Solucionario serway cap 3

3Vectors

CHAPTER OUTLINE

3.1 Coordinate Systems3.2 Vector and Scalar Quantities3.3 Some Properties of Vectors3.4 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 −2�D

1 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 s� , 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 fi rst means head end fi rst.(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 defi ned. 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= ( ) °. cos m , y2 3 80 120= ( ) °. sin m , and

x y2 2 1 90 3 29, . , . m( ) = −( )

(b) d x y= + = + =( ) ( ) . . .∆ ∆2 2 2 24 07 2 04 4 55 m m

P3.3 The x distance out to the fl y is 2.00 m and the y distance up to the fl y is 1.00 m.

(a) We can use the Pythagorean theorem to fi nd the distance from the origin to the fl y.

distance m m m2= + = ( ) + ( ) = =x y2 2 2 22 00 1 00 5 00. . . 22 24. m

(b) θ = ⎛⎝⎜

⎞⎠⎟ = °−tan .1 1

226 6 ;

�r = °2 24 26 6. , . m

P3.4 We have 2 00 30 0. .= °r cos

r =°

=2 00

30 02 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 r2 2 2 2

and its angle is

tan−

−⎛⎝

⎞⎠ =1 180

y

x° − θ .

(b) ( ) ( )− + − =2 2 22 2x y r This point is in the third quadrant if x y, ( ) is in the fi rst 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 2x y r+ − = This point is in the fourth quadrant if x y, ( ) is in the fi rst 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 0100

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° above the 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 thesemi-circle (ACB).

s r= ( ) = =1

22 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 offi cers, we estimate the total vector height as

~ . m m10 0 03 10 15 2( ) + ( ) ~ 103 m upward .

P3.11 To fi nd these vector expressions graphically,we draw each set of vectors. Measurements ofthe results are taken using a ruler and protractor. (Scale: 1 0 5 unit 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

AR1

C

C C

R 3R

AA

B

B

100 m

FIG. P3.12

P3.13 The scale drawing for the graphicalsolution should be similar to thefi gure to the right. The magnitude anddirection of the fi nal displacementfrom the starting point are obtainedby measuring d and θ on the drawingand applying the scale factor used inmaking 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 fl oor is level. Take the x axis in the direction of the fi rst 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 ˆj( )m

to give (a) displacement magnitude (202 + 152)1� 2 m = 25.0 m

at (b) tan−1(15�20) = 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 ˆj( )m

to give (a) displacement magnitude (602 + 152)1�2 m = 61.8 m

at (b) tan−1(15�60) = 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

Ay

x

.

So

φ =⎛

⎝⎜⎞

⎠⎟= = ( ) =− −tan tan

.

.tan .1 140 0

25 01 60 5

A

Ay

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 m �s.

We can satisfy this requirement simply by taking v ≥ (28 m �s)�cos 8.5° = 28.3 m �s.

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 23 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

3 001 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

m�d ˆ . 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 3582 2 2 2

. m

md x y= ( ) + ( ) = ( ) + −( ) =∑ ∑ .

taan.θ

θ

=( )( ) = − = −

= −

=

∑∑

y

x

12 5

3580 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

dDC 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

Ax

y

= = −= =

150 75 0

150 120 130

cos120 cm

cm

°

°

.

sin

RR

Rx

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

cm�B 190 49 7 196

49 7

190

2 2

1

.

tan.θ ⎠⎠ = −14 7. .°

P3.24 (a) See fi gure 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. j

�C = + ⎛

⎝⎞⎠ =−25 0 16 0 6 401. . tan .at

4

5at 38.7°

� � �D A B i j i j= − = + − + = −2 00 6 00 3 00 2 00 1 00. ˆ . ˆ . ˆ . ˆ . ii j+ 8 00. ˆ

�D = −( ) + ( )

−⎛⎝⎜

⎞⎠⎟

−1 00 8 008 00

1 002 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) θ A B+−= −⎛

⎝⎞⎠ = − =tan .1 6

271 6 288° °

θ A B−

−= ⎛⎝

⎞⎠ =tan .1 2

426 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 .° ° 880 j( ) 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 . î 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 29 20 2 30 9 48. . . m .

The direction is θ =−

⎛⎝

⎞⎠ =arctan

.

.

2 30

9 20166° .

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 72 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 x2 2

4 64at tan m at 78.6 N of1− ( ) .� ° 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

.

° °�

ˆ ˆ . ˆ . ˆ . ˆ î j i j i j

F

+ − + = +( )=

104 20 7 77 3 39 3 181 N�

339 3 181 185

181

39 377 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 zˆ ˆ ˆ . ˆ . ˆ . ˆ

.

4 00 6 00 3 00

4 000 6 00 3 00 7 81

4 00

7 8159

2 2 2

1

+ + =

= ⎛⎝

⎞⎠ =−

. . .

cos.

.α ..

cos.

2

6 01

° is the angle with the axisx

β = − 00

7 8139 8

..⎛

⎝⎞⎠ = ° is the angle with the axy iis

is the angle wγ = ⎛⎝

⎞⎠ =−cos

.

..1 3 00

7 8167 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 . î j k+ −( ) cm.

(a) The magnitude is d = ( ) + ( ) + −( ) =4 80 8 50 3 70 10 42 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

Ai j k= = + −

42 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 1032 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. j( ) 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 fl y 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 3 54 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 23 00 2 00 3 61. . .

tan.

..θ = =

−( ) = −A

Ay

x

2 00

3 000 667, 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 fi rst 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 i representing east and j representing

north, its total displacement is:

41 0 60 0 3 00 41 0. cos . . ˆ . sin

km

hh

km

h°⎛

⎝⎞⎠ ( ) −( ) +i 660 0 3 00 25 0 1 50. . ˆ . . ˆ°⎛

⎝⎞⎠ ( ) + ⎛

⎝⎞⎠ ( )

=

hkm

hhj j

661 5 144. ˆ ˆkm km−( ) +i j

with magnitude 61 5 144 1572 2

. km km km( ) + ( ) = .

P3.43 (a) R

Rx

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 ˆj

(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

E27.0º

FIG. P3.40

A

y

x

B

45º

C

45ºO

FIG. P3.43


56 Chapter 3

*P3.44 (a) Taking components along i and j, 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 fi nd −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 fi nish is 16 i + 12 j − (5 i + 3 j) = (11 i + 9 j) meters.

The displacement from the starting point to A is f (11 i + 9 j) meters.

(a) The position vector of point A is 5 i + 3 j + f (11 i + 9 j ) = (5 + 11f ) i + (3 + 9f ) j meters .

(b) For f = 0 we have the position vector (5 + 0) i + (3 + 0) j meters.

This is reasonable because it is the location of the starting point, 5 i + 3 j meters.

(c) For f = 1 = 100%, we have position vector (5 + 11) i + (3 + 9) j meters = 16 i + 12 j meters .

This is reasonable because we have completed the trip and this is the position vector of the endpoint.

*P3.46 We note that − =i west and − =j 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 themagnitudes 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 9 872 2

. ˆ . ˆ

. .

b b

b b at south of west

b at

tan.

..

−

=

1 9 87

7 4412 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 directionsof

�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 , θ2

, and θ2

.

The magnitude of �R is then R A= ⎛

⎝⎜⎞⎠⎟2

2cos

θ.

This can be seen from applying the law of cosinesto 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 22

2−( ) = ⎛⎝⎜

⎞⎠⎟cos sinθ θ

gives the magnitude of �D as D A= ⎛

⎝⎜⎞⎠⎟2

2sin

θ.

The problem requires that R D= 100 .

Thus, 22

2002

A Acos sinθ θ⎛

⎝⎜⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

. This gives tan .θ2

0 010⎛⎝⎜

⎞⎠⎟ = and θ = 1 15. ° .

P3.48 Let θ represent the angle between the directionsof

�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 , θ2

, and θ2

.

The magnitude of �R is then R A= ⎛

⎝⎜⎞⎠⎟2

2cos

θ. This can

be seen by applying the law of cosines to the isoscelestriangle 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

22−( ) = ⎛

⎝⎜⎞⎠⎟cos sinθ θ

gives the magnitude of �D as D A= ⎛

⎝⎜⎞⎠⎟2

2sin

θ.

The problem requires that R nD= or cos sinθ θ2 2

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟n giving θ = ⎛

⎝⎜⎞⎠⎟

−211tann

.

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 fi rst 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 fi rst 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 2 292 2 2

. . . .( ) + ( ) + ( ) = km .

A

BRθ

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 3352 2( ) + ( ) = m m. From v = distance

∆t, the time for each child’s run is

Inge:distance m h km s

∆t = =( )( )( )

v335 1 3 600

12 km m hs

Olaf:m s

3.

( )( )( ) =

= ⋅1 000 1

101

420∆t333 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 θxx

xxR R= ⇒ =

⎛

⎝⎜

⎞

⎠⎟ =−� �

R Rf1 57 7° rrom + x

cos cos .θ θyy

yyR R

= ⇒ =⎛

⎝⎜

⎞

⎠⎟ =−� �

R Rf1 74 5° rrom + y

cos cos .θ θzz

zzR R

= ⇒ =⎛

⎝⎜

⎞

⎠⎟ =−� �

R Rf1 36 7° rrom + z

N

B28º

A

C

69º

37º1

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 60 802 2cos sin coθ θ2 1/2

cm ss sin ( )( ) cos2 2 2 280 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

1/2cm = 0000 600 cm1/2−[ ]9 cosθ

(b) For θ = 270° , cosθ = −1 and the expression takes on its maximum value,

[10 000 + 9 600] 1�2 cm = 140 cm .

(c) For θ = 90° , cosθ = +1 and the expression takes on its minimum value, [10 000 − 9 600]

1�2 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.� �9 8

0

0 380t dt 22 m s m s

1.2

s2

s

ttˆ . ˆ

ˆ

..

i j� �0

0 38 2

0

0 38

9 82

−

= ii jm s s m ss2� �( ) −( ) − ( ) −⎛

0 38 0 9 80 38 0

2

2

. . ˆ .

⎝⎝⎜⎞⎠⎟

= −0 456 0 708. ˆ . î jm m

P3.56 Choose the + x axis in the direction of the fi rstforce, and the y axis at 90° counterclockwisefrom the x axis. Then each force will have onlyone 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 7 872 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. . .° ° °+ = .

R35.0º

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

13057 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 00ˆ 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 m �s, 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 1100 j .

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 . ˆj.

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 . ii j− 40 0. ˆ .

You cover one-third, moving to

� � �r r r i j i3 2 23 45 0 30 0

1

355 0 40 0= + = + + − −∆ . ˆ . ˆ . ˆ . jj 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. j .

You traverse one-quarter of it, moving to

� � � �r r r r i j i4 3 3

1

426 7 16 7

1

413 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. j

of which you cover one-fi fth 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 . ii 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 rA 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

�

then to � � � � � � � � �r r r r r r r r rA B C D A B C A B

+ +( )+

− + +( )= +

3

3

4

� ++ +� �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

�== + + + +� � � � �

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 . î 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 y2 2 2 2 25 00+ = + = . .

From A Bx x= − , it is seen that

A Bx x2 2= .

Therefore, A A B Bx y x y2 2 2 2+ = + gives

A By y2 2= .

Then, A By y= and Equation [2] gives

A By y= = 3 00. .

Defi ning θ as the angle between either �A or

�B and the y axis, it is seen that

cos.

..θ = = = =

A

A

B

By y 3 00

5 000 600 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 fl oor. 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) 5.00 i + 4.00 j, 6.40 at 38.7°, –1.00 i + 8.00 j, 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 i + 4.50 j)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 211tan− ⎛

⎝⎞⎠n

P3.50 25.4 s

P3.52 (a) 2.00, 1.00, 3.00 (b) 3.74 (c) θx = 57.7°, θ

y = 74.5°, θ

z = 36.7°

P3.54 (a) (10 000 − 9 600 cos θ )1�2 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. ˆm �s 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



Solucionario serway cap 3

Engineering