
cm
°
°
.
sin
RR
R
x
y
= =
= =
140 35 0 115
140 35 0 80 3
cos .
sin . .
°
°
cm
cmm
Therefore,
B i j i j= − −( )[ ] + −[ ] = −(115 75 80 3 130 190 49 7ˆ . ˆ ˆ . ˆ))
= + =
= −⎛
⎝
⎞−
cm
cmB 190 49 7 196
49 7
190
2 2
1
.
tan
.
θ
⎠⎠
= −14 7. .°
P3.24 (a) See ﬁgure to the right.
(b) C A B i j i j i= + = + + − =2 00 6 00 3 00 2 00 5 00. ˆ . ˆ . ˆ . ˆ . ˆ ++ 4 00. ˆj
C = + ⎛
⎝
⎞
⎠
=−
25 0 16 0 6 401
. . tan .at
4
5
at 38.7°
D A B i j i j= − = + − + = −2 00 6 00 3 00 2 00 1 00. ˆ . ˆ . ˆ . ˆ . ˆii j+ 8 00. ˆ
D = −( ) + ( )
−
⎛
⎝⎜
⎞
⎠⎟
−
1 00 8 00
8 00
1 00
2 2 1
. . tan
.
.
at
D = −( ) =8 06 180 82 9 8 06 97 2. . . .at at° ° °
x
FIG. P3.23
FIG. P3.24

52 Chapter 3
P3.25 (a) A B i j i j i j+( )= −( )+ − −( )= −3 2 4 2 6ˆ ˆ ˆ ˆ ˆ ˆ
(b) A B i j i j i j−( )= −( )− − −( )= +3 2 4 4 2ˆ ˆ ˆ ˆ ˆ ˆ
(c) A B+ = + =2 6 6 322 2
.
(d) A B− = + =4 2 4 472 2
.
(e) θA B+
−
= −⎛
⎝
⎞
⎠
= − =tan .1 6
2
71 6 288° °
θA B−
−
= ⎛
⎝
⎞
⎠
=tan .1 2
4
26 6°
*P3.26 We take the x axis along the slope uphill. Students, get used to this choice! The y axis is perpen-
dicular to the slope, at 35° to the vertical. Then the displacement of the snow makes an angle of
90° − 35° − 20° = 35° with the x axis.
(a) Its component parallel to the surface is 5 m cos 35° = 4.10 m toward the top of the hill .
(b) Its component perpendicular to the surface is 5 m sin 35° = 2.87 m .
P3.27 d j1 3 50= −( ). ˆ m
d i j i2 8 20 45 0 8 20 45 0 5 80 5= + = +. cos . ˆ . sin . ˆ . ˆ .° ° 880ˆj( ) m
d i3 15 0= −( ). ˆ m
R d d d i j= + + = − +( ) + −( )1 2 3 15 0 5 80 5 80 3 50. . ˆ . . ˆ == − +( )9 20 2 30. ˆ . ˆi j m
(or 9.20 m west and 2.30 m north)
The magnitude of the resultant displacement is
R = + = −( ) + ( ) =R Rx y
2 2 2 2
9 20 2 30 9 48. . . m .
The direction is θ =
−
⎛
⎝
⎞
⎠
=arctan
.
.
2 30
9 20
166° .
P3.28 Refer to the sketch
R A C= + + = − − +
= −
B i j i
i
10 0 15 0 50 0
40 0 15
. ˆ . ˆ . ˆ
. ˆ .. ˆ
. . .
/
0
40 0 15 0 42 7
2 2 1 2
j
R = ( ) + −( )⎡⎣ ⎤⎦ = yards
P3.29 East North
x y
0 m 4.00 m
1.41 1.41
–0.500 –0.866
+0.914 4.55
R = + =x y y x
2 2
4 64at tan m at 78.6 N of1−
( ) .ր ° E
A = 0.10
B = 15 0.
C = 50 0.
R
FIG. P3.28

Vectors 53
P3.30 A i j= − +8 70 15 0. ˆ . ˆ and B i j= −13 2 6 60. ˆ . ˆ
A B C− + =3 0: 3 21 9 21 6
7 30 7 20
C B A i j
C i j
= − = −
= −
. ˆ . ˆ
. ˆ . ˆ
or Cx = 7 30. cm ; Cy = −7 20. cm
P3.31 (a) F F F
F i
= +
= ( ) + ( )
1 2
120 60 0 120 60 0cos . ˆ sin . ˆ° ° jj i j
F
− ( ) + ( )
=
80 0 75 0 80 0 75 0
60 0
. cos . ˆ . sin . ˆ
.
° °
ˆˆ ˆ . ˆ . ˆ . ˆ ˆi j i j i j
F
+ − + = +( )
=
104 20 7 77 3 39 3 181 N
339 3 181 185
181
39 3
77 8
2 2
1
.
tan
.
.
+ =
= ⎛
⎝
⎞
⎠
=−
N
θ °
(b) F F i j3 39 3 181= − = − −( ). ˆ ˆ N
P3.32 A = 3 00. m, θA = °30 0. B = 3 00. m, θB = °90 0.
A Ax A= = ° =cos . cos . .θ 3 00 30 0 2 60 m A Ay A= = =sin . sin . .θ 3 00 30 0 1 50° m
A i j i j= + = +( )A Ax y
ˆ ˆ . ˆ . ˆ2 60 1 50 m
Bx = 0, By = 3 00. m so B j= 3 00. ˆ m
A B i j j i j+ = +( )+ = +(2 60 1 50 3 00 2 60 4 50. ˆ . ˆ . ˆ . ˆ . ˆ)) m
P3.33 B i j k i j k
B
= + + = + +
=
B B Bx y z
ˆ ˆ ˆ . ˆ . ˆ . ˆ
.
4 00 6 00 3 00
4 000 6 00 3 00 7 81
4 00
7 81
59
2 2 2
1
+ + =
= ⎛
⎝
⎞
⎠
=−
. . .
cos
.
.
α ..
cos
.
2
6 01
° is the angle with the axisx
β = − 00
7 81
39 8
.
.⎛
⎝
⎞
⎠
= ° is the angle with the axy iis
is the angle wγ = ⎛
⎝
⎞
⎠
=−
cos
.
.
.1 3 00
7 81
67 4° iith the axisz
P3.34 (a) D A B C i j= + + = −2 2ˆ ˆ
D = + = =2 2 2 83 3152 2
. m at θ °
(b) E A B C i j= − − + = − +6 12ˆ ˆ
E = + = =6 12 13 4 1172 2
. m at θ °
P3.35 (a) C A B i j k= + = − −( )5 00 1 00 3 00. ˆ . ˆ . ˆ m
C = ( ) + ( ) + ( ) =5 00 1 00 3 00 5 92
2 2 2
. . . .m m
(b) D A B i j k= − = − +( )2 4 00 11 0 15 0. ˆ . ˆ . ˆ m
D = ( ) + ( ) + ( ) =4 00 11 0 15 0 19 0
2 2 2
. . . .m m

P3.36 Let the positive x-direction be eastward, the positive y-direction be vertically upward, and the
positive z-direction be southward. The total displacement is then
d i j j k= +( ) + −( ) =4 80 4 80 3 70 3 70 4. ˆ . ˆ . ˆ . ˆ .cm cm 880 8 50 3 70ˆ . ˆ . ˆi j k+ −( ) cm.
(a) The magnitude is d = ( ) + ( ) + −( ) =4 80 8 50 3 70 10 4
2 2 2
. . . .cm cm .
(b) Its angle with the y axis follows from cos
.
.
θ =
8 50
10 4
, giving θ = 35 5. ° .
P3.37 (a) A i j k= + −8 00 12 0 4 00. ˆ . ˆ . ˆ
(b) B
A
i j k= = + −
4
2 00 3 00 1 00. ˆ . ˆ . ˆ
(c) C A i j k= − = − − +3 24 0 36 0 12 0. ˆ . ˆ . ˆ
P3.38 The y coordinate of the airplane is constant and equal to 7 60 103
. × m whereas the x coordinate is
given by x ti= v where vi is the constant speed in the horizontal direction.
At t = 30 0. s we have x = ×8 04 103
. , so vi = =8 40 2680 m 30 s m s. The position
vector as a function of time is
P i j= ( ) + ×( )268 7 60 103
m s mtˆ . ˆ.
At t = 45 0. s, P i j= × + ×⎡
⎣
⎤
⎦1 21 10 7 60 104 3
. ˆ . ˆ m. The magnitude is
P = ×( ) + ×( ) = ×1 21 10 7 60 10 1 43 104 2 3 2 4
. . .m m
and the direction is
θ =
×
×
⎛
⎝⎜
⎞
⎠⎟ =arctan
.
.
.
7 60 10
1 21 10
32 2
3
4
° above tthe horizontal .
P3.39 The position vector from radar station to ship is
S i j i= +( ) = −17 3 136 17 3 136 12 0. sin ˆ . cos ˆ . ˆ° ° km 112 4. ˆj( ) km.
From station to plane, the position vector is
P i j k= + +( )19 6 153 19 6 153 2 20. sin ˆ . cos ˆ . ˆ° ° km,
or
P i j k= − +( )8 90 17 5 2 20. ˆ . ˆ . ˆ km.
(a) To ﬂy to the ship, the plane must undergo displacement
D S P i j k= − = + −( )3 12 5 02 2 20. ˆ . ˆ . ˆ km .
(b) The distance the plane must travel is
D = = ( ) + ( ) + ( ) =D 3 12 5 02 2 20 6 31
2 2 2
. . . .km km .
54 Chapter 354 Chapter 3

Vectors 55
55
P3.40 (a) E i j= ( ) + ( )17 0 27 0 17 0 27 0. cos . ˆ . sin . ˆcm cm° °
E i j= +( )15 1 7 72. ˆ . ˆ cm
(b) F i j= −( ) + ( )17 0 27 0 17 0 27 0. sin . ˆ . cos . ˆcm cm° °
F i j= − +( )7 72 15 1. ˆ . ˆ cm Note that we do not need to
explicitly identify the angle with the positive x axis.
(c) G i j= +( ) + ( )17 0 27 0 17 0 27 0. sin . ˆ . cos . ˆcm cm° °
G i j= + +( )7 72 15 1. ˆ . ˆ cm
P3.41 Ax = −3 00. , Ay = 2 00.
(a) A i j i j= + = − +A Ax y
ˆ ˆ . ˆ . ˆ3 00 2 00
(b) A = + = −( ) + ( ) =A Ax y
2 2 2 2
3 00 2 00 3 61. . .
tan
.
.
.θ = =
−( )
= −
A
A
y
x
2 00
3 00
0 667, tan . .−
−( ) = − °1
0 667 33 7
θ is in the 2nd
quadrant, so θ = −( ) =180 33 7 146° ° °+ . .
(c) Rx = 0, Ry = −4 00. , R A B= + thus B R A= − and
B R Ax x x= − = − −( ) =0 3 00 3 00. . , B R Ay y y= − = − − = −4 00 2 00 6 00. . . .
Therefore, B i j= −3 00 6 00. ˆ . ˆ .
P3.42 The hurricane’s ﬁrst displacement is
41 0
3 00
.
.
km
h
h
⎛
⎝⎜
⎞
⎠⎟ ( ) at 60 0. ° N of W, and its second
displacement is
25 0
1 50
.
.
km
h
h
⎛
⎝⎜
⎞
⎠⎟ ( ) due North. With ˆi representing east and ˆj representing
north, its total displacement is:
41 0 60 0 3 00 41 0. cos . . ˆ . sin
km
h
h
km
h
°⎛
⎝
⎞
⎠
( ) −( )+i 660 0 3 00 25 0 1 50. . ˆ . . ˆ°⎛
⎝
⎞
⎠
( ) + ⎛
⎝
⎞
⎠
( )
=
h
km
h
hj j
661 5 144. ˆ ˆkm km−( )+i j
with magnitude 61 5 144 157
2 2
. km km km( ) + ( ) = .
P3.43 (a) R
R
x
y
= + =
=
40 0 45 0 30 0 45 0 49 5
40 0
. cos . . cos . .
. sin
° °
445 0 30 0 45 0 20 0 27 1
49 5 27 1
. . sin . . .
. ˆ .
° °− + =
= +R i ˆˆj
(b) R = ( ) + ( ) =
= ⎛
⎝
⎞
⎠
=−
49 5 27 1 56 4
27 1
49 5
2 2
1
. . .
tan
.
.
θ 228 7. °
F
y
x
27.0º
G
27.0º
E
27.0º
FIG. P3.40
A
y
x
B
45º
C
45º
O
FIG. P3.43

56 Chapter 3
*P3.44 (a) Taking components along î and ˆj, we get two equations:
6 00 8 00 26 0 0. . .a b− + =
and
− + + =8 00 3 00 19 0 0. . .a b .
We solve simultaneously by substituting a = 1.33 b − 4.33 to find −8(1.33 b − 4.33) + 3 b + 19 = 0
or 7.67b = 53.67 so b = 7.00 and a = 1.33(7) − 4.33.
Thus
a b= =5 00 7 00. , . .
Therefore,
5 00 7 00 0. .A B C+ + = .
(b) In order for vectors to be equal, all of their components must be equal. A vector equa-
tion contains more information than a scalar equation.
*P3.45 The displacement from the start to the finish is 16 î + 12 ˆj − (5 î + 3 ˆj) = (11 î + 9 ˆj) meters.
The displacement from the starting point to A is f(11 î + 9 ˆj) meters.
(a) The position vector of point A is 5 î + 3 ˆj + f(11 î + 9 ˆj) = (5 + 11f ) î + (3 + 9f ) ˆj meters .
(b) For f = 0 we have the position vector (5 + 0) î + (3 + 0) ˆj meters.
This is reasonable because it is the location of the starting point, 5 î + 3 ˆj meters.
(c) For f = 1 = 100%, we have position vector (5 + 11) î + (3 + 9) ˆj meters = 16 î + 12 ˆj meters .
This is reasonable because we have completed the trip and this is the position vector of the
endpoint.
*P3.46 We note that − =î west and − =ˆj south. The given mathematical representation of the trip can be
written as 6 3 4 3. b west b at 40 south of west b at+ +° 550 south of east b south° + 5 .
(a) (b) The total odometer distance is the sum of the
magnitudes of the four displacements:
6 3 4 3 5 18 3. .b b b b b+ + + = .
(c) R i= − − +( ) + − − −( )6 3 3 06 1 93 2 57 2 30 5. . . ˆ . . ˆb b jj
i j= − −
= ( ) + ( )
7 44 9 87
7 44 9 87
2 2
. ˆ . ˆ
. .
b b
b b at south of west
b at
tan
.
.
.
−
=
1 9 87
7 44
12 4 553.0 south of west
b at 233 counter
°
°= 12 4. cclockwise from east
R
= 1 block
EW
S
N
f
i

Vectors 57
Additional Problems
P3.47 Let θ represent the angle between the directions
of A and B. Since A and B have the same magnitudes,
A, B, and R A B= + form an isosceles triangle
in which the angles are 180° −θ, θ
2
, and θ
2
.
The magnitude of R is then R A=
⎛
⎝⎜
⎞
⎠⎟2
2
cos
θ
.
This can be seen from applying the law of cosines
to the isosceles triangle and using the fact that B A= .
Again, A, −B, and D A B= − form an isosceles triangle with apex angle θ.
Applying the law of cosines and the identity
1 2
2
2
−( ) =
⎛
⎝⎜
⎞
⎠⎟cos sinθ
θ
gives the magnitude of D as D A=
⎛
⎝⎜
⎞
⎠⎟2
2
sin
θ
.
The problem requires that R D= 100 .
Thus, 2
2
200
2
A Acos sin
θ θ⎛
⎝⎜
⎞
⎠⎟ =
⎛
⎝⎜
⎞
⎠⎟ . This gives tan .
θ
2
0 010
⎛
⎝⎜
⎞
⎠⎟ = and θ = 1 15. ° .
P3.48 Let θ represent the angle between the directions
of A and B. Since A and B have the same
magnitudes, A, B, and R A B= + form an isosceles
triangle in which the angles are 180° −θ, θ
2
, and
θ
2
.
The magnitude of R is then R A=
⎛
⎝⎜
⎞
⎠⎟2
2
cos
θ
. This can
be seen by applying the law of cosines to the isosceles
triangle and using the fact that B A= .
Again, A, –B, and D A B= − form an isosceles triangle with
apex angle θ. Applying the law of cosines and the identity
1 2
2
2
−( ) =
⎛
⎝⎜
⎞
⎠⎟cos sinθ
θ
gives the magnitude of D as D A=
⎛
⎝⎜
⎞
⎠⎟2
2
sin
θ
.
The problem requires that R nD= or cos sin
θ θ
2 2
⎛
⎝⎜
⎞
⎠⎟ =
⎛
⎝⎜
⎞
⎠⎟n giving θ =
⎛
⎝⎜
⎞
⎠⎟
−
2
11
tan
n
.
The larger R is to be compared to D, the smaller the angle between A and B becomes.
P3.49 The position vector from the ground under the controller of the ﬁrst airplane is
r i j1 19 2 25 19 2 25= ( )( ) + ( )( ). cos ˆ . sin ˆkm km° ° ++ ( )
= + +( )
0 8
17 4 8 11 0 8
. ˆ
. ˆ . ˆ . ˆ .
km
km
k
i j k
The second is at
r i j2 17 6 20 17 6 20= ( )( ) + ( )( ). cos ˆ . sin ˆkm km° ° ++ ( )
= + +( )
1
16 5 6 02 1 1
.1 km
km
ˆ
. ˆ . ˆ . ˆ .
k
i j k
Now the displacement from the ﬁrst plane to the second is
r r i j k2 1 0 863 2 09 0 3− = − − +( ). ˆ . ˆ . ˆ km
with magnitude
0 863 2 09 0 3 2 29
2 2 2
. . . .( ) + ( ) + ( ) = km .
A
B
R
θ
θ
θ
/2
A
D
-B
FIG. P3.47
FIG. P3.48

58 Chapter 3
P3.50 Take the x axis along the tail section of the snake. The displacement from tail to head is
240 420 240 180 105 180m m mˆ cos ˆ sii i+ −( ) −( ) −° ° nn ˆ ˆ ˆ75 287°j i j= −m 174 m .
Its magnitude is 287 174 335
2 2
( ) + ( ) =m m. From v =
distance
∆t
, the time for each child’s run is
Inge:
distance m h km s
∆t = =
( )( )( )
v
335 1 3 600
12 km m h
s
Olaf:
m s
3.
( )( )( )
=
=
⋅
1 000 1
101
420
∆t
333 m
s= 126 .
Inge wins by 126 101 25 4− = . s .
P3.51 Let A represent the distance from island 2 to island 3.
The displacement is A = A at 159°. Represent the
displacement from 3 to 1 as B = B at 298°. We have
4.76 km at 37 0° + + =A B .
For x components
4 76 37 159 298 0
3 80
. cos cos cos
.
km
km
( ) + + =
−
° ° °A B
00 934 0 469 0
8 10 1 99
. .
. .
A B
B A
+ =
= − +km
For y components
4 76 37 159 298 0
2 86
. sin sin sin
.
km
km
( ) + + =
+
° ° °A B
00 358 0 883 0. .A B− =
(a) We solve by eliminating B by substitution:
2 86 0 358 0 883 8 10 1 99 0
2 86
. . . . .
.
km km+ − − +( ) =A A
kkm km
km
+ + − =
=
=
0 358 7 15 1 76 0
10 0 1 40
7
. . .
. .
.
A A
A
A 117 km
(b) B = − + ( ) =8 10 1 99 7 17 6 15. . . .km km km
P3.52 (a) Rx = 2 00. , Ry = 1 00. , Rz = 3 00.
(b) R = + + = + + = =R R Rx y z
2 2 2
4 00 1 00 9 00 14 0 3 74. . . . .
(c) cos cos .θ θx
x
x
xR R
= ⇒ =
⎛
⎝
⎜
⎞
⎠
⎟ =−
R R
f1
57 7° rrom + x
cos cos .θ θy
y
y
yR R
= ⇒ =
⎛
⎝
⎜
⎞
⎠
⎟ =−
R R
f1
74 5° rrom + y
cos cos .θ θz
z
z
zR R
= ⇒ =
⎛
⎝
⎜
⎞
⎠
⎟ =−
R R
f1
36 7° rrom + z
N
B28º
A
C
69º
37º
1
2
3
E
FIG. P3.51

Vectors 59
P3.53 v i j i= + = +( ) +v vx y
ˆ ˆ cos . ˆ sin .300 100 30 0 100 30 0° °°( )
= +( )
=
ˆ
ˆ . ˆ
j
v i j
v
387 50 0
390
mi h
mi h at 7.337 N of E°
*P3.54 (a) A j i j= − = +60 80 80cm and B cos sinθ θ(( ) cm
so A B i j+ = + −( )80 80 60cos sinθ θ centimeters and
A B+ = ( ) + −( )⎡⎣ ⎤⎦ =80 80 60 80
2 2
cos sin coθ θ2 1/2
cm ss sin ( )( )cos2 2 2 2
80 2 80 60 60θ θ θ+ − +⎡⎣ ⎤⎦
1/2
cm
Now sin2
θ + cos2
θ = 1 for all θ, so we have
A B+ = + −⎡⎣ ⎤⎦80 60 2 80 60 102 2
( )( )cosθ
1/2
cm = 0000 600 cm
1/2
−[ ]9 cosθ
(b) For θ = 270° , cosθ = −1 and the expression takes on its maximum value,
[10 000 + 9 600] 1ր2 cm = 140 cm .
(c) For θ = 90° , cosθ = +1 and the expression takes on its minimum value, [10 000 − 9 600]
1ր2 cm = 20.0 cm .
(d) They do make sense. The maximum value is attained when A and B are in the same direction,
and it is 60 cm + 80 cm. The minimum value is attained when A and B are in opposite direc-
tions, and it is 80 cm − 60 cm.
*P3.55 ∆r i j= −( ) =∫ 1.2 m s m s 1.2
s
ˆ . ˆ
.
ր ր9 8
0
0 380
t dt 22 m s m s
1.2
s
2
s
t
tˆ . ˆ
ˆ
.
.
i jր ր
0
0 38
2
0
0 38
9 8
2
−
= ii jm s s m s
s2
ր ր( ) −( )−
( ) −⎛
0 38 0 9 8
0 38 0
2
2
. . ˆ .
⎝⎝⎜
⎞
⎠⎟ = −0 456 0 708. ˆ . î jm m
P3.56
Choose the + x axis in the direction of the first
force, and the y axis at 90° counterclockwise
from the x axis. Then each force will have only
one nonzero component.
The total force, in newtons, is then
12 0 31 0 8 40 24 0 3 60 7 00. ˆ . ˆ . ˆ . ˆ . ˆ . î j i j i j+ − − = ( )+ (( ) N .
The magnitude of the total force is
3 60 7 00 7 87
2 2
. . .( ) + ( ) =N N
and the angle it makes with our + x axis is given by tan
.
.
θ =
( )
( )
7 00
3 60
, θ = °62 8. .
Thus, its angle counterclockwise from the horizontal is 35 0 62 8 97 8. . .° ° °+ = .
R
35.0º
y
24 N
horizontal
31 N
8.4 N
12 N
x
FIG. P3.56

60 Chapter 3
P3.57 d i
d j
d i
1
2
3
100
300
150 30 0 150
=
= −
= − ( ) −
ˆ
ˆ
cos . ˆ° ssin . ˆ ˆ . ˆ
cos .
30 0 130 75 0
200 60 04
°
°
( ) = − −
= −
j i j
d (( ) + ( ) = − +
= +
ˆ sin . ˆ ˆ î j i j
R d d
200 60 0 100 173
1
°
22 3 4
2
130 202
130 202
+ + = − −( )
= −( ) + −(
d d i j
R
ˆ ˆ m
)) =
= ⎛
⎝
⎞
⎠
=
=
−
2
1
240
202
130
57 2
180
m
φ
θ
tan . °
++ =φ 237°
P3.58 d
dt
d t
dt
r i j j
j=
+ −( ) = + − = −( )
4 3 2
0 0 2 2 00
ˆ ˆ ˆ
ˆ . ˆm s jj
The position vector at t = 0 is 4 3ˆ î j+ . At t = 1 s, the position is 4 1ˆ î j+ , and so on. The object is
moving straight downward at 2 mրs, so
d
dt
r represents its velocity vector .
P3.59 (a) You start at point A: r r i j1 30 0 20 0= = −( )A . ˆ . ˆ m.
The displacement to B is
r r i j i j iB A− = + − + = +60 0 80 0 30 0 20 0 30 0. ˆ . ˆ . ˆ . ˆ . ˆ 1100ˆj.
You cover half of this, 15 0 50 0. ˆ . î j+( ) to move to
r i j i j i2 30 0 20 0 15 0 50 0 45 0 30 0= − + + = +. ˆ . ˆ . ˆ . ˆ . ˆ . ˆˆj.
Now the displacement from your current position to C is
r r i j i jC − = − − − − = −2 10 0 10 0 45 0 30 0 55 0. ˆ . ˆ . ˆ . ˆ . îi j− 40 0. ˆ .
You cover one-third, moving to
r r r i j i3 2 23 45 0 30 0
1
3
55 0 40 0= + = + + − −∆ . ˆ . ˆ . ˆ . ˆjj i j( )= +26 7 16 7. ˆ . ˆ .
The displacement from where you are to D is
r r i j i j iD − = − − − = −3 40 0 30 0 26 7 16 7 13 3. ˆ . ˆ . ˆ . ˆ . ˆ 446 7. ˆj.
You traverse one-quarter of it, moving to
r r r r i j i4 3 3
1
4
26 7 16 7
1
4
13 3= + −( ) = + + −D . ˆ . ˆ . ˆ 446 7 30 0 5 00. ˆ . ˆ . ˆj i j( )= + .
The displacement from your new location to E is
r r i j i j iE − = − + − − = −4 70 0 60 0 30 0 5 00 100. ˆ . ˆ . ˆ . ˆ ˆ ++ 55 0. ˆj
of which you cover one-fifth the distance, − +20 0 11 0. ˆ . î j, moving to
r r i j i j4 45 30 0 5 00 20 0 11 0 10 0+ = + − + =∆ . ˆ . ˆ . ˆ . ˆ . îi j+16 0. ˆ .
The treasure is at 10 0. m, 16.0 m( ) .
FIG. P3.57
continued on next page

Vectors 61
(b) Following the directions brings you to the average position of the trees. The steps we took
numerically in part (a) bring you to
r r r
r r
A B A
A B
+ −( ) =
+⎛
⎝⎜
⎞
⎠⎟
1
2 2
then to
r r r r r r r rA B C A B A B C
+( ) +
− +( ) =
+ +
2
2
3 3
ր
then to
r r r r r r r r rA B C D A B C A B
+ +( ) +
− + +( ) =
+
3
3
4
ր ++ +r rC D
4
and last to
r r r r r r r r rA B C D E A B C D+ + +( ) +
− + + +( )
4
4
5
ր
==
+ + + +r r r r rA B C D E
5
.
This center of mass of the tree distribution is the same location whatever order we take
the trees in.
P3.60 (a) Let T represent the force exerted by each child. The
x component of the resultant force is
T T T T T Tcos cos cos . .0 120 240 1 0 5 0 5+ + = ( )+ −( )+ −(° ° )) = 0
The y component is
T T T T Tsin sin sin . .0 120 240 0 0 866 0 866 0+ + = + − = .
Thus,
∑ =F 0
(b) If the total force is not zero, it must point in some direction. When each child moves one
space clockwise, the whole set of forces acting on the tire turns clockwise by that angle
so the total force must turn clockwise by that angle, 360°
N
. Because each child exerts
the same force, the new situation is identical to the old and the net force on the tire must
still point in the original direction. But the force cannot have two different directions.
The contradiction indicates that we were wrong in supposing that the total force is not
zero. The total force must be zero.
FIG. P3.60

62 Chapter 3
P3.61 Since
A B j+ = 6 00. ˆ,
we have
A B A Bx x y y+( ) + +( ) = +ˆ ˆ ˆ . ˆi j i j0 6 00
giving
A Bx x+ = 0 or A Bx x= − [1]
and
A By y+ = 6 00. . [2]
Since both vectors have a magnitude of 5.00, we also have
A A B Bx y x y
2 2 2 2 2
5 00+ = + = . .
From A Bx x= − , it is seen that
A Bx x
2 2
= .
Therefore, A A B Bx y x y
2 2 2 2
+ = + gives
A By y
2 2
= .
Then, A By y= and Equation [2] gives
A By y= = 3 00. .
Deﬁning θ as the angle between either A or B and the y axis, it is seen that
cos
.
.
.θ = = = =
A
A
B
B
y y 3 00
5 00
0 600 and θ = °53 1. .
The angle between A and B is then φ θ= =2 106° .
P3.62 (a) From the picture, R i j1 = +a bˆ ˆ and R1
2 2
= +a b .
(b) R i j k2 = + +a b cˆ ˆ ˆ ; its magnitude is
R1
2 2 2 2 2
+ = + +c a b c .
ANSWERS TO EVEN PROBLEMS
P3.2 (a) 2 17 1 25. , .m m( ); −( )1 90 3 29. , .m m (b) 4.55 m
P3.4 y = 1.15; r = 2.31
P3.6 310 km at 57° S of W
P3.8 9.5 N at 57°
P3.10 (a) ~105
m vertically upward (b) ~103
m vertically upward
P3.12 See the solution; the sum of a set of vectors is not affected by the order in which the vectors are
added.
FIG. P3.62
FIG. P3.61

Vectors 63
P3.14 We assume that the shopping cart stays on the level floor. There are two possibilities. If both of
the turns are right or both left, the net displacement is (a) 25.0 m (b) at 36.9°. If one turn is right
and one is left, we have (a) 61.8 m (b) at 14.0°.
P3.16 1.31 km north; 2.81 km east
P3.18 (a) 5.00 blocks at 53.1° N of E (b) 13.0 blocks
P3.20 − +25 0 43 3. ˆ . ˆm mi j
P3.22 788 48 0mi at north of east. °
P3.24 (a) see the solution (b) 5.00î + 4.00ˆj, 6.40 at 38.7°, –1.00î + 8.00ˆj, 8.06 at 97.2°
P3.26 (a) 4.10 m toward the top of the hill (b) 2.87 m
P3.28 42.7 yards
P3.30 C i j= −7 30 7 20. ˆ . ˆcm cm
P3.32 A B+ = (2.60 î + 4.50 ˆj)m
P3.34 (a) 2.83 m at θ = 315° (b) 13.4 m at θ = 117°
P3.36 (a) 10.4 cm; (b) 35.5°
P3.38 1 43 104
. × m at 32.2° above the horizontal
P3.40 (a) 15 1 7 72. ˆ . î j+( ) cm (b) − +( )7 72 15 1. ˆ . î j cm (c) + +( )7 72 15 1. ˆ . î j cm
P3.42 157 km
P3.44 (a) a = 5.00 and b = 7.00 (b) For vectors to be equal, all of their components must be equal.
A vector equation contains more information than a scalar equation.
P3.46 (a) see the solution (b) 18.3 b (c) 12.4 b at 233° counterclockwise from east
P3.48 2
11
tan− ⎛
⎝
⎞
⎠n
P3.50 25.4 s
P3.52 (a) 2.00, 1.00, 3.00 (b) 3.74 (c) θx
= 57.7°, θy
= 74.5°, θz
= 36.7°
P3.54 (a) (10 000 − 9 600 cos θ )1ր2
cm (b) 270° ; 140 cm (c) 90° ; 20.0 cm (d) They do make sense.
The maximum value is attained when A and B are in the same direction, and it is 60 cm + 80 cm.
The minimum value is attained when A and B are in opposite directions, and it is 80 cm − 60 cm.
P3.56 We choose the x axis to the right at 35° above the horizontal and the y axis at 90° counterclock-
wise from the x axis. Then each vector has only a single nonzero component. The resultant is
7.87 N at 97.8° counterclockwise from a horizontal line to the right.
P3.58 −( )2 00. ˆmրs j; its velocity vector
P3.60 (a) zero (b) see the solution
P3.62 (a) R i j1 = +a bˆ ˆ; R1
2 2
= +a b (b) R i j k2 = + +a b cˆ ˆ ˆ; R2
2 2 2
= + +a b c

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