Chapter 31. THINK In this problem we’re given the magnitude and direction of a vector in two dimensions, and asked to calculate its x- and y-components. EXPRESS The x- and the y- components of a vector a lying in the xy plane are given by ax a cos , ay a sin where a | a | ax2 a y2 is the magnitude and tan 1 (ay / ax ) is the angle between a and the positive x axis. Given that 250 , we see that the vector is in the third quadrant, and we expect both the x- and the y-components of a to be negative. ANALYZE (a) The x component of a is ax a cos (7.3 m) cos 250 2.50 m , (b) and the y component is ay a sin (7.3 m)sin 250 6.86 m 6.9 m. The results are depicted in the figure below: LEARN In considering the variety of ways to compute these, we note that the vector is 70° below the – x axis, so the components could also have been found from ax (7.3 m) cos 70 2.50 m, ay (7.3 m)sin 70 6.86 m. Similarly, we note that the vector is 20° to the left from the – y axis, so one could also achieve the same results by using ax (7.3 m)sin 20 2.50 m, ay (7.3 m) cos 20 6.86 m. 82 83 As a consistency check, we note that ax2 ay2 ( 2.50 m)2 ( 6.86 m)2 7.3 m and tan 1 ay / ax tan 1[( 6.86 m) /(2.50 m)] 250 , which are indeed the values given in the problem statement. 2. (a) With r = 15 m and = 30°, the x component of r is given by rx = rcos = (15 m) cos 30° = 13 m. (b) Similarly, the y component is given by ry = r sin = (15 m) sin 30° = 7.5 m. 3. THINK In this problem we’re given the x- and y-components a vector A in two dimensions, and asked to calculate its magnitude and direction. EXPRESS A vector A can be represented in the magnitude-angle notation (A, ), where A Ax2 Ay2 is the magnitude and Ay Ax tan 1 is the angle A makes with the positive x axis. Given that Ax = 25.0 m and Ay = 40.0 m, the above formulas can be readily used to calculate A and . ANALYZE (a) The magnitude of the vector A is A Ax2 Ay2 ( 25.0 m)2 (40.0 m) 2 47.2 m (b) Recalling that tan = tan ( + 180°), tan–1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°. Noting that the vector is in the second quadrant (by the signs of its x and y components) we see that 122° is the correct answer. The results are depicted in the figure to the right. LEARN We can check our answers by noting that the x- and the y- components of A can be written as Ax A cos , Ay A sin . 84 CHAPTER 3 Substituting the results calculated above, we obtain Ax (47.2 m) cos122 25.0 m, Ay (47.2 m)sin122 40.0 m which indeed are the values given in the problem statement. 4. The angle described by a full circle is 360° = 2 rad, which is the basis of our conversion factor. 2 rad 0.349 rad . 360 2 rad (b) 50.0 50.0 0.873 rad . 360 2 rad (c) 100 100 1.75 rad . 360 360 (d) 0.330 rad = 0.330 rad 18.9 . 2 rad 360 (e) 2.10 rad = 2.10 rad 120 . 2 rad 360 (f) 7.70 rad = 7.70 rad 441 . 2 rad (a) 20.0 20.0 5. The vector sum of the displacements d storm and d new must give the same result as its originally intended displacement do (120 km)jˆ where east is i , north is j . Thus, we write dstorm (100 km) ˆi , dnew A ˆi B ˆj. (a) The equation dstorm dnew do readily yields A = –100 km and B = 120 km. The magnitude of d new is therefore equal to | dnew | A2 B 2 156 km . (b) The direction is tan–1 (B/A) = –50.2° or 180° + ( –50.2°) = 129.8°. We choose the latter value since it indicates a vector pointing in the second quadrant, which is what we expect here. The answer can be phrased several equivalent ways: 129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west. 6. (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. 85 7. (a) The vectors should be parallel to achieve a resultant 7 m long (the unprimed case shown below), (b) anti-parallel (in opposite directions) to achieve a resultant 1 m long (primed case shown), (c) and perpendicular to achieve a resultant 32 42 5 m long (the double-primed case shown). In each sketch, the vectors are shown in a “head-to-tail” sketch but the resultant is not shown. The resultant would be a straight line drawn from beginning to end; the beginning is indicated by A (with or without primes, as the case may be) and the end is indicated by B. 8. We label the displacement vectors A , B , and C (and denote the result of their vector sum as r ). We choose east as the ˆi direction (+x direction) and north as the ˆj direction (+y direction). All distances are understood to be in kilometers. (a) The vector diagram representing the motion is shown next: A ( 3.1 km) ˆj B (2.4 km) ˆi C ( 5.2 km) ˆj (b) The final point is represented by r A B C (2.4 km)iˆ (2.1 km)ˆj whose magnitude is r 2.4 km 2.1 km 2 (c) There are two possibilities for the angle: 2 3.2 km . 0 k) 10. what we found in part (b).0 (1. It should be noted that many graphical calculators have polar rectangular “shortcuts” that automatically produce the correct answer for angle (measured counterclockwise from the +x axis). We may phrase the angle.0 4. (b) a b [4. or 221 .0)kˆ (5.0 (1. in unit vector notation. Adding the two vectors gives r a b (ax bx )iˆ (ay by )ˆj rx ˆi ry ˆj ANALYZE (a) Given that a (4.0)] ˆi [(3.1 m 3. 2. a second vector b can be expressed as b bx ˆi by ˆj .4 m 4.0)kˆ (3.0 4.4 km tan 1 We choose the latter possibility since r is in the third quadrant. respectively. c (5.0 ˆi 4.8 m. EXPRESS In two dimensions. (a) rx cx d x 7.4 m 12 m .0iˆ 2. ˆ m. (b) ry cy d y 3. a vector a can be written as. y. we find the x and the y components of r to be .0 k) ˆ m.0) 1.0]jˆ (1. a ax ˆi a y ˆj . Similarly. which we note is the opposite of ˆ m. The resultant r is not shown in our sketch.0 ˆj 3.0 m)ˆj .0jˆ 5.3 m 2. All distances in this solution are understood to be in meters.0 m)ˆj and b (13.0)]iˆ [(3. then.0) 1. it would be an arrow directed from the “tail” of A to the “head” of C . as 221° counterclockwise from East (a phrasing that sounds peculiar. (a) a b [4.0 k) (c) The requirement a b c 0 leads to c b a .86 CHAPTER 3 2. at best) or as 41° south from west or 49° west from south.0 ˆj 3. We want to find the magnitude and direction of the resulting vector.8 m 2. and (c) rz cz d z 6.8 m . The x. 9.0] ˆj (1. 11.1 km 41.0 m 5.0 ˆi 4. Thus. and z components of r c d are.0 m)iˆ (7. THINK This problem involves the addition of two vectors a and b .0 m)iˆ (3. 0 m) + (–13 m) = –9. A (50 km) ˆi B (30 km) ˆj C (25 km) cos 60 ˆi + (25 km )sin 60 ˆj . We choose east as the ˆi direction (+x direction) and north as the ˆj direction (+y direction). Thus r (9. 9. Thus. Indeed. we find the angle is 132° (measured counterclockwise from +x axis).0 m) + (7. and C (and denote the result of their vector sum as r ).87 rx = ax + bx = (4. LEARN The addition of the two vectors is depicted in the figure below (not to scale). 12. we expect r to be in the second quadrant.0 m tan 48 or 132 .0 m) = 10.0 m)2 (10 m)2 13 m. since rx 0 and ry 0 .0 m Since the x component of the resultant is negative and the y component is positive. (b) The magnitude of r is r | r | rx2 ry2 (9. B . We note that the angle between C and the x axis is 60°. characteristic of the second quadrant.0 m ry = ay + by = (3.0m) ˆi (10m) ˆj . (c) The angle between the resultant and the +x axis is given by ry rx tan 1 1 10.0 m. We label the displacement vectors A . the magnitude of the overall displacement is given by ( 140 m)2 (30 m) 2 143 m.5km)2 (51.74 km. 14. Although the resultant r is shown in our sketch. 13.88 CHAPTER 3 (a) The total displacement of the car from its initial position is represented by r A B C (62. .5 km) 40°. in unit vector notation. THINK This problem involves the addition of two vectors a and b in two dimensions. 15. EXPRESS In two dimensions. (d) The angle is given by tan 1 (30 /( 140)) 12 .7 km) ˆj which means that its magnitude is r (62.7 km/62. magnitude and direction of the resulting vector. which is to say that it points 40° north of east. we have which gives bx 80 m. or 168 measured counterclockwise from the +x axis).7 km)2 81 km. With d = 3. We’re asked to find the components.5 km) ˆi (51. it would be a direct line from the “tail” of A to the “head” of C . (b) The angle (counterclockwise from +x axis) is tan–1 (51.40 km and = 35. We find the components and then add them (as scalars. (a) Summing the x components. not vectors). (c) Using the Pythagorean theorem. 20 m + bx – 20 m – 60 m = 140 m. a ax ˆi ay ˆj (a cos )iˆ (a sin )ˆj . which implies cy =110 m.0° we find d cos + d sin = 4. we have 60 m – 70 m + cy – 70 m = 30 m. (b) Summing the y components. a vector a can be written as. (which would be 12 measured clockwise from the –x axis. (d) b a (5.0 ˆj) m (2. the resultant r lies in the first quadrant.0 m) ˆj.1 m tan 82. (d) The angle between r and the +x-axis is ry rx tan 1 1 12. From the figure. the y component of r is ry a sin 1 b sin(1 2 ) (10 m)sin 30 (10 m)sin(30 105) 12.0 ˆj) m (5. (c) The magnitude of r is r | r | (1.2 m.0iˆ 2. This is what we expect.2 m.59 m LEARN As depicted in the figure.0 m)ˆj .0 m) ˆi (6. the x component of r is rx a cos1 b cos(1 2 ) (10 m) cos30 (10 m) cos(30 105) 1.0 ˆj) m (3.0iˆ 4. .59 m)2 (12.0 m) ˆi (2.89 Similarly.1 m. Note that the magnitude of r can also be calculated by using law of cosine ( a . (a) a b (3.2 m. 1.0iˆ 2. (b) The magnitude of a b is | a b | (8.1 m)2 12.5 . (c) The angle between this vector and the +x axis is tan–1[(2.0 m)2 (2. 1 and 1 2 (since the angles are measured from the +x-axis) and the resulting vector is r a b [a cos 1 b cos(1 2 )]iˆ [a sin 1 b sin(1 2 )]jˆ rx ˆi ry ˆj ANALYZE (a) Given that a b 10 m. a second vector b can be expressed as b bx ˆi by ˆj (b cos )iˆ (b sin )ˆj .0 m)] = 14°. 16. 1 30 and 2 105.0 m)2 8.0 m)/(8.0 ˆj) m (8. b and r form an isosceles triangle): r a 2 b2 2ab cos(180 2 ) (10 m)2 (10 m)2 2(10 m)(10 m) cos 75 12. we have.59 m (b) Similarly.0iˆ 4. 3 m. a (50 m) cos(30)iˆ (50 m) sin(30) ˆj b (50 m) cos (195) ˆi (50 m) sin (195) ˆj c (50 m) cos (315) ˆi (50 m) sin (315) ˆj a b c (30. which is to say that it is 37.4 m) ˆi (23. The vector is 72° clockwise from the axis defined by ˆi .” In this solution.4)] ˆj (127 ˆi 2. The magnitude of this result is | a b c | (127 m)2 (2.0 m)] = –72°.5° counterclockwise from +x.6 m)2 1. (f) The angle between this vector and the +x axis is tan-1[( –6. The magnitude of this result is (30.90 CHAPTER 3 (e) The magnitude of the difference vector b a is | b a | (2.5°. (d) The angle between the vector described in part (c) and the +x axis is tan 1 (2. Where the length unit is not displayed.9) ( 35.2 .6 m/127 m) 1. (a) Using unit-vector notation.3 m) ˆj.30 102 m. we employ the “traditional” methods (such as Eq. (b) The two possibilities presented by a simple calculation for the angle between the vector described in part (a) and the +x direction are tan–1[(–23.4 m)] = –37. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular polar “shortcuts.60 ˆj) m in unit-vector notation.4] ˆi [25 ( 12. Thus. (c) We find a b c [43. 17.5° clockwise from the +x axis. the angle is –37.3 m)2 38 m .5°) = 142.5°.3 (48.5°.3) 35. 3-6). The former possibility is the correct answer since the vector is in the fourth quadrant (indicated by the signs of its components). This is equivalent to 322.0 m)2 (6.0 m)/(2.2 m)/(30.0 m)2 6. . and 180° + ( –37.4 m)2 (23. the unit meter should be understood. 00 m)2 = 5.9° = 209°. d is given by d a b c ( 40.01 km)iˆ (5.0 .4 ˆj) m . (a) The x and y components of B are given by Bx = Cx – Ax = (15.4 ˆi 47. The desired result is the displacement vector.4)) 50.0 m) cos 40° = –23. We choose the latter possibility as the correct one since it indicates that B is in the third quadrant (indicated by the signs of its components). in units of km.8 km). the resultant is (5.0° = 200°. We choose the latter possibility as the correct one since it indicates that d is in the second quadrant (indicated by the signs of its components). too. its magnitude is | B | (23. (f) The two possibilities presented by a simple calculation for the angle between the vector described in part (e) and the +x axis are tan 1 (47.3 m)2 (12. that the answer can be equivalently stated as 151 . and 180° +28.3 m)] = 28.39 m. This consists of the sum of two displacements: during the whiteout. we should note that the angle between C and the +x axis is 180° + 20.4 m)2 (47. B (7.8 km)(cos50 ˆi sin50 ˆj) (5. If we wish to use Eq. 50 .0 m) sin 200° – (12.4 /(40.6 m .00/5.385 m (b) The angle is tan1(2. which has a magnitude of (40.0 m) cos 200° – (12.00) 21.8 m. or B (7.98 km)ˆj and the unknown C .8 m)2 26.8º (left of forward). 90º (measured counterclockwise from the +x axis).91 (e) Using unit-vector notation.6 km)jˆ . where ˆj is the unit vector along the positive y axis (north). Thus.3 m.6 km).00 ^j) m . and 180 (50.00 m)2 (2. (a) With ^i directed forward and ^j directed leftward.4 m)2 62 m. or A (5. A B C . (5. 18. We note. The magnitude is given by the Pythagorean theorem: 5.8 m)/( –23. Consequently. 20. A = (5. (b) The two possibilities presented by a simple calculation for the angle between B and the +x axis are tan–1[( –12.0 m) sin 40° = –12.0) 130 .9°. . 3-5 in an unmodified fashion. By =Cy – Ay = (15.00 ^i + 2. 19. (a) Allowing for the different angle units used in the problem statement. y) specifications for each “dart” are to be interpreted as ( x.01 km) ˆi (0.0 70. (b) The magnitude of the vector sum found in part (a) is (1.73 i 4.0) is (140)2 (20. y) descriptions of the corresponding displacement vectors.29 i 4. which gives bx = –70.38 km)2 5.60 m)2 6.38 km) /( 5. The magnitude is ( 5.0 140.70 j F 1.72 m . the angle of the overall displacement is given by + tan 1[( 20. so Eq.38 km) ˆj . the unit meter should be understood.0 km.00 j E F G H 1.92 CHAPTER 3 (a) The desired displacement is given by C A B ( 5. However.0 cm. 21.83 j G 1. we see that the (x.28 i 6. –20.0 80.0) /(140)] or 188° counterclockwise from the +x axis (or 172 counterclockwise from the +x axis). (b) The angle is tan 1[( 0.3 . Wherever the length unit is not displayed in the solution below. we arrive at E 3.28 m)2 (6. We use this to write the vectors in unit-vector notation before adding them. (a) Along the x axis. (b) Along the y axis we have 40. 3-5 applies directly. a very differentlooking approach using the special capabilities of most graphical calculators can be imagined. . Angles are given in ‘standard’ fashion.0 which yields cy = 80. Reading carefully.01 km)] 4. (c) The magnitude of the final location (–140 .73 j H 5.0 cm. south of due west.0 cy 70. we have (with the centimeter unit understood) 30.0 bx 20.01 km)2 ( 0.45 i 3.0 20. (d) Since the displacement is in the third quadrant. 22. We combine the different parts of this problem into a single exposition.0)2 141 cm.20 i 3.60 j. 3-6. Thus (since C = 5. we express the situation (described in the problem ^ ^ ^ ^ statement) as A + B = (3A) j . 3.41) (2.10j) ˆ m . If the angle between C and the y axis is tan 1 (3/ 4) 36. Expressing B and D in unit-vector ˆ m and (2. and then finding the difference between that location and the oasis ( B ).93 (c) Its angle measured counterclockwise from the +x axis is tan–1(6. 23. Since i j we may use the Pythagorean theorem to express B in terms of the magnitudes of the other two vectors: B = (3A)2 + A2 A= 1 B = 2. The resultant (along the y axis. 26. The vector equation is R A B C D .63j) length unit is not displayed in the solution below.0 (with x axis).28) = 79.2 m . (b) Using Eq.2.0) we find B = 3.87 . Where the notation. then it should be clear that (referring to the magnitudes of the vectors) B 2C sin( / 2) . As a vector addition problem. The strategy is to find where the camel is ( C ) by adding the two consecutive displacements described in the problem.41) so B C (25 0) (23. the unit meter should be understood.87iˆ 4.60/1.69 m. (a) Adding corresponding components.0° = 1.0°.72 m)2 5.72 m 56. 79.0 90) = (23.5 45) which is efficiently implemented using a vector-capable calculator in polar mode. respectively.25 4.0 m. The distance is therefore 2.25 4.38 rad. 10 25.72 m) ˆj . (d) Using the conversion factor rad = 180 .69iˆ 3. where A = A i and B = 7. we obtain R (3.18 m)iˆ ( 4. with the same magnitude as C ) forms (along with C ) a side of an isosceles triangle (with B forming the base). we have (1.18 m)2 (4. Using the magnitude-angle notation C = (24 15) + (8. 24.18 m tan 1 .6 km. the magnitude is | R | (3. (c) The angle is 4. (b) The angle of d A is tan 1 (d A.19 m) ˆi (0. Let A represent the first part of Beetle 1’s trip (0. converting the result to polar coordinates. the displacement for point B is . Therefore. the angle is then 180 56.0 iˆ sin50. Let l0 2. For their respective second parts: B is 0. the displacement vector for point A is d A w v i h l0 (cos 60ˆi sin60 ˆj) l0 ˆj l0 (cos120ˆi sin120 ˆj) l0 ˆj (2 3)l0 ˆj.0 cm) 7. The final position of Beetle 1 is A B (0. and (b) d2 = d3 = 2 ^i + 4 ^j. (c) Similarly.0ˆj) (1.16) 79 .72g b318 b g 5. 29.23 m)jˆ (a) We find D A B C (0.60 m)(cos50.5 cm . where C (1.84 m.8 m)(cos30 ˆi sin30 ˆj) (1.40 m) ˆj. y / d A.50 m east or 0.80 m at 30º north of east and D is the unknown. . The nest is located at the endpoint of segment w.83 m)ˆj .0 124 . (a) Using unit-vector notation.5 m)iˆ (0. Thus.83/ 0. Solving the simultaneous equations yields the answers: (a) d1 = 4 d3 = 8 ^i + 16 ^j . The equation relating these is A B C D .6 m at 50º north of east). x ) tan 1 () 90 . (b) The angle is tan 1 ( 0.16 m)iˆ ( 0. 28. 4.5 ˆi ) and C represent the first part of Beetle 2’s trip intended voyage (1. the magnitude of d A is | d A | (2 3)(2. we obtain .0 cm be the length of each segment. which is interpreted to mean 79º south of east (or 11º east of south).69 124 27. and the magnitude is D = 0.03 m)iˆ (1.94 CHAPTER 3 If measured counterclockwise from +x-axis. 0 m)] = 53°. (a) The magnitude of a is a (4.0 m)2 (8. .0 m)2 5.3) 8. (d) The direction of d B is d B. the same result as in part (e) (exactly. we employ the “traditional” methods (such as Eq.0 m)2 ( 3.0 m) 8.0 m) 2 10 m.0 m 6.0 m) ˆi [8.” In this solution.0 m]jˆ (10 m)iˆ (5.0 m 4.95 dB w v j p o l0 (cos 60ˆi sin 60 ˆj) l0 ˆj l0 (cos 60ˆi sin60 ˆj) l0 (cos 30ˆi sin30 ˆj) l0 ˆi (2 3 / 2)l0 ˆi (3 / 2 3)l0 ˆj.0 m)/(10 m)] = 27°.0 m)] ˆj (2. (g) b a (6. (d) The angle between b and the +x axis is tan–1[(8. (c) The magnitude of b is b (6. The magnitude of this vector is | a b | (10 m)2 (5.0 cm)(4.0 m)/(6.0 m. which is. the magnitude of d B is | d B | l0 (2 3 / 2)2 (3/ 2 3)2 (2. (e) a b (4.0 m (3.0 m)2 (11 m)2 11 m. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular polar “shortcuts. The magnitude of this vector is | b a | (2. y 1 3/ 2 3 1 tan tan (1.13) 48 . The vector is 37° clockwise from the axis defined by i .6 cm .0 m)ˆj. not just to 2 significant figures) (this curious coincidence is made possible by the fact that a b ). interestingly.0 m)2 11 m. d 2 3/2 B. we round to two significant figures in our results.0 m) iˆ (11 m) ˆj. x B tan 1 30. (f) The angle between the vector described in part (e) and the +x axis is tan–1[(5. (b) The angle between a and the +x axis is tan–1 [(–3.0 m)/(4. 3-6).0 m) ˆi [( 3.0 m)] = –37°. Therefore. 0°) = 38.0 m) 8. 0.0 m 6. 0.0 m] ˆj (2. a) with the position vector a ˆi a kˆ . the vector along the line is the difference a ˆi a ˆj a kˆ . the vector along the line is the difference a ˆi aˆj a kˆ . Thus. the diametrically opposite point is (0. . The magnitude of this vector is | a b | ( 2. which corresponds to the position vector a î.0 m) ˆi (11 m) ˆj. the angle between them is 180°. the point diametrically opposite the origin (0.96 CHAPTER 3 (h) The angle between the vector described in part (g) and the +x axis is tan–1[(11 m)/(2. (k) Since a b ( 1)(b a ) .0. a. Thus.” (b) From the point (a. a. and 180° + 80° = 260°.0°. (c) The angle relative to the new coordinate system is ´ = (56. The latter possibility is the correct answer (see part (k) for a further observation related to this result).0° we find ax = a cos = 9.1 m. (c) If the starting point is (0.0 m)] = 80°.4 m. (j) The two possibilities presented by a simple calculation for the angle between the vector described in part (i) and the +x direction are tan–1 [(–11 m)/(–2. 31. ay = a sin = 14.0° – 18.0 m)] = 80°. Thus. 32.51 m. 0) with the corresponding position vector a ˆj .5 m.0 m and = 56. they point in opposite (anti-parallel) directions.0 m)2 ( 11 m)2 11 m . (b) Similarly. the diametrically opposite point is (a.0) has position vector a i a j a k and this is the vector along the “body diagonal. ax a cos 13. ay = a sin ´ = 10. (a) With a = 17. 0). (a) As can be seen from Figure 3-30.0 m) ˆi [( 3. a) with the position vector a j a k . (d) Similarly. (i) a b (4. a) with the position vector a kˆ .0) (4. 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 diagonal vectors and any of the cube sides adjacent to them. Examining the figure.0) ˆi (5. (a) a b = (axby a y bx ) kˆ since all other terms vanish. Thus 54. It is ˆ We may think of it as the sum of the vector a i parallel to the x axis and a ˆi a ˆj a k. 0.0) (5.0)(2. Since the magnitude of the perpendicular component is a 2 a 2 a 2 and the magnitude of the parallel component is a. 3-23 and Eq. 3-27.0) (2. the vector a b points in the ˆi ˆj kˆ .0 4. or the +z direction.0)]kˆ 2. (f) The length of any of the diagonals is given by 33.97 (d) If the starting point is (a.0)(4.0) = 46 . (c) a b (3. Thus. a 2 a 2 a 2 a 3.0) = 12 since the angle between them is 90º. The tangent of the angle between the vector and the x axis is the perpendicular component divided by the parallel component. (a) a b = (3.0) = 26.0) ˆj (a + b ) b = (5. the vector along the line is the difference a ˆi a ˆj a kˆ . (b) a b axbx a yby yields (3. due to the fact that neither a nor b have any z components. where a b .0)(4. . (b) Using the Right-Hand Rule. (f) The vector points in the +z direction. Consequently.7 .0 2.0) + (5.0) + (9. or the z direction. tan a 2 / a 2 . as in part (a). 34. the vector a j a k perpendicular to the x axis. the diametrically opposite point is (0.0)(2. (e) Consider the vector from the back lower left corner to the front upper right corner.0)(4. 0) with the corresponding position vector a ˆi a ˆj . (c) | a c | = | a ( a b )| = | a b )| = (d) The vector a b points in the ˆi ˆj kˆ . we obtain [(3. We apply Eq. (e) | b c | = | b ( a b )| = | b a ) | = | a b ) | = 12. we see that a + b + c = 0.0 kˆ . a. We apply Eq.0)(3. and dcross d1 d2 .30) sin(320º – 85.0)(5.0kˆ . 36.0) (3.0 ˆj bˆ . we rewrite the given expression as 4( dplane · dcross ) where dplane = d1 + d2 and in the plane of d1 and d 2 .50)(7.0)( 2.0) (3. (a) The scalar or dot product is (4.0 ˆi 4.0)] kˆ 5iˆ 11jˆ 9kˆ 38.0)(2.0) (1.0º) = – 18.0) 9.50)(7.0)2 (4.0) (6.8. Thus.3-27. In this case. a (b + c ) [(3. (c) Finally. Here we briefly sketch the method.0) ( 2. we will construct a b unit-vector and “dot” it (take the scalar product of it) with a . ^ (b) The vector or cross product is in the k direction (by the right-hand rule) with magnitude |(4. (a) We note that b c 8.0 ˆi 5. Using the fact that we obtain ˆi ˆj k. First. this makes a good exercise for getting familiar with those features.0) ( 2.0 ˆj 6.0)] ˆi [( 2.0)] ˆj .0º)| = 26.0 k. Noting that dcross is perpendicular to the plane of d1 and d 2 .0) (5. a (b c ) = (3. (2. In this solution.0) = 21. we see that the answer must be 0 (the scalar or dot product of perpendicular vectors is zero).0)2 (4. 3-23 and Eq.0)(1. we make the desired unitvector by b 2. If a vector-capable calculator is used.0) 2 35.0)(4.9 . (b) We note that b + c = 1.98 CHAPTER 3 (d) Several approaches are available.0)(1.0)( 2.0) (3. |b | (2.0 ˆj + 3. ˆ ˆj kˆ ˆi.8 . a (b c ) (3.0) ( 2.0)(3. [(3.0.30)cos(320º – 85. 37.0)2 We therefore obtain (3.0) ab a bˆ 5.0 ˆi 2. kˆ ˆi ˆj .0) (3.0) ( 2. ˆ Thus.0) (3.0) ( 8. Next. d (1.04 ˆj 4.00 ˆi 8.00ˆj 44.70 m) 2 1.0jˆ 34.04 m) ˆj (4.01 m) kˆ d1 (4.0iˆ 16. Thus.01 m) 2 4.21 m)2 (0.00kˆ 3.04 m) 2 (4.0k.70 m) kˆ .48 k) (c) The magnitudes of d1 and d 2 are d1 (2. A B AB cos .40 m.50 m)(cos 63 ˆj sin 63 k) ˆ (1.333 .00iˆ 3.70k) ˆ = 2.01)(1.21iˆ 0.43 ˆi 4.21 m) ˆi (0.0)] 540.01k) ˆ + (2.04)(1.00iˆ 4.40 m)(cos 30 ˆi sin 30 k) 2 (a) The dot product of d1 and d 2 is ˆ (1. making use of ˆi ˆi = ˆj ˆj = kˆ kˆ = 1 ˆi ˆj = ˆj kˆ = kˆ ˆi = 0 we have 3C 2 A B 3 7.00) (44. 40.00 and A B 14.04 ˆj 4.0) (0) (34.70)iˆ (4.00jˆ 2.50 m d 2 (1.0 ˆj 34.00) (16. we have cos A B AB With A 6.21)ˆj (2. The displacement vectors can be written as (in meters) ˆ (2. B 7.0)+( 8.21)(k) ˆ m2 . d1 d2 (2.00kˆ ˆ 44. or 70.00jˆ 4. (1.01k) (b) The cross product of d1 and d 2 is ˆ (1.86 ˆj 2.70 k) ˆ d1 d 2 (2.04)(0.5 .0 . the angle between the two vectors is . 39.00 .01k) ˆ (0.21iˆ 0. cos 0.0 kˆ 3[(7.0 iˆ 16.99 2 A B 2 2. From the definition of the dot product between A and B .81 m2 .70k) ˆ = (4. 0) (2.74.0) 0.0) 2 3. THINK The angle between two vectors can be calculated using the definition of scalar product. The angle between them is found to be cos (3.0)ˆj (3.0)kˆ .81 m2 1 cos cos 63.5. LEARN As the name implies. ANALYZE Given that a (3.0) 2 (3.0) 2 (3.40 m) d1d 2 1 41.50 m)(1.20) (3.0) 2 5.0) (3. It can be regarded as the product between the magnitude of one of the vectors and the scalar component of the second vector along the direction of the first one.0)ˆj (3.0)iˆ (3.926.20 b | b | bx2 by2 bz2 (2.0) (3.0)kˆ and b (2. the scalar product (or dot product) between two vectors is a scalar quantity.0)iˆ (1. (5. (4. as illustrated below (see also in Fig.0) (3.74) or = 22°. EXPRESS Since the scalar product of two vectors a and b is a b ab cos axbx a yby az bz . the angle can be readily calculated.0) (1.0) 2 (3. the angle between them is given by cos axbx a y by az bz ab axbx a yby az bz cos 1 ab .100 CHAPTER 3 d1 d 2 2. the magnitudes of the vectors are a | a | ax2 a y2 az2 (3.0) 2 (1. Once the magnitudes and components of the vectors are known. 3-18 of the text): a b ab cos (a)(b cos ) . (b) Similarly. (d) and the y-component is by = b sin 30° = (4. d1 4.0 ˆj d1x ˆi d1 y ˆj. (c) (d1 d2 ) d2 d1 d2 d22 8.101 42.0) (5. 43. Therefore the component of the d1 vector along the direction of the d2 vector is 6.00 m. (d) Note that the magnitude of the d1 vector is 16+25 = 6. b and c on the xy-plane. the three vectors can be written as a ax ˆi b b ˆi b ˆj (b cos )iˆ (b sin )ˆj x y ˆ c cx ˆi c y ˆj [c cos( 90)]iˆ [c sin( 90)]j.0)2 33. which implies that the angle between c and the +x axis is + 90°. we note that c b .00 m. (e) The x-component of c is cx = c cos 120° = (10. d2 3.0)]k=31 kˆ (b) The scalar (dot) product gives d1 d2 d1x d2 x d1 y d2 y (4. EXPRESS From the figure.6. ANALYZE (a) The x-component of a is ax = a cos 0° = a = 3. THINK In this problem we are given three vectors a .0)2 (4.0 m) cos 120° = –5.0 ˆi+5. (c) The x-component of b is bx = b cos 30° = (4.0)(4. In unit-vector notation. in unit of meters.0)cos = 8.0)(4.5.0 ˆj d2 x ˆi d2 y ˆj (a) The vector (cross) product gives ˆ d1 d2 (d1x d2 y d1 y d2 x )kˆ [(4.0) 8. The two vectors are written as. the y-componnet of a is ay = a sin 0° = 0. the dot product is (6.0.4. Now.0 ˆi+4.4cos 1. .46 m.00 m) cos 30° = 3.4)(5. The above expressions allow us to evaluate the components of the vectors. and asked to calculate their components.00 m) sin 30° = 2.00 m. Dividing both sides by 32 and taking the inverse cosine yields = 75.0)( 3.0) (5.0 ( 3.0)(3. LEARN This exercise shows that given two (non-parallel) vectors in two dimensions.0 Bx 2.0 Bx ) Since we are told that Bx = By.00 m) q (3. 45.0 ˆj 4. we find Bz = –4.13 ˆj B B ˆi B ˆj 7.0 Bz ) 12 2 (2.00 m p (3. q = 4. Substituting the values found above.20 ˆj. we find p = –6.33 (note that it’s easiest to solve for q first).46 m) 8. Applying Eq. Thus.66 m. Inserting this value into the first equation.0.0 k. the third equation leads to By = –3. Solving these equations.0 Bz 6.0 m) sin 120° = 8. our answer is ˆ B 3. 3-23. The numbers p and q have no units.72 ˆi 9.14 ˆi 6.66 m q (2. F qv B (where q is a scalar) becomes Fx ˆi Fy ˆj Fz kˆ q vy Bz vz By ˆi q vz Bx vx Bz ˆj q vx By vy Bx kˆ which — plugging in values — leads to three equalities: 4.67.00 m).0 2 (4. The two vectors are given by A 8. x y . (h) Similarly. (g) The fact that c pa qb implies c cx ˆi cy ˆj p(ax ˆi) q(bx ˆi by ˆj) ( pax qbx )iˆ qby ˆj or cx pax qbx .00(cos130 ˆi sin130 ˆj) 5. we have 5.0 By ) 20 2 (6.102 CHAPTER 3 (f) and the y-component is cy = c sin 30° = (10.0.0 ˆi 3. the third vector can always be written as a linear combination of the first two. cy qby . 44.0 By 4. The angle measured from the +z axis is = cos1(3. by = (4.6k) (c) We note that the azimuthal angle is undefined for a vector along the z axis.71 m and cy ay by = 0 + 3. (a) Let c a b .0 m) cos 35° = 3.28 m. Then ax = 5.14)(7.13)2 (3. then the answer is 90.72iˆ 9. The vectors are shown on the diagram. and = 0.14)2 (6.14 103 kˆ 4 A 3B 12 A B 12(5.71m 2 3. and A B is perpendicular to that plane.54 .00 k = –5. ^ (e) Clearly. ay = 0. (f) The Pythagorean theorem yields magnitude A (5.2 m.54) = 69.14 ^i + 6. Then cx ax bx = 5.28 m .20 ˆj) 5[(5. Thus. (b) In unit vector notation ˆ 1.4. just as it was in the problem statement ( A is the projection onto the xy plane of the new vector created in part (e)). The azimuthal angle is = 130.72) (6. The magnitude of c is c cx2 cy2 2.14 ˆi 6.4.0 m) sin 35° = –2.13 ˆj) (7.00/8.29 m = 2.103 (a) The dot product of 5A B is 5 A B 5(5.13)(9.28 m = 3. our result is “1.20)] 83.13 ˆj) (7.00 k^ .28m 4.14iˆ 6.14103. 46. bx = –(4.” (d) Since A is in the xy plane.29 m.00)2 8. not defined.20 ˆj) 12(94.00 m 2. 2 .13 ^j + 3.72 ˆi 9. The x axis runs from west to east and the y axis runs from south to north. A + 3.0 m. 8°. The magnitude of c is c cx2 c 2y 8.72 ˆi 9.00(cos130 ˆi sin130 ˆj) 5.2° and 155.29 m There are two solutions: –24.28 m 4. As the diagram shows.28 tan 50.13 ˆj B B ˆi B ˆj 7. Let c b a.0 m .4° + 180° = 230. (c) The vector b a is found by adding a to b . x y (a) The angle between the negative direction of the y axis ( ˆj ) and the direction of A is . Noting that the given 130 is measured counterclockwise from the +x axis.5 50.50.29 m cy by ay 3. the two vectors can be written as A 8. 47. 7. The components are cx bx ax 2. The vector c a b is 24° north of west.00 m 7.4°) is rejected because it would point in a direction opposite to c .29 m 5.20 ˆj. (d) The tangent of the angle that c makes with the +x axis (east) is tan cy cx 3.104 CHAPTER 3 (b) The angle that c a b makes with the +x axis is cy 1 3.14 ˆi 6.71 cx tan 1 The second possibility ( = 50. 2. the second solution is correct.28 m. The result is shown on the diagram to the right. Therefore.45) = 2. which leads to .14 ˆi 6.13)2 A cos1 6.2 m.6) (4.53 .5) (3.6)2 3.00k) y axis ( ˆj ) and the direction of the above vector is 15.6°.2)2 (1.6°.4°) = (5.5)2 4.4° (the other possibility. Where the length unit is not displayed.00 Alternatively.14)2 (6.6° –90° = –63.4° )= (5. a b axbx a y by ab cos (3.50)2 (4.72 ˆi 9.53) cos which leads to = 57° (the inverse cosine is double-valued as is the inverse tangent.42 ˆj 94.6. (c) The vector can be simplified as ˆ (5. (b) Since the y axis is in the xy plane. and the answer is simply given by 270130 = 140.39 ˆi 15. we know the angle for c is 26. (b) Since the angle (measured from +x) for a is tan–1(1.6 cos1 48. 97.20 ˆj 3. 26.105 A (ˆj) 6. and A B is perpendicular to that plane.13 ˆj) (7.0)( –0.00 k) ˆ A ( B 3.6° + 90° = 116. (d) And we know the angle for d to be 26.42 99.5 m.2) (0. the unit meter is understood.58 and b | b | (0.0.6/3. cx = c cos (–63.61kˆ ˆ | 97. but we know this is the right solution since both vectors are in the same quadrant). one may say that the y direction corresponds to an angle of 270. The angle between the negative direction of the Its magnitude is | A ( B 3. (c) Also. Now.13 cos1 140.00 k) 18.58) (4.0)(0.13 1 cos (5.1. (a) We first note that the magnitudes of the vectors are a | a | (3.6° + 90° would lead to a cx < 0). then the answer is 90. 8. cy = c sin (–63.89) = – 4.2) = 26.50) (1. 89) = 4.0 km tan 1 The angle is related to by 90 .9 north of due west). . LEARN This problem could also be solved by first expressing the vectors in unit-vector ˆ c (90. The angle between b and the +x-axis is 90. b c a (50. We want to find the displacement vector between two locations.5 m.0 km)iˆ (90. dy = d sin 116.45) = –2. This gives notation: a (50. (e) Finally. EXPRESS The situation is depicted in the figure below.0 km north).0 km tan 1 west of north (which is equivalent to 60.1 . We look for a vector b such that c a b .0)( –0. 50. ANALYZE (a) Using the Pythagorean theorem. THINK This problem deals with the displacement of a sailboat. the distance traveled by the sailboat is b (50. 49.0 km east) and c represent the intended voyage (90. Let a represent the first part of his actual voyage (50.2 m.1 90.0 km)2 (90. (b) The direction is 50.0 km)i.6° = (5.6°) = (5.0 km)ˆj .0 km 29.106 CHAPTER 3 dx = d cos(116.0)(0.0 km)2 103 km.0 km 119.0 km)ˆj . (b) The magnitude of the vertical component of AB is |AD| sin 52.107 50. it is rendered effectively two-dimensional by referring measurements to its well-defined plane of the fault.0 ˆi 3.0 kˆ d3 4. the magnitude is d1d2 / 4. 52. The three vectors are d1 4.0 m) 2 27.0° = 13. (c) d1 d2 0 since ˆi ˆj = 0. The ¼ factor does not affect the result. (g) The magnitude of the vector in (e) is d1d 2 . (e) d1 d2 d1d2 (ˆj ˆi) = d1d2 kˆ .0ˆj+2.0 ˆi 2.0 m) 2 (22.8m.0ˆj 6. (d) d1 (d2 / 4) (d1 d2 ) / 4 0 . The two vectors are perpendicular to each other. (b) The vector d1 /(4) (d1 / 4)ˆj points in the +y direction. The two vectors d1 and d 2 are given by d1 d1 ˆj and d2 d2 ˆi.4 m. (a) The magnitude of the net displacement is | AB | | AD |2 | AC |2 (17.0 kˆ d 2 1. (j) The direction of d1 (d2 / 4) (d1d2 / 4)kˆ is in the +z-direction. (h) The magnitude of the vector in (f) is d1d 2 . Although we think of this as a three-dimensional movement.0 ˆi 5. The minus sign (with the “4”) does affect the direction: (–y) = + y. (a) The vector d2 / 4 (d2 / 4) ˆi points in the +x direction. as in part (c). (f) d2 d1 d2 d1 (iˆ ˆj) = d1d2 kˆ . in the +z-direction. in the z-direction.0 kˆ . 51.0ˆj+3. (i) Since d1 (d2 / 4) (d1d2 / 4)kˆ . ANALYZE (a) Given that a | a | 10 .0 m 0.108 CHAPTER 3 (a) r d1 d2 d3 (9. THINK This problem involves finding scalar and vector products between two vectors a and b .0) ( 6.0)2 (5. (b) Similarly. the magnitude of the vector (cross) product of the two vectors is . From (c).543 | r | 12. and uˆ is the unit vector in the direction of d 2 .0 m)iˆ (6.0) (5.0 m)2 (6. 3-20 and 3-24 to calculate the scalar and vector products between two vectors: a b ab cos | a b | ab sin .0) 12 d d1 1 2 = 1 2 3. (c) The component of d1 along the direction of d 2 is given by d d1 uˆ = d1cos where is the angle between d1 and d 2 . 3-24).0) 2 (3.2 m.0)2 ( 6.0 m)kˆ . d2 14 ( 1.0)(2.0) cos 60 30. This gives the magnitude of the perpendicular component (and is consistent with what one would get using Eq. Using the properties of the scalar (dot) product.9 m. the scalar (dot) product of a and b is a b ab cos (10) (6.0 m) 2 ( 7.0 and 60 .2 m)2 8. we have d d d d (4.2 m.0 m) 2 12. The angle between r and the z-axis is given by cos r kˆ 7.0)(3. b | b | 6. or a full specification in terms of unit vectors) is sought then more computation is needed. 53.9 m which implies 123 .0)2 (2.0 m)ˆj ( 7.0) 2 d1d 2 (d) Now we are looking for d such that d12 (4.0)( 1. EXPRESS We apply Eqs. (b) The magnitude of r is | r | (9.0)2 77 d 2 d2 . we have d 77 m2 ( 3. but if more information (such as the direction. (g) The sum of x-components is dx = d1x + d2x + d3x = –2.00 m.83 m. (e) The x-component of d 3 is d3x = d3 cos 3 = 3.0 .00 m and direction 3 = 60°. Also.0) sin 60 52. vector d1 has magnitude d1 = 4. respectively.83 m. d 2 has magnitude d2 = 5. 2 (b) a · c = a · ( a b ) = a = (c) Similarly.83 m + 5.83 m + 0 + 5.00 m (with the unit meter) and direction 1 = 225°. b · c = 9. Thus. We choose +x east and +y north and measure all angles in the “standard” way (positive ones are counterclockwise from +x).00 m = 5. (a) a · b = 0 since the angle between them is 90º. 55. (i) The magnitude of the resultant displacement is . LEARN When two vectors a and b are parallel ( 0 ). However. when they are perpendicular ( 90 ).00 m + 3. 54. it is clear that a + b + c = 0. and vector d 3 has magnitude d3 = 6. (h) The sum of y-components is dy = d1y + d2y + d3y = –2. (f) The y-component of d 3 is d3y = d3 sin 3 = 5.00 m. (b) The y-component of d1 is d1y = d1 sin 1 = –2.20 m = 2.17 m.37 m. we have a b ab cos 0 and | a b | ab sin ab . their scalar and vector products are a b ab cos ab and | a b | ab sin 0 .20 m.109 | a b | ab sin (10) (6. (a) The x-component of d1 is d1x = d1 cos 1 = –2. (c) The x-component of d 2 is d2x = d2 cos 2 = 5. (d) The y-component of d 2 is d2y = d2 sin 2 = 0.00 m and direction 2 = 0°. where a b . From the figure. A B (4. with the unit meter understood. we have A B (6.110 CHAPTER 3 d d x2 d y2 (5. it has the same magnitude (5.0)ˆj. if they are measured counterclockwise from the +x axis. 56.0sin 25.00 cos 250 ˆi 8.37 m)2 5.2 m .0 m) 9.63 m)2 10.0)iˆ (7. That is. respectively. we have P 10.0sin 100 ˆj R 8. of the previous (net) displacement. R.17) = 24. 3-5 directly. and S are 100°. the new displacement is the negative.0)iˆ (1.00sin 250 ˆj S 9. Thus.00sin 310 ˆj P Q R S (10. (c) The angle is tan–1 (1. (j) And its angle is = tan–1 (2. 250°.0 ˆi 10. we should note that the angles for Q.69 m) but points in the opposite direction (25° south of west).0 m )iˆ (1.0)ˆj Solving the simultaneous equations gives A and B .0 cos 100 ˆi 12.69 m.24° measured counterclockwise from the +x axis. which (recalling our coordinate choices) means it points at about 25° north of east.0 m)2 (1. EXPRESS From the problem statement. or opposite.0 cos 25.00 cos 310 ˆi 9. THINK This problem deals with addition and subtraction of two vectors. 57. .0 ˆj Q 12.37/5.63 m/10.17 m)2 (2.6°. If we wish to use Eq. and 310°.63 m)ˆj (b) The magnitude of the vector sum is (10. (k) and (l) This new displacement (the direct line home) when vectorially added to the previous (net) displacement must give zero. (a) Using unit-vector notation. and its magnitude is B | B | Bx2 By2 (5.0)2 4. the direction of c is south. the unit meter should be understood. Where the length unit is not displayed in the solution.5 m.0)(2.0 and B (14. where the ‘prime’ notation indicates that b g b g the description is in terms of the new coordinates. Thus.5 m) = 7.0)iˆ (4. The vector can be written as d (2. If we convert B to the magnitude-angle notation (as A already is) we have B 14. The results are summarized in the figure to the right.71 m) ˆj. In the magnitude-angle notation.1 LEARN The vector B is B (5.0)ˆj .5 m)jˆ . Converting these results to (x. A 12.19 m) ˆi (7. y) representations.7 (appropriate notation especially if we are using a vector capable calculator in polar mode). (a) The magnitude of the vector a 4. (c) The magnitude of the vector c = 3. where we have taken ˆj to be the unit vector pointing north. (d) The direction of the vector c = 3. we obtain (a) A (9.0 40. Thus. rotating the axis by +20° amounts to subtracting that angle from the angles previously specified.4 13.0 d is (4.4 33.0d is (3.5 m) = 10 m.0)2 (3. The magnitude of A is A | A | Ax2 Ay2 (1.0d is the opposite of the direction of d . 59. Reference to Figure 3-18 (and the accompanying material in that section) is helpful.0)(2.111 ANALYZE Adding the above equations and dividing by 2 leads to A (1.0)iˆ ( 3.7 ) .0)ˆj .8 .0d is the same as the direction of d (north). .0)2 5. 58.0)2 (4. (b) The direction of the vector a = 4. 0 6. 3-20 with Eq. In this case.0 m) ˆi (3.0 3. We briefly illustrate both methods.0 14 1 cos 120 .0)kˆ ˆ ˆ 7.0 2. we make the desired unit-vector by ˆ ˆ +3. The three vectors given are a 5.0 b 2.0 4. B (14. 60.0 3.0 2. 3-23 leads to r k 7.9 .0k. Eq.0 4. it is efficient to “dot” it (take the scalar product of it) with a unit-vector in that direction.0 kˆ kˆ kˆ (a) The vector equation r a b c is r [5.112 CHAPTER 3 (b) Similarly. We note that if .0 ˆi 3.0 3.0i+2.0)2 (3.0 ( 2. The two vectors can be found be solving the simultaneous equations. which leads to a 3c 9 ˆi 12 ˆj .0) We therefore obtain 5.0) 4.0)2 = 14.0)2 + ( 7.0) (3.0]iˆ (4.0 2. ˆ =11i+5. (b) Plugging this result back in.0 ˆj 2.0 ˆj 3.0 ˆj 6.0j bˆ . we find b c 3i 4j . ab a bˆ 2 2. that r = |r | = (11.41 m) ˆj . (a) If we add the equations. (c) To find the component of a vector in a certain direction. 61. 2 2 2 |b | 2.0)ˆj ( 6.0 c 4.0kˆ b 2.0j Noting (b) We find the angle from +z by “dotting” (taking the scalar product) r with k.0)2 (d) One approach (if all we require is the magnitude) is to use the vector cross product.0 (2.0 ˆi 4. another (which supplies more information) is to subtract the result in part (c) (multiplied by b ) from a .0 (2. as the problem suggests. we obtain 2a 6c .0 2.0)2 + (5.0 ˆi 2. the first putt must aim in the direction 69° north of east.0 kˆ 2 ˆ d3 2. we compute (2.65 m y : d1 sin 1 d 2 sin 2 d3 sin 3 1. respectively: x : d1 cos 1 d 2 cos 2 d3 cos 3 0.66 (with the unit meter and three significant figures assumed) and direction 1 = 90°.113 a cos (where is the angle between a and b ) gives ab (the component along b ) then we expect a sin to yield the orthogonal component: a sin a b b 7. We choose +x east and +y north and measure all angles in the “standard” way (positive ones counterclockwise from +x.0 kˆ d 2.7 m)2 1.3 (alternatively.0 2. (b) The angle (understood in the sense described above) is tan–1 (1.35 i 4. The second method proceeds as follows: b g c b gh cb g b gh a ab b 5. negative ones clockwise).0 ˆi 3. (a) The magnitude of the direct displacement (the vector sum d1 + d2 + d3 ) is (0.35 j + 6. We add the x and y components.7 m. To find the magnitude of this part. one might compute form part (c) and proceed more directly).65i 6.0 ˆi 3.65) = 69°. The three vectors are d1 3.47)2 7.83 and direction 2 = –45°.0 ˆj 2.7/0. Thus. vector d1 has magnitude d1 = 3.0 353 .0 ˆj 1.0 ˆi 4.0 k.65)2 (6. Also.8 m.3 which agrees with the first method. k = 2.0 2. That is. 63.0 ˆj 2. and vector d 3 has magnitude d3 = 0.35)2 ( 2. d 2 has magnitude d2 = 1.47 k This describes the perpendicular part of a completely.35j 2. 62.91 and direction 3 = –135°.65 m)2 (1. . 0 kˆ .0 ˆj 2.0 m 2 . 3-27 then leads to ˆ (0 ˆi 1.114 CHAPTER 3 (a) Since d2 d3 0 ˆi 1.70 m)2 (4.0 ˆj 2. . we obtain d | d | w2 l 2 h2 (3.0 ˆj 3.0 52 m3.0 kˆ ) m 2 64.0 ˆj 2.0 ˆj 2.0 ˆj 2.0 k) = (11iˆ + 9. we have ˆ (0 ˆi 1. 3-27. (c) We found d2 + d3 in part (a).0 3.0 k) 30 18 4.0 ˆi 3.0 ˆj + 3.0 k. THINK This problem deals with the displacement and distance traveled by a fly from one corner of a room to the diagonally opposite corner. we obtain d2 d3 10 ˆi 6. ANALYZE (a) The magnitude of the displacement from one corner to the diagonally opposite corner is d | d | w2 l 2 h2 Substituting the values given.0 k) ˆ d1 (d 2 d3 ) (3.0 ˆj 3. The displacement vector is threedimensional.30 m)2 (3.0 ˆi 3.0 ˆi 3.0 ˆj 3. EXPRESS The displacement of the fly is illustrated in the figure below: A coordinate system such as the one shown (above right) allows us to express the displacement as a three-dimensional vector.00 m)2 6. ˆ (10 ˆi 6.0 k) ˆ d1 (d 2 d3 ) (3.42 m. (b) Using Eq.0 + 6. Use of Eq. ˆ Thus.0 k) 0 3.0 k) ˆ d1 (d 2 d3 ) (3. The shortest walking distance between the lower left back of the room and the upper right front corner is the dotted straight line shown on the diagram. the y component of the displacement is 4.30 m. Since a straight line is the shortest distance between two points.70 m) ˆi ( 4.00 m 2 (4. for example. Thus. and the z axis to be upward (as shown in the figure above).30 m) 2 7. however. Pretend there is a hinge where the front wall of the room joins the floor and lay the wall down as shown (above right). (e) We take the x axis to be out of the page.00 m . The fly might. the magnitude of the displacement.115 (b) The displacement vector is along the straight line from the beginning to the end point of the trip. Its displacement would be the same but the path length would be w h 11. the y axis to be to the right. Its length is smin w h 2 l2 3.30 m) ˆj (3. we write the length of the path as s y 2 w2 (l y)2 h2 . and the z component is 3. the length of the path cannot be less than d. (f) Suppose the path of the fly is as shown by the dotted lines on the diagram (below left). crawl along the edges of the room. Then the x component of the displacement is w = 3.0 m. the displacement vector can be written as ˆ d (3. LEARN To show that the shortest path is indeed given by smin . The condition for minimum is given by . (d) The path length is the same as the magnitude of the displacement if the fly flies along the displacement vector.70 m.70 m 3.00 m)k. (c) The length of the path of the fly can be greater than d.96 m. 66. (c) Since cos 90° = 0. and k^ directed upward (in order to have a right-handed coordinate system). b a points in the –z-direction. . then (by the same reasoning as in the previous part) a (b / d ) 0 . 3-20. Other examples include (40 ^i + 20 ^j + 25 k^ ) m and (40i^ – 20 ^j – 25 k^ ) m (with slightly different interpretations for the unit vectors). however. which gives l2 l2 2 smin w2 1 h 1 ( w h) 2 l 2 . 2 2 ( w h ) ( w h ) Any other path would be longer than 7. Since the coefficient of j is positive. (a) We are asked to consider b b j d d FG IJ H K in the case d > 0. We note that b a a b is true in this case and quite generally. with ^i directed antiparallel to the first path. then the coefficient is negative and the vector points in the –y direction. then the vector points in the +y direction. The vectors can be written as a a ˆi and b bˆj where a. (b) Using the Pythagorean theorem. b 0. (b) If. using Eq. then a b 0 . A little algebra shows that the condition is satisfied when y lw /(w h) . a b points in the +z-direction. ^j directed anti-parallel to the second path. (d) Since b / d is along the y axis. (a) This is one example of an answer: (40 ^i – 20 ^j + 25 k^ ) m. (e) By the right-hand rule. d < 0. 65. (f) By the same rule. Note that the product of the components is positive in each example.116 CHAPTER 3 ds dy y y 2 w2 ly (l y )2 h2 0.96 m.7 m 45 m. we have (40 m)2 (20 m)2 = 44. With a careful drawing. (b) (k) (c) ˆj ( ˆj)= 1. (e) (ˆi) ( ˆj)= kˆ (upward). 3-24 gives | a b | ab where a is the magnitude of a . Eq. Since R = 0. not just some “on the paper” or “on the blackboard” representation of it. | a b | | b a | ab . It is displaced horizontally by half the circumference of the wheel. (h) Also. we find that a (b / d ) has magnitude ab/d. ˆ (a) ˆi k=0 since ˆi kˆ ˆ ( ˆj)=0 since kˆ ˆj . where R is the radius of the wheel. Students sometimes confuse “north” with “up”. ˆ ( ˆj)= iˆ (west). so it might be necessary to emphasize that these are being treated as the mutually perpendicular directions of our real world. or R. these questions are basic to the definitions of the scalar (dot) and vector (cross) products. We note that the set of choices for unit vector directions has correct orientation (for a right-handed coordinate system). The point P is displaced vertically by 2R. one should find that the resultant vector has length 29. . (i) With d > 0.117 (g) Since sin 90° = 1. (j) The vector a (b / d ) points in the +z direction. The resultant (not shown) would be a straight line from start (Bank) to finish (Walpole). (d) kˆ ˆj= ˆi (west). A sketch of the displacements is shown. Once the terminology is clear. 69.450 m. 67.5 km at 35° west of south. (f) (k) 68. 900 ˆj . labeled r . which introduces some indefiniteness in our result.5. 300 m rx .414 m and the vertical component of the displacement is 0. However. the distances are understood to be in meters. The components of the total displacement are rx = Ax + Bx = 125 + 175 = 300 ry = Ay + By = 216.118 CHAPTER 3 the horizontal component of the displacement is 1. The components of B are Bx = 175 and By = 0. Thus. If the x axis is horizontal and the y axis is vertical.414 ˆi + 0. yet that angle computation seemed to yield something exact. (b) The angle the resultant displacement makes with the +x axis is ry 216.900 m. the vector displacement (in meters) is r 1. The displacement has a magnitude of r R 2R 2 2 R 2 4 1. the components of A are Ax = 250 cos60° = 125 and Ay = 250 sin60° = 216. and the total (“final”) displacement vector. (a) The magnitude of the resultant displacement is | r | rx2 ry2 (300 m)2 (216. shows the displacement vectors for the two segments labeled A and B .68m and an angle of 2R 1 2 tan 1 tan 32. there has to be some uncertainty in the observation that the wheel rolled half of a revolution. We observe that the angle between A and the x axis is 60°. In physics there are no “exact” measurements.5 R above the floor.5 + 0 = 216. We take east to be the +x direction and north to be the +y direction. Where the units are not explicitly shown. 70.5 m)2 370m.5.5 m tan 1 tan 1 36. The diagram of her walk. 0d can be written as 2.0(3.0| = 2.0 m. (d) The minus sign carried by this scalar factor reverses the direction.40 m)(cos 225 ˆi sin 225 ˆj) (0.0d 5. where ˆj is the unit vector pointing south. The new magnitude is 6.30 m) ˆi (0.28 m .28 m) ˆi (0. The vector d (measured in meters) can be represented as d (3. (b) The new direction of 5d is the same as the old: south.0) affects the magnitude.52 m )ˆj. Therefore. The ant’s trip consists of three displacements: d1 (0.0d is 15 m.0 m)(ˆj) . (c) The total distance walked is d = 250 m + 175 m = 425 m.28 m . 71. (d) The total distance walked than the magnitude of the resultant displacement. The vector 2.0) affects the magnitude but not the direction. respectively. The magnitude of 5.50 m . . 5.40 m)sin 225 0. (d) The y component of d 2 is d2 y 0 m . (a) The x component of d1 is d1x (0.50 m) ˆi 2 d3 (0.0 m) ˆj.28 m) ˆj d (0. (c) The absolute value of the scalar factor (|2. (b) The y component of d1 is d1 y (0.0 m ˆj) (15 m) ˆj. 72. We have taken the positive x and y directions to correspond to east and north.0d (6.40 m) cos 225 0.119 The direction is 36 north of due east. (a) The positive scalar factor (5. where the angle is measured with respect to the positive x axis. (c) The x component of d 2 is d2 x 0. or north. so the new direction is ˆj .60 m)(cos 60 ˆi sin 60 ˆj) (0. is greater The diagram shows why: A and B are not collinear. 24 m tan 25 (north of east) 0.0)(2.0) (2.0) = 46 .0 4. (c) a b = (3. (a) a b (axby a ybx ) kˆ since all other terms vanish.57 m.0) (5. In this case.0) (4. (k) The distance the ant has to travel is | dnet | 0. (f) The y component of d 3 is d3 y (0. y d1 y d2 y d3 y (0.57 m.24 m)2 0.0)(2.0 2.0) ˆi (5. (0. we will construct a b unit-vector and “dot” it (take the scalar product of it) with a . x d net. (i) The magnitude of the net displacement is 2 2 dnet dnet.0kˆ . 73.0)(4.0) + (5. (d) Several approaches are available. y 1 0.60 m)sin 60 0. we obtain ((3.120 CHAPTER 3 (e) The x component of d 3 is d3 x (0.0) ˆj (a + b ) b = (5.50 m) (0. we make the desired unitvector by . We apply Eq.0))kˆ 2.52 m .30 m) 0. (l) The direction the ant has to travel is 25 (south of west) .52 m d net. In this solution.52 m)2 (0.30 m .52 m.52 m) 0.28 m) (0.0) + (9. due to the fact that neither a nor b have any z components. x d1x d2 x d3 x (0.60 m) cos 60 0. the displacement would be d net . 3-23 and Eq. (h) The y component of the net displacement d net is dnet.0) = 26. 3-27. (b) a b axbx + a yby yields (3.24 m. Consequently. y (j) The direction of the net displacement is d net.28 m) (0 m) (0. (g) The x component of the net displacement d net is dnet.0)(4. x tan 1 If the ant has to return directly to the starting point. b gb g b gb g b gb g (b) The vector (cross) product is a b a y bz az by ˆi + a z bx axbz ˆj + axby a y bx kˆ 1. and use the following fundamental products: ˆi ˆj ˆj ˆi kˆ ˆj kˆ kˆ ˆj ˆi kˆ ˆi ˆi kˆ ˆj (a) “north cross west” = ˆj (ˆi) kˆ = “up.0)(4.0) ab a bˆ 5. 2.40 sin 48° = 1. (3.937 145 .45 ˆj 2.0) (5.” .937. The two vectors a and b are given by ˆ 1.40 cos 48° = 0.85 kˆ a 3.51iˆ + 2.20(cos 63 ˆj sin 63 k) ˆ 0.0)(2.0)2 (4.81. and k upward.0 ˆj (2.67 ˆj 1.0 ˆi 4.20 1.40(cos 48 ˆi sin 48 k) The components of a are ax = 0.97.04.937 0 ˆj 0 1. 3. We orient i eastward.5 .97 48. (c) The angle between a and b is given by a b 1 cos ab cos 1 2.85 104 . 0 2. by = 0.20 cos 63° = 1. ay = 3. and az = 3.451. (a) The scalar (dot) product is therefore a b axbx a yby azbz 0 0.45.85.0)2 (4.0)2 . The components of b are bx = 1. (2.937iˆ 1. j northward.45 0.0)2 74.937 kˆ 1.121 b bˆ |b | We therefore obtain 2.20 sin 63° = 2.40 75.04 0 ˆi + 2.36kˆ .85 0. and bz = 1.04 kˆ b 1. Using the Pythagorean theorem. The area of a triangle is half the product of its base and altitude.” (c) “east cross up” = ˆi (k) (d) “west dot west” = ( ˆi) ( ˆi ) 1 . Then the altitude is b sin and the area is A 12 ab sin 12 | a b | . 76.6 m. (a) The displacement is d (1300 m)iˆ ( 2200 m)ˆj ( 410 m)kˆ .122 CHAPTER 3 ˆ ( ˆj) 0 .4 . (b) “down dot south” = (k) ˆ ˆj = “south. The net displacement is zero since his final position matches his initial position.6 .90)2 (2. (b) The displacement for the return portion is d (1300 m)iˆ ( 2200 m)ˆj and the magnitude is d (1300 m)2 ( 2200 m)2 2. The base is the side formed by vector a. 77. Substituting the values given.4)sin 46 8.0 36. We are also told that the angle (measured in the ‘standard’ sense) for A is 0° and the angle for C is 90°.70)sin 63. Let A denote the magnitude of . Since c is perpendicular to a the magnitude of a c is ac.0 m and C = 2A. we obtain | a (b a ) | a 2b sin (3. j northward.0 A 2 4 A2 which leads to A = 8 / 5 = 3. we have A 1 1 ab sin (4. (e) “south cross south” = ( ˆj) ( ˆj) 0 .56 103 m . Substituting the values given. B A2 C 2 8. The magnitude of a (b a ) is consequently | a (b a ) | ac a 2b sin . Then the magnitude of c is c = ab sin .3)(5. We orient i eastward. 79. The vector equation A is A B = C where B = 8. 2 2 . similarly for the other vectors. which makes this a right triangle (when drawn in a “head-to-tail” fashion) where B is the size of the hypotenuse. and k upward. 78. Let c b a .