Cap.3 Ricardo

7/28/2019 Cap.3 Ricardo

1/62

1. The vectors should be parallel to achieve a resultant 7 m long (the unprimed case shown below), antipar-allel (in opposite directions) to achieve a resultant 1 m long (primed case shown), and perpendicular to

achieve a resultant

32 + 42 = 5 m long (the double-primed case shown). In each sketch, the vectorsare shown in a head-to-tail sketch but the resultant is not shown. The resultant would be a straightline drawn from beginning to end; the beginning is indicated by A (with or without primes, as the casemay be) and the end is indicated by B.

b

A-

b

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B

b

A

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b

B

b

A

-b

6

B


2/62

2. A sketch of the displace-ments is shown. The re-

sultant (not shown) wouldbe a straight line fromstart (Bank) to finish(Walpole). With a carefuldrawing, one should findthat the resultant vectorhas length 29.5 km at 35

west of south.

West East

South

Bank

@

@

@

@

@

@

@

@

@

@R

H

H

H

H

H

H

H

H

H

H

H

H

H

H

HY

B

B

B

B

B

B

BN

Walpole


3/62

3. The x component of a is given by ax = 7.3cos250 = 2.5 and the y component is given by ay =

7.3sin250 = 6.9. In considering the variety of ways to compute these, we note that the vector is

70 below thex axis, so the components could also have been found from ax =

7.3cos70 and

ay = 7.3sin70. In a similar vein, we note that the vector is 20 from the y axis, so one could use

ax = 7.3sin20 and ay = 7.3cos20

to achieve the same results.


4/62

4. The angle described by a full circle is 360 = 2 rad, which is the basis of our conversion factor. Thus,

(20.0) 2 rad360

= 0.349 rad

and (similarly) 50.0 = 0.873 rad and 100 = 1.75 rad. Also,

(0.330 rad)360

2 rad= 18.9

and (similarly) 2.10 rad = 120 and 7.70 rad = 441.


5/62

5. The textbooks approach to this sort of problem is through the use of Eq. 3-6, and is illustrated in SampleProblem 3-3. However, most modern graphical calculators can produce the results quite efficiently using

rectangular polar shortcuts.

(a)

(25)2 + 402 = 47.2 m

(b) Recalling that tan() = tan( + 180), we note that the two possibilities for tan1 (40/ 25)are 58 and 122. Noting that the vector is in the third quadrant (by the signs of its x and ycomponents) we see that 122 is the correct answer. The graphical calculator shortcuts mentionedabove are designed to correctly choose the right possibility.


6/62

6. The x component ofr is given by 15 cos30 = 13 m and the y component is given by 15 sin 30 = 7.5 m.


7/62

7. The point P is displaced vertically by 2R, where R is the radius of the wheel. It is displaced horizontallyby half the circumference of the wheel, or R. Since R = 0.450 m, the horizontal component of the

displacement is 1.414 m and the vertical component of the displacement is 0.900 m. If the x axis ishorizontal and the y axis is vertical, the vector displacement (in meters) is r = (1.414 i + 0.900 j). Thedisplacement has a magnitude of

|r| =

(R)2 + (2R)2 = R2 + 4 = 1.68 m

and an angle of

tan1

2R

R

= tan1

2

= 32.5

above the floor. In physics there are no exact measurements, yet that angle computation seemed toyield something exact. However, there has to be some uncertainty in the observation that the wheelrolled half of a revolution, which introduces some indefiniteness in our result.


8/62

8. Although we think of this as a three-dimensional movement, it is rendered effectively two-dimensionalby referring measurements to its well-defined plane of the fault.

(a) The magnitude of the net displacement is

| AB| =

|AD|2 + |AC|2 =

172 + 222 = 27.8 m .

(b) The magnitude of the vertical component of AB is |AD| sin52.0 = 13.4 m.


9/62


10/62

10. We label the displacement vectors A, B and C (and denote the result

of their vector sum as r). Wechoose eastas the i direction (+xdirection) and north as the j di-rection (+y direction). All dis-tances are understood to be inkilometers. We note that the an-gle between C and the x axis is60. Thus,

-

6

?

eastwest

north

south

e

A- e

6 B

e

C

A = 50 iB = 30 jC = 25 cos (60) i + 25 sin (60) j

r = A + B + C = 62.50 i + 51.65 j

which means

that its magnitude is

|r| =

62.502 + 51.652 81 km .

and its angle (counterclockwise from +x axis) is tan1 (51.65/62.50) 40, which is to say that it points40 north of east. Although the resultant r is shown in our sketch, it would be a direct line from thetail of A to the head of C.


11/62

11. The diagram shows the displacement vectors for the two segments of her walk, labeled A and B, andthe total (final) displacement vector, labeled r. We take east to be the +x direction and north to be

the +y direction. We observe that the angle between A and the x axis is 60. Where the units are notexplicitly shown, the distances are understood

to be in meters. Thus, the componentsof A are Ax = 250cos60 = 125 andAy = 250sin30

= 216.5. The compo-

nents of B are Bx = 175 and By = 0.The components of the total displace-ment are rx = Ax + Bx = 125 + 175 =300 and ry = Ay + By = 216.5 + 0 =216.5. ....

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......................North

East

A

B

r

(a) The magnitude of the resultant displacement is

|r| =

r2x

+ r2y

=

3002 + 216.52 = 370 m .

(b) The angle the resultant displacement makes with the +x axis is

tan1ry

rx

= tan1

216.5

300

= 36 .

(c) The total distance walked is d = 250 + 175 = 425 m.

(d) The total distance walked is greater than the magnitude of the resultant displacement. The diagram

shows why: A and B are not collinear.


12/62

12. We label the displacement vectors A, B and C (and denote the result

of their vector sum as r). Wechoose eastas the i direction (+xdirection) and north as the j di-rection (+y direction). All dis-tances are understood to be inkilometers. Therefore,

-

6

?

eastwest

north

south

e

A

6

e

B

e

C

?

A = 3.1 jB = 2.4 iC = 5.2 j

r = A + B + C = 2.1 i 2.4 j

which means

that its magnitude is|r| =

(2.1)2 + (2.4)2 3.2 km .

and the two possibilities for its angle are

tan12.4

2.1

= 41, or 221 .

We choose the latter possibility since r is in the third quadrant. It should be noted that many graphicalcalculators have polar rectangular shortcuts that automatically produce the correct answer for angle(measured counterclockwise from the +x axis). We may phrase the angle, then, as 221 counterclockwisefrom East (a phrasing that sounds peculiar, at best) or as 41 south from west or 49 west from south.

The resultant r is not shown in our sketch; it would be an arrow directed from the tail of A to thehead of C.


13/62

13. We write r = a + b. When not explicitly displayed, the units here are assumed to be meters. Thenrx = ax + bx = 4.0 13 = 9.0 and ry = ay + by = 3.0 + 7.0 = 10. Thus r = (9.0 m) i + (10 m)j . The

magnitude of the resultant is

r =

r2x

+ r2y

=

(9.0)2 + (10)2 = 13 m .

The angle between the resultant and the +x axis is given by tan1 (ry/rx) = tan1 10/(9.0) which is

either 48 or 132. Since the x component of the resultant is negative and the y component is positive,characteristic of the second quadrant, we find the angle is 132 (measured counterclockwise from +xaxis).


14/62

14. The x, y and z components (with meters understood) ofr are

rx = cx + dx = 7.4 + 4.4 = 12ry = cy + dy = 3.8 2.0 = 5.8

rz = cz + dz = 6.1 + 3.3 = 2.8 .


15/62

15. The vectors are shown on the diagram. The x axis runs from west

to east and the y axis runfrom south to north. Thenax = 5.0 m, ay = 0, bx =(4.0m)sin35 = 2.29m,and by = (4.0m)cos35

=3.28m.

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a

ba +b

E

N

50.5

35

(a) Let c = a+b. Then cx = ax +bx = 5.0 m2.29m = 2.71 m and cy = ay +by = 0+ 3.28m = 3.28m.The magnitude ofc is

c =

c2x

+ c2y

=

(2.71m)2 + (3.28m)2 = 4.3 m .

(b) The angle that c = a +b makes with the +x axis is

= tan1cy

cx= tan1

3.28 m

2.71 m= 50.4 .

The second possibility ( = 50.4 + 180 = 126) is rejected because it would point in a directionopposite to c.

(c) The vector ba is found by adding a to b. The result is shown

on the diagram to theright. Let c = b a. Then cx = bx

ax =

2.29 m

5.0 m =7.29m and cy = by ay = 3.28 m. The magni-

tude ofc is c =

c2x

+ c2y

= 8.0 m . ...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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a

ba +b

W

N

(d) The tangent of the angle that c makes with the +x axis (east) is

tan =cy

cx=

3.28 m

7.29 m= 4.50, .

There are two solutions: 24.2 and 155.8. As the diagram shows, the second solution is correct.The vector c = a +b is 24 north of west.


16/62

16. All distances in this solution are understood to be in meters.

(a) a +b = (3.0 i + 4.0j) + (5.0 i 2.0j) = 8.0 i + 2.0j .(b) The magnitude ofa +b is

|a +b| =

8.02 + 2.02 = 8.2 m .

(c) The angle between this vector and the +x axis is tan1(2.0/8.0) = 14.

(d) b a = (5.0 i 2.0j) (3.0 i + 4.0j) = 2.0 i 6.0j .

(e) The magnitude of the difference vector b a is

|b a| =

2.02 + (6.0)2 = 6.3 m .

(f) The angle between this vector and the +x axis is tan1(6.0/2.0) = 72. The vector is 72

clockwise from the axis defined by i.


17/62

17. All distances in this solution are understood to be in meters.

(a) a +b = (4.0 + (

1.0)) i + ((

3.0) + 1.0)j + (1.0 + 4.0) k = 3.0 i

2.0j + 5.0 k .

(b) ab = (4.0 (1.0)) i + ((3.0) 1.0)j + (1.0 4.0) k = 5.0 i 4.0j 3.0 k .

(c) The requirement ab + c = 0 leads to c = b a, which we note is the opposite of what we found

in part (b). Thus, c = 5.0 i + 4.0j + 3.0 k .


18/62

18. Many of the operations are done efficiently on most modern graphical calculators using their built-invector manipulation and rectangular

polar shortcuts. In this solution, we employ the traditional

methods (such as Eq. 3-6).

(a) The magnitude ofa is

42 + (3)2 = 5.0 m.(b) The angle between a and the +x axis is tan1(3/4) = 37. The vector is 37 clockwise from the

axis defined by i .

(c) The magnitude ofb is

62 + 82 = 10 m.

(d) The angle between b and the +x axis is tan1(8/6) = 53.

(e) a +b = (4 + 6) i + ((3) + 8)j = 10 i + 5j , with the unit meter understood. The magnitude of thisvector is

102 + 52 = 11 m; we rounding to two significant figures in our results.

(f) The angle between the vector described in part (e) and the +x axis is tan1(5/10) = 27.

(g) b

a = (6

4) i + (8

(

3))j = 2 i + 11j , with the unit meter understood. The magnitude of thisvector is 22 + 112 = 11 m, which is, interestingly, the same result as in part (e) (exactly, not justto 2 significant figures) (this curious coincidence is made possible by the fact that a b).

(h) The angle between the vector described in part (g) and the +x axis is tan1(11/2) = 80.

(i) a b = (4 6) i + ((3) 8)j = 2 i 11j , with the unit meter understood. The magnitude ofthis vector is

(2)2 + (11)2 = 11 m.

(j) The two possibilities presented by a simple calculation for the angle between the vector described inpart (i) and the +x direction are tan1(11/2) = 80, and 180 + 80 = 260. The latter possibilityis the correct answer (see part (k) for a further observation related to this result).

(k) Since ab = (1)(ba), they point in opposite (antiparallel) directions; the angle between themis 180.


19/62

19. Many of the operations are done efficiently on most modern graphical calculators using their built-invector manipulation and rectangular

polar shortcuts. In this solution, we employ the traditional

methods (such as Eq. 3-6). Where the length unit is not displayed, the unit meter should be understood.

(a) Using unit-vector notation,

a = 50 cos (30) i + 50 sin (30)j

b = 50 cos (195) i + 50 sin(195)j

c = 50 cos (315) i + 50 sin(315)j

a +b + c = 30.4 i 23.3 j .

The magnitude of this result is

30.42 + (23.3)2 = 38 m.(b) The two possibilities presented by a simple calculation for the angle between the vector described

in part (a) and the +x direction are tan1(

23.2/30.4) =

37.5, and 180 + (

37.5) = 142.5.

The former possibility is the correct answer since the vector is in the fourth quadrant (indicatedby the signs of its components). Thus, the angle is 37.5, which is to say that it is roughly 38clockwise from the +x axis. This is equivalent to 322.5 counterclockwise from +x.

(c) We find ab+c = (43.3 (48.3)+35.4) i (25 (12.9)+(35.4))j = 127 i + 2.6j in unit-vectornotation. The magnitude of this result is

1272 + 2.62 1.3 102 m.

(d) The angle between the vector described in part (c) and the +x axis is tan1(2.6/127) 1.(e) Using unit-vector notation, d is given by

d = a +b c= 40.4 i + 47.4j

which has a magnitude of(40.4)2 + 47.42 = 62 m.

(f) The two possibilities presented by a simple calculation for the angle between the vector describedin part (e) and the +x axis are tan1(47.4/(40.4)) = 50, and 180+ (50) = 130. We choosethe latter possibility as the correct one since it indicates that d is in the second quadrant (indicatedby the signs of its components).


20/62

20. Angles are given in standard fashion, so Eq. 3-5 applies directly. We use this to write the vectors inunit-vector notation before adding them. However, a very different-looking approach using the special

capabilities of most graphical calculators can be imagined. Where the length unit is not displayed in thesolution below, the unit meter should be understood.

(a) Allowing for the different angle units used in the problem statement, we arrive at

E = 3.73 i + 4.70jF = 1.29 i 4.83jG = 1.45 i + 3.73jH = 5.20 i + 3.00j

E+ F+ G + H = 1.28 i + 6.60j .

(b) The magnitude of the vector sum found in part (a) is

1.282 + 6.602 = 6.72 m. Its angle measured

counterclockwise from the +x axis is tan1(6.6/1.28) = 79 = 1.38 rad.


21/62

21. It should be mentioned that an efficient way to work this vector addition problem is with the cosinelaw for general triangles (and since a, b and r form an isosceles triangle, the angles are easy to figure).

However, in the interest of reinforcing the usual systematic approach to vector addition, we note thatthe angle b makes with the +x axis is 135 and apply Eq. 3-5 and Eq. 3-6 where appropriate.

(a) The x component ofr is 10cos30 + 10cos 135 = 1.59 m.

(b) The y component ofr is 10sin30 + 10sin 135 = 12.1 m.

(c) The magnitude ofr is

1.592 + 12.12 = 12.2 m.

(d) The angle between r and the +x direction is tan1 (12.1/1.59) = 82.5.


22/62

22. If we wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle between C and the+x axis is 180 + 20 = 200.

(a) The x component of B is given by Cx Ax = 15cos200 12cos40 = 23.3 m, and the y

component of B is given by Cy Ay = 15sin200 12sin40 = 12.8 m. Consequently, its

magnitude is

(23.3)2 + (12.8)2 = 26.6 m.

(b) The two possibilities presented by a simple calculation for the angle between B and the +x axisare tan1((12.8)/(23.3)) = 28.9, and 180 + 28.9 = 209. We choose the latter possibility

as the correct one since it indicates that B is in the third quadrant (indicated by the signs of itscomponents). We note, too, that the answer can be equivalently stated as 151.


23/62

23. We consider A with (x, y) components given by (A cos , A sin ). Similarly, B (B cos , B sin ). Theangle (measured from the +x direction) for their vector sum must have a slope given by

tan =A sin + B sin

A cos + B cos .

The problem requires that we now consider the orthogonal direction, where tan + 90 = cot . If this(the negative reciprocal of the above expression) is to equal the slope for their vector difference, then wemust have

A cos + B cos

A sin + B sin =

A sin B sin

A cos B cos .

Multiplying both sides by A sin + B sin and doing the same with A cos B cos yields

A2 cos2 B2 cos2 = A2 sin2 B2 sin2 .

Rearranging, using the cos2

+ sin

2

= 1 identity, we obtain

A2 = B2 = A = B .

In a later section, the scalar (dot) product of vectors is presented and this result can be revisited witha more compact derivation.


24/62

24. If we wish to use Eq. 3-5 directly, we should note that the angles for Q,R and S are 100, 250 and 310,respectively, if they are measured counterclockwise from the +x axis.

(a) Using unit-vector notation, with the unit meter understood, we have

P = 10 cos (25) i + 10 sin(25)j

Q = 12 cos (100) i + 12 sin (100)j

R = 8cos(250) i + 8 sin(250)j

S = 9cos(310) i + 9 sin(310)j

P+ Q + R + S = 10.0 i + 1.6j

(b) The magnitude of the vector sum is

102 + 1.62 = 10.2 m and its angle is tan1 (1.6/10) 9.2measured counterclockwise from the +x axis. The appearance of this solution would be quitedifferent using the vector manipulation capabilities of most modern graphical calculators, althoughthe principle would be basically the same.


25/62

25. Without loss of generality, we assume a points along the +x axis, and that b is at measured counter-clockwise from a. We wish to verify that

r2 = a2 + b2 + 2ab cos

where a = |a| = ax (well call it a for simplicity) and b = |b| =

b2x

+ b2y

. Since r = a + b then

r = |r| =

(a + bx)2 + b2y. Thus, the above expression becomes

(a + bx)2 + b2y

2= a2 +

b2x

+ b2y

2+ 2ab cos

a2 + 2abx + b

2

x+ b2

y= a2 + b2

x+ b2

y+ 2ab cos

which makes a valid equality since (the last term) 2ab cos is indeed the same as 2abx (on the left-handside). In a later section, the scalar (dot) product of vectors is presented and this result can be revisited

with a somewhat different-looking derivation.


26/62

26. The vector equation is R = A + B + C+ D. Expressing B and D in unit-vector notation, we have1.69 i + 3.63j and 2.87 i + 4.10j, respectively. Where the length unit is not displayed in the solution

below, the unit meter should be understood.

(a) Adding corresponding components, we obtain R = 3.18 i + 4.72j.

(b) and (c) Converting this result to polar coordinates (using Eq. 3-6 or functions on a vector-capablecalculator), we obtain

(3.18, 4.72) (5.69 124)

which tells us the magnitude is 5.69 m and the angle (measured counterclockwise from +x axis) is124.


27/62

27. (a) There are 4 such lines, one from each of the corners on the lower face to the diametrically oppositecorner on the upper face. One is shown on the diagram. Using an xyz coordinate system as shown

(with the origin at the back lower left corner) The position vector of the starting point of thediagonal shown is a i and the position vector of the ending point is a j + a k, so the vector alongthe line is the difference a j + a k a i.

The point diametrically oppositethe origin has position vectora i +a j +a k and this is the vec-tor along the diagonal. Anothercorner of the bottom face is ata i+a j and the diametrically op-posite corner is at a k, so anothercube diagonal is a k a i a j.The fourth diagonal runs from a j

to a i + a k, so the vector alongthe diagonal is a i + a k a j.

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............. ............. ............. ............. ............. ............. ............. ............. ............. ..........

.............

.............

.

.

.

.

.

.

.

.

.

.

..

.

.

..

.

.

.

.

.

.

.

.............

.............

.............

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a

a

a

x

y

z

(b) Consider the vector from the back lower left corner to the front upper right corner. It is a i +

a j + a k. We may think of it as the sum of the vector a i parallel to the x axis and the vectora j + a k perpendicular to the x axis. The tangent of the angle between the vector and the x axisis the perpendicular component divided by the parallel component. Since the magnitude of theperpendicular component is

a2 + a2 = a

2 and the magnitude of the parallel component is a,

tan =a

2/a =

2. Thus = 54.7. The angle between the vector and each of the other two

adjacent sides (the y and z axes) is the same as is the angle between any of the other diagonalvectors and any of the cube sides adjacent to them.

(c) The length of any of the diagonals is given bya2 + a2 + a2 = a

3.


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28. Reference to Figure 3-18 (and the accompanying material in that section) is helpful. If we convert B to the

magnitude-angle notation (as A already is) we have B = (14.4 33.7) (appropriate notation especially

if we are using a vector capable calculator in polar mode). Where the length unit is not displayed in thesolution, the unit meter should be understood. In the magnitude-angle notation, rotating the axis by+20 amounts to subtracting that angle from the angles previously specified. Thus, A = (12.0 40.0)

and B = (14.4 13.7), where the prime notation indicates that the description is in terms of the newcoordinates. Converting these results to (x, y) representations, we obtain

A = 9.19 i

+ 7.71j

B = 14.0 i

+ 3.41j

with the unit meter understood, as already mentioned.


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29. We apply Eq. 3-20 and Eq. 3-27.

(a) The scalar (dot) product of the two vectors is a b = ab cos = (10)(6.0)cos60

= 30.

(b) The magnitude of the vector (cross) product of the two vectors is |ab| = ab sin = (10)(6.0)sin60 =52.


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30. We consider all possible products and then simplify using relations such as i k = 0 and i i = 1. Thus,

ab =

ax i + ay j + az k

bx i + byj + bz k

= axbx i i + axby i j + axbz i k + aybxj i + aybj j j +

= axbx(1) + axby(0) + axbz(0) + aybx(0) + aybj(1) +

which is seen to reduce to the desired result (one might wish to show this in two dimensions beforetackling the additional tedium of working with these three-component vectors).


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31. Since ab cos = axbx + ayby + azbz,

cos = axbx + ayby + azbz

ab.

The magnitudes of the vectors given in the problem are

a = |a| =

(3.0)2 + (3.0)2 + (3.0)2 = 5.2

b = |b| =

(2.0)2 + (1.0)2 + (3.0)2 = 3.7 .

The angle between them is found from

cos =(3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0)

(5.2)(3.7)= 0.926 .

The angle is = 22.


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32. We consider all possible products and then simplify using relations such as i i = 0 and the importantfundamental products

i j = j i = k

j k = k j = i

k i = i k = j .

Thus,

ab =ax i + ayj + az k

bx i + by j + bz k

= axbx i i + axby i j + axbz i k + aybxj i + aybj j j +

= axbx(0) + axby( k) + axbz(j) + aybx( k) + aybj(0) +

which is seen to simplify to the desired result.


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33. The area of a triangle is half the product of its base and altitude. The base is the side formed by vector

a. Then the altitude is b sin and the area is A = 12ab sin = 1

2|ab|.


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34. Applying Eq. 3-23, F = qv B (where q is a scalar) becomes

Fx i + Fyj + Fz k = q(vyBz vzBy) i + q(vzBx vxBz) j + q(vxBy vyBx) k

which plugging in values leads to three equalities:

4.0 = 2 (4.0Bz 6.0By)

20 = 2 (6.0Bx 2.0Bz)

12 = 2 (2.0By 4.0Bx)

Since we are told that Bx = By , the third equation leads to By = 3.0. Inserting this value into thefirst equation, we find Bz = 4.0. Thus, our answer is

B = 3.0 i 3.0j 4.0 k .


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35. Both proofs shown below utilize the fact that the vector (cross) product ofa and b is perpendicular to

both a and b. This is mentioned in the book, and is fundamental to its discussion of the right-hand rule.

(a) (b a) is a vector that is perpendicular to a, so the scalar product ofa with this vector is zero.This can also be verified by using Eq. 3-30, and then (with suitable notation changes) Eq. 3-23.

(b) Let c = ba. Then the magnitude ofc is c = ab sin. Since c is perpendicular to a the magnitude

ofac is ac. The magnitude ofa (ba) is consequently |a (ba)| = ac = a2b sin . This toocan be verified by repeated application of Eq. 3-30, although it must be admitted that this is muchless intimidating if one is using a math software package such as MAPLE or Mathematica.


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36. If a vector capable calculator is used, this makes a good exercise for getting familiar with those features.Here we briefly sketch the method. Eq. 3-30 leads to

2 A B = 2

2 i + 3j 4 k

3 i + 4j + 2 k

= 44 i + 16j + 34 k .

We now apply Eq. 3-23 to evaluate 3 C

2 A B

:

3

7 i 8j

44 i + 16j + 34 k

= 3 ((7)(44) + (8)(16) + (0)(34)) = 540 .


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37. From the figure, we note that c b, which implies that the angle between c and the +x axis is 120.

(a) Direct application of Eq. 3-5 yields the answers for this and the next few parts. ax = a cos0

=a = 3.00 m.

(b) ay = a sin0 = 0.

(c) bx = b cos30 = (4.00m)cos30 = 3.46 m.

(d) by = b sin30 = (4.00m)sin30 = 2.00 m.

(e) cx = c cos120 = (10.0m)cos120 = 5.00 m.

(f) cy = c sin30 = (10.0m)sin120 = 8.66 m.

(g) In terms of components (first x and then y), we must have

5.00 m = p(3.00 m) + q(3.46m)

8.66 m = p(0) + q(2.00m) .

Solving these equations, we find p = 6.67

(h) and q= 4.33 (note that its easiest to solve for q first). The numbers p and q have no units.


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38. We apply Eq. 3-20 with Eq. 3-23. Where the length unit is not displayed, the unit meter is understood.

(a) We first note that a = |a| = 3.22 + 1.62 = 3.58 m and b = |b| = 0.52 + 4.52 = 4.53 m. Now,a b = axbx + ayby = ab cos

(3.2)(0.5) + (1.6)(4.5) = (3.58)(4.53) cos

which leads to = 57 (the inverse cosine is double-valued as is the inverse tangent, but we knowthis is the right solution since both vectors are in the same quadrant).

(b) Since the angle (measured from +x) for a is tan1 (1.6/3.2) = 26.6, we know the angle for cis 26.6 90 = 63.4 (the other possibility, 26.6 + 90 would lead to a cx < 0). Therefore,cx = c cos63.4 = (5.0)(0.45) = 2.2 m.

(c) Also, cy = c sin63.4 = (5.0)(0.89) = 4.5 m.(d) And we know the angle for d to be 26.6 + 90 = 116.6, which leads to dx = d cos 116.6 =

(5.0)(0.45) = 2.2 m.(e) Finally, dy = d sin116.6 = (5.0)(0.89) = 4.5 m.


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39. The solution to problem 27 showed that each diagonal has a length given by a

3, where a is the lengthof a cube edge. Vectors along two diagonals are b = a i + a j + a k and c =

a i + a j + a k. Using

Eq. 3-20 with Eq. 3-23, we find the angle between them:

cos =bxcx + bycy + bzcz

bc=a2 + a2 + a2

3a2=

1

3.

The angle is = cos1(1/3) = 70.5.


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40. (a) The vector equation r = abv is computed as follows: (5.0 (2.0)+ 4.0)i + (4.0 2.0 + 3.0)j +

((6.0) 3.0 + 2.0)k. This leads to r = 11 i + 5.0j 7.0 k.

(b) We find the angle from +z by dotting (taking the scalar product) r with k. Noting that r =|r| =

112 + 52 + (7)2 = 14, Eq. 3-20 with Eq. 3-23 leads to

r k = 7.0 = (14)(1) cos = = 120 .

(c) To find the component of a vector in a certain direction, it is efficient to dot it (take the scalarproduct of it) with a unit-vector in that direction. In this case, we make the desired unit-vector by

b =b

|b|=

2 i + 2j + 3 k

(2)2 + 22 + 32.

We therefore obtain

ab = a b =(5)(2) + (4)(2) + (6)(3)

(2)2 + 22 + 32 = 4.9 .

(d) One approach (if we all we require is the magnitude) is to use the vector cross product, as theproblem suggests; another (which supplies more information) is to subtract the result in part (c)

(multiplied by b) from a. We briefly illustrate both methods. We note that ifa cos (where is

the angle between a and b) gives ab (the component along b) then we expect a sin to yield theorthogonal component:

a sin =|ab|

b= 7.3

(alternatively, one might compute form part (c) and proceed more directly). The second methodproceeds as follows:

a ab b = (5.0 2.35)i + (4.0 (2.35))j + ((6.0) (3.53))k= 2.65 i + 6.35j 2.47 k

This describes the perpendicular part of a completely. To find the magnitude of this part, wecompute

2.652 + 6.352 + (2.47)2 = 7.3

which agrees with the first method.


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41. The volume of a parallelepiped is equal to the product of its altitude and the area of its base. Take thebase to be the parallelogram formed by the vectors b and c. Its area is bc sin, where is the angle

between b and c. This is just the magnitude of the vector (cross) product b c. The altitude of theparallelepiped is a cos , where is the angle between a and the normal to the plane ofb and c. Sincebc is normal to that plane, is the angle between a and bc. Thus, the volume of the parallelepipedis V= a|b c| cos = a (b c).


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42. We apply Eq. 3-30 and Eq. 3-23.

(a) ab = (axby aybx) k since all other terms vanish, due to the fact that neither a nor b have anyz components. Consequently, we obtain ((3.0)(4.0) (5.0)(2.0)) k = 2.0 k .

(b) a b = axbx + ayby yields (3)(2) + (5)(4) = 26.(c) a +b = (3 + 2) i + (5 + 4)j , so that

a +b

b = (5)(2) + (9)(4) = 46.

(d) Several approaches are available. In this solution, we will construct a b unit-vector and dot it(take the scalar product of it) with a. In this case, we make the desired unit-vector by

b =b

|b|=

2 i + 4j22 + 42

.

We therefore obtain

ab = a b = (3)(2) + (5)(4)22 + 42

= 5.8 .


43/62

43. We apply Eq. 3-30 and Eq.3-23. If a vector capable calculator is used, this makes a good exercise forgetting familiar with those features. Here we briefly sketch the method.

(a) We note that b c = 8 i + 5j + 6 k. Thus, a b c

= (3)(8) + (3)(5) + (2)(6) = 21.

(b) We note that b + c = 1 i 2j + 3 k. Thus, a b + c

= (3)(1) + (3)(2) + (2)(3) = 9.

(c) Finally, ab + c

= ((3)(3)(2)(2)) i+((2)(1)(3)(3))j+((3)(2)(3)(1)) k = 5 i11j9 k.


44/62

44. The components ofa are ax = 0, ay = 3.20cos63 = 1.45, and az = 3.20sin63

= 2.85. The components

ofb are bx = 1.40cos48 = 0.937, by = 0, and bz = 1.40sin48 = 1.04.

(a) The scalar (dot) product is therefore

a b = axbx + ayby + azbz = (0)(0.937) + (1.45)(0) + (2.85)(1.04) = 2.97 .

(b) The vector (cross) product is

ab = (aybz azby) i + (azbx axbz) j + (axby aybx) k

= ((1.45)(1.04) 0)) i + ((2.85)(0.937) 0)) j + (0 (1.45)(0.94)) k

= 1.51 i + 2.67j 1.36 k .

(c) The angle between a and b is given by

= cos1

a b

ab

= cos1

2.96

(3.30)(1.40)

= 48 .


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45. We observe thati i

=

i

i sin0 vanishes because sin 0 = 0. Similarly, jj = k k = 0. When the

unit vectors are perpendicular, we have to do a little more work to show the cross product results. First,the magnitude of the vector ij is

ij

=

i

j sin90

which equals 1 because sin 90 = 1 and these are all units vectors (each has magnitude equal to 1). This

is consistent with the claim that i j = k since the magnitude of k is certainly 1. Now, we use theright-hand rule to show that ij is in the positive z direction. Thus ij has the same magnitude anddirection as k, so it is equal to k. Similarly, k i = j and j k = i. If, however, the coordinate systemis left-handed, we replace k k in the work we have shown above and get

i i = jj = k k = 0 .

just as before. But the relations that are different are

ij = k k i = j j k = i .


46/62

46. (a) By the right-hand rule, A B points upward if A points north and B points west. If A and B havemagnitude = 1 then, by Eq. 3-27, the result also has magnitude equal to 1.

(b) Since cos 90 = 0, the scalar dot product between perpendicular vectors is zero. Thus, A B = 0 isA points down and B points south.

(c) By the right-hand rule, A B points south if A points east and B points up. If A and B have unitmagnitude then, by Eq. 3-27, the result also has unit magnitude.

(d) Since cos 0 = 1, then A B = AB (where A is the magnitude of A and B is the magnitude of B).If, additionally, we have A = B = 1, then the result is 1.

(e) Since sin 0 = 0, A B = 0 if both A and B point south.


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47. Let A denote the magnitude of A; similarly for the other vectors. The vector equation is A + B = Cwhere B = 8.0 m and C = 2A. We are also told that the angle (measured in the standard sense) for

A is 0 and the angle for C is 90, which makes this a right triangle (when drawn in a head-to-tailfashion) where B is the size of the hypotenuse. Using the Pythagorean theorem,

B =A2 + C2 = 8.0 =

A2 + 4A2

which leads to A = 8/

5 = 3.6 m.


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48. We choose +x east and +y north and measure all angles in the standard way (positive ones are

counterclockwise from +x). Thus, vector d1 has magnitude d1 = 4 (with the unit meter and three

significant figures assumed) and direction 1 = 225. Also, d2 has magnitude d2 = 5 and direction

2 = 0, and vector d3 has magnitude d3 = 6 and direction 3 = 60

.

(a) The x-component of d1 is d1 cos 1 = 2.83 m.

(b) The y-component of d1 is d1 sin 1 = 2.83 m.

(c) The x-component of d2 is d2 cos 2 = 5.00 m.

(d) The y-component of d2 is d2 sin 2 = 0.

(e) The x-component of d3 is d3 cos 3 = 3.00 m.

(f) The y-component of d3 is d3 sin 3 = 5.20 m.

(g) The sum ofx-components is

2.83 + 5.00 + 3.00 = 5.17 m.

(h) The sum ofy-components is 2.83 + 0 + 5.20 = 2.37 m.

(i) The magnitude of the resultant displacement is

5.172 + 2.372 = 5.69 m.

(j) And its angle is = tan1(2.37/5.17) = 24.6 which (recalling our coordinate choices) means itpoints at about 25 north of east.

(k) and () This new displacement (the direct line home) when vectorially added to the previous(net) displacement must give zero. Thus, the new displacement is the negative, or opposite, of theprevious (net) displacement. That is, it has the same magnitude (5.69 m) but points in the oppositedirection (25 south of west).


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49. Reading carefully, we see that the (x, y) specifications for each dart are to be interpreted as (x,y)descriptions of the corresponding displacement vectors. We combine the different parts of this problem

into a single exposition. Thus, along the x axis, we have (with the centimeter unit understood)

30.0 + bx 20.0 80.0 = 140.0 ,

and along y axis we have40.0 70.0 + cy 70.0 = 20.0 .

Hence, we find bx = 70.0 cm and cy = 80.0 cm. And we convert the final location (140,20) intopolar coordinates and obtain (141 172), an operation quickly done using a vector capable calculatorin polar mode. Thus, the ant is 141 cm from where it started at an angle of 172, which means 172

clockwise from the +x axis or 188 counterclockwise from the +x axis.


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50. We find the components and then add them (as scalars, not vectors). With d = 3.40 km and = 35.0

we find d cos + d sin = 4.74 km.


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51. We choose +x east and +y north and measure all angles in the standard way (positive ones counter-

clockwise from +x, negative ones clockwise). Thus, vector d1 has magnitude d1 = 3.66 (with the unit

meter and three significant figures assumed) and direction 1 = 90. Also, d2 has magnitude d2 = 1.83

and direction 2 = 45, and vector d3 has magnitude d3 = 0.91 and direction 3 = 135. We add thex and y components, respectively:

x : d1 cos 1 + d2 cos 2 + d3 cos 3 = 0.651 m

y : d1 sin 1 + d2 sin 2 + d3 sin 3 = 1.723 m .

(a) The magnitude of the direct displacement (the vector sum d1+d2+d3 ) is

0.6512 + 1.7232 = 1.84 m.

(b) The angle (understood in the sense described above) is tan1(1.723/0.651) = 69. That is, the firstputt must aim in the direction 69 north of east.


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52. (a) We write b = bj where b > 0. We are asked to consider

bd

=

bd

j

in the case d > 0. Since the coefficient ofj is positive, then the vector points in the +y direction.

(b) If, however, d < 0, then the coefficient is negative and the vector points in the y direction.

(c) Since cos 90 = 0, then a b = 0, using Eq. 3-20.

(d) Since b/d is along the y axis, then (by the same reasoning as in the previous part) a (b/d) = 0.

(e) By the right-hand rule, a b points in the +z direction.

(f) By the same rule, ba points in the z direction. We note that ba = ab is true in this caseand quite generally.

(g) Since sin 90

= 1, Eq. 3-27 gives |a b| = ab where a is the magnitude ofa. Also, |a b| = |b a|so both results have the same magnitude.

(h) and (i) With d > 0, we find that a (b/d) has magnitude ab/d and is pointed in the +z direction.


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53. (a) With a = 17.0 m and = 56.0 we find ax = a cos = 9.51 m.

(b) And ay = a sin = 14.1 m.

(c) The angle relative to the new coordinate system is = 5618 = 38. Thus, ax

= a cos = 13.4 m.

(d) And ay

= a sin = 10.5 m.


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54. Since cos0 = 1 and sin 0 = 0, these follows immediately from Eq. 3-20 and Eq. 3-27.


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55. (a) The magnitude of the vector a = 4.0d is (4.0)(2.5) = 10 m.

(b) The direction of the vectora

= 4.

0d

is the same as the direction ofd

(north).(c) The magnitude of the vector c = 3.0d is (3.0)(2.5) = 7.5 m.

(d) The direction of the vector c = 3.0d is the opposite of the direction of d. Thus, the direction ofcis south.


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56. The vector sum of the displacements dstorm and dnew must give the same result as its originally intendeddisplacement do = 120j where east is i, north is j, and the assumed length unit is km. Thus, we write

dstorm = 100 i and dnew = A i + Bj .

(a) The equation dstorm + dnew = do readily yields A = 100 km and B = 120 km. The magnitude ofdnew is therefore

A2 + B2 = 156 km.

(b) And its direction is tan1(B/A) = 50.2 or 180 + (50.2) = 129.8. We choose the latter valuesince it indicates a vector pointing in the second quadrant, which is what we expect here. Theanswer can be phrased several equivalent ways: 129.8 counterclockwise from east, or 39.8 westfrom north, or 50.2 north from west.


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57. (a) The height is h = d sin , where d = 12.5 m and = 20.0. Therefore, h = 4.28 m.

(b) The horizontal distance is d cos = 11.7 m.


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58. (a) We orient i eastward, j northward, and k upward. The displacement in meters is consequently

1000 i + 2000j 500 k .

(b) The net displacement is zero since his final position matches his initial position.


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59. (a) If we add the equations, we obtain 2a = 6c, which leads to a = 3c = 9 i + 12j .

(b) Plugging this result back in, we findb

=c

= 3i + 4

j.


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60. (First problem in Cluster 1)The given angle = 130 is assumed to be measured counterclockwise from the +x axis. Angles (if

positive) in our results follow the same convention (but if negative are clockwise from +x).

(a) With A = 4.00, the x-component of A is A cos = 2.57.

(b) The y-component of A is A sin = 3.06.

(c) Adding A and B produces a vector we call R with components Rx = 6.43 and Ry = 1.54.Using Eq. 3-6 (or special functions on a calculator) we present this in magnitude-angle notation:R = (6.61 167).

(d) From the discussion in the previous part, it is clear that R = 6.43 i 1.54j .

(e) The vector C is the difference of A and B. In unit-vector notation, this becomes

C= A B = 2.57 i 3.06j 3.86 i 4.60jwhich yields C= 1.29 i + 7.66j .

(f) Using Eq. 3-6 (or special functions on a calculator) we present this in magnitude-angle notation:C= (7.77 80.5).

(g) We note that C is the constant in all six pictures. Remembering that the negative of a vector

simply reverses it, then we see that in form or another, all six pictures express the relation C= A B.


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61. (Second problem in Cluster 1)

(a) The dot (scalar) product of 3 A and B is found using Eq. 3-23:

3 A B = 3(4.00cos130) (3.86) + 3 (4.00sin130) (4.60) = 12.5 .

(b) We call the result D and combine the scalars ((3)(4) = 12). Thus, D = (4 A) (3B) becomes, usingEq. 3-30,

12 A B = 12((4.00cos130) (4.60) (4.00sin130) (3.86)) k

which yields D = 284 k .

(c) Since D has magnitude 284 and points in the +z direction, it has radial coordinate 284 and angle-measured-from-z-axis equal to 0. The angle measured in the xy plane does not have a well-definedvalue (since this vector does not have a component in that plane)

(d) Since A is in the xy plane, then it is clear that A D. The angle between them is 90.

(e) Calling this new result G we have

G = (4.00cos130) i + (4.00sin130)j + (3.00)k

which yields G = 2.57 i + 3.06j + 3.00 k .

(f) It is straightforward using a vector-capable calculator to convert the above into spherical coordi-nates. We, however, proceed the hard way, using the notation in Fig. 3-44 (where is in the xyplane and is measured from the z axis):

|G| = r =

(2.57)2 + 3.062 + 3.002 = 5.00

= tan1(4.00/3.00) = 53.1

= 130 given in problem 60 .


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62. (Third problem in Cluster 1)

(a) Looking at the xy plane in Fig. 3-44, it is clear that the angle to A (which is the vector lying

in the plane, not the one rising out of it, which we called G in the previous problem) measuredcounterclockwise from the y axis is 90 +130 = 220. Had we measured this clockwise we wouldobtain (in absolute value) 360 220 = 140.

(b) We found in part (b) of the previous problem that A B points along the z axis, so it is perpendicularto the y direction.

(c) Let u = j represent the y direction, and w = 3 k is the vector being added to B in this problem.

The vector being examined in this problem (well call it Q) is, using Eq. 3-30 (or a vector-capablecalculator),

Q = A

B + w

= 9.19 1 + 7.71j + 23.7 k

and is clearly in the first octant (since all components are positive); using Pythagorean theorem,its magnitude is Q = 26.52. From Eq. 3-23, we immediately find u Q = 7.71. Since u has unitmagnitude, Eq. 3-20 leads to

cos1

u Q

Q

= cos1

7.71

26.52

which yields a choice of angles 107 or 107. Since we have already observed that Q is in the firstoctant, the the angle measured counterclockwise (as observed by someone high up on the +z axis)

from the y axis to Q is 107.

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