NELSON SENIOR MATHS SPECIALIST 11

NELSON SENIOR MATHS SPECIALIST 11 FULLY WORKED SOLUTIONS CHAPTER 1: Basic vectors Exercise 1.01: Two-dimensional vectors Concepts and techniques 1 Speed has no direction. B 2 The magnitude of P(5, 0) is 2 2 5 + 0 = 5 E 3 South so 270 E 4 (5, 225 ) B y 225 o x 5 Cengage Learning Australia 2014 ISBN 9780170251501 1

6 7 8 a (3, 4) = 2 2 3 + 4 = 5 b (5, 12) = 2 2 5 + 12 = 13 c (24, 7) = 2 2 24 7 25 + = d (1, 1) = 2 2 1 1 2 + = e (2, 7) = 2 2 2 + 7 = 53 f (3, 6) = 2 2 3 + 6 = 45 = 3 5 g (a, 4) = 2 a + 16 h (6, b) = 2 36 + b Cengage Learning Australia 2014 ISBN 9780170251501 2

9 a b c d e f = 53.1 3 1 4 1 12 = 112.6 5 = 343.7 24 1 7 = 225 1 1 1 = 105.9 2 1 7 1 6 = 296.6 3 10 a b c Cengage Learning Australia 2014 ISBN 9780170251501 3

d e 11 a (3, 4) = 5 b ( 5, 5) = 7.07 c ( 7, 10) = 3.49 d (4, 12) = 12.65 e ( 10, 16) = 18.87 12 a (8, 15) = 17 b (12, 5) = 13 c ( 9, 12) = 15 d ( 7, 12) = 13.9 e (8, 8) = 11.3 1 4 = 306.9 (5, 306.9 ) 3 1 5 = 135 (7.07, 135 ) 5 1 10 = 235 7 (12.21, 235.0 ) 1 12 = 71.6 4 (12.65, 71.6 ) 1 16 = 122 10 (18.87, 122.0 ) 1 15 = 61.9 8 (17, 61.9 ) 1 5 = 22.6 (13, 22.6 ) 12 1 12 = 233.1 9 (15, 233.1 ) 1 12 = 239.7 7 (13.9, 239.7 ) 1 8 = 135 (11.3, 315 ) 8 Cengage Learning Australia 2014 ISBN 9780170251501 4

13 a A(3, 4) to B(5, 11) AB = (5 4, 11 4) = (2, 7) AB = 53 = 74.05 2 1 7 so AB (7.3, 74.1 ) b R( 2, 1) to B( 7, 2) RB = ( 5, 3) RB = 34 1 3 = 239.04 5 so RB (5.8, 211.0 ) c M( 3, 8) to B( 9, 1) MB = ( 6, 9) MB = 117 = 3 13 = 123.69 6 1 9 so MB (10.8, 123.7 ) Cengage Learning Australia 2014 ISBN 9780170251501 5

Reasoning and communication 14 a u has a magnitude of 8 and is in the direction 60 south of east b c d u + u = 0 Cengage Learning Australia 2014 ISBN 9780170251501 6

Exercise 1.02: Addition of vectors Concepts and techniques 1 We do not know disces travelled so all that can be said is north of east. For example: D or or 2 The forces cancel each other out. B 3 a 5 + 9 14 east b c d e 20 + 12 + 25 57 east 60 37 23 east 14 + 25 18 7 west 48 28 + 17 + 36 73 east 4 a 80 20 = 60 N up b c d e 17 26 = 9 N down 48 + 35 57 = 26 N up 93 110 86 = 103 N down 110 N east as the other two forces balance each other out. Cengage Learning Australia 2014 ISBN 9780170251501 7

5 Given w = (6, 220 ), x = (10, 60 ), y = (8, 100 ) and z = (9, 50 ), a w + y = (7.21, 146.1 ) b y + z = (4.50, 12.6 ) c x + z = (10.93, 9.3 ) d w + z = (10.82, 276.3 ) e y + x = (16.93, 77.7 ) 6 Given a = (7, 160 ), b = (15, 40 ), c = (18, 120 ) and d = (11, 50 ), a a + b = (13, 67.8 ) b c + b = (6.4, 292.7 ) c d + a = (6.1, 274.7 ) d a + c = (20.4, 220.3 ) e c + d = (24.1, 265.4 ) Cengage Learning Australia 2014 ISBN 9780170251501 8

Reasoning and communication 7 A 24 km 45 o B 12 km O C 11 km OB = 2 2 12 + 24 = 724 = 26. 83 ABO = 1 12 = 26. 57 24 OBC = 45 26. 57 = 18.43 AOB = 90 26. 57 = 63.43 (needed later) B 26.83 km 18.43 o 11 km C O Using the cosine law, OC 2 = 26.83 2 + 11 2 2 26.83 11 cos (18.43 ) OC = 16.76 Using the sine law BOC = 11.98 ( O ) (. ) sin sin 18 43 = 11 16. 76 Bearing of ship is 63.43 + 11.98 = 75.41 The ship s result displacement: disce of 16.8 km on a bearing of 075 (or as a vector, (16.8, 15 ) where the angle is measured from the positive x axis.) Cengage Learning Australia 2014 ISBN 9780170251501 9

8 Total force F = 400 j + 300 cos (35 ) j + 300 sin (35 ) i = 175.072 i + 645.745 j F 645.746 N x 172.073 N 2 2 F = 175.02... + 645.745... 668.28 645.745... x = 1 172.072... x = 75.079 The total force acting is about 668 N in the direction 14.9 from the vertical. Cengage Learning Australia 2014 ISBN 9780170251501 10

9 Q 800 m 800 m 60 o P 105 o 1000 m 1000 m 45 o R Using the cosine law, QR 2 = 800 2 + 1000 2 2 800 1000 cos (105 ) QR = 1433.2 Using the sine law ( R) ( ) sin sin 105 = 800 1433. 2 PRQ = 32.63 Direction of Sam s final position from the positive x-axis anticlockwise is 270 + (32.63 + 45 ) = 347.63 Sam s result displacement is (1433, 348 ) Cengage Learning Australia 2014 ISBN 9780170251501 11

10 120 o 5km 7km 45 o 5km Q 105 o 7km P R PQR = 45 + 60 = 105 Using the cosine law, PR 2 = 5 2 + 7 2 2 5 7 cos (105 ) PR = 9.6 Using the sine law ( ) ( ) sin P sin 105 = 7 96. QPR = 44.8 Bearing of ship s final position is 45 + 44.8 and disce is 9.6 km. So disce and direction from starting point is 9.6 km on a bearing of 090. Cengage Learning Australia 2014 ISBN 9780170251501 12

11 16km 24km 45 o B 135 o 16km C 24km A Using the cosine law, AC 2 = 24 2 + 16 2 2 24 16 cos (135 ) AC = 37.08 Using the sine law ( A) ( ) sin sin 135 = 16 37. 08 BAC = 17.76 Bearing of ship is 45 + 17.76 = 62.76 The ship s result displacement: disce of 37.08 km on a bearing of 063 (or as a vector, (37.08, 27 ) where the angle is measured from the positive x axis.) Cengage Learning Australia 2014 ISBN 9780170251501 13

12 45 o 20 A 20 40 o 12 C 12 85 o B Using the cosine law, AC 2 = 20 2 + 12 2 2 20 12 cos (85 ) AC = 22.41 Using the sine law ( A) ( ) sin sin 85 = 12 22. 41 BAC = 32.24 Bearing of the bird is 90 + 32.24 + 45 = 167.24 The result velocity of the bird : velocity is 22.41 ms 1 on a bearing of 167 Cengage Learning Australia 2014 ISBN 9780170251501 14

Exercise 1.03: Component and polar forms of vectors Concepts and techniques 1 a b c d = 2 1 7 = 1 1 9 = 3 1 6 = 5 1 8 74.1 83.7 63.4 58.0 e f g h 1 9 = 2 1 4 = 6 1 7 = 8 1 5 = 12 102.5 146.3 221.2 337.4 2 a ( 3, 4) = 2 2 ( 3) + 4 = 5 b (12, 5) = 13 c ( 6, 2) = 40 = 6. 32 d (11, 4) = 137 = 11. 70 e (7, 9) = 130 = 11. 40 f (13, 8) = 233 = 15. 26 g ( 7, 12) = 193 = 13. 89 h (13, 2) = 174 = 13. 15 Cengage Learning Australia 2014 ISBN 9780170251501 15

3 a (3, 4) = 5 1 4 = 306.87 (5, 306.9 ) 3 b 5 = 5 2 2 ( 5) 5 + =12.21 = 135 (7.1, 135 ) 5 1 5 c ( 7, 10) = 149 = 12.21 1 10 = 235.0 (12.2, 235 ) 7 d 4 12 = 160 = 12.65 1 12 = 71.6 (12.7, 71.6 ) 4 e ( 10, 16) = 356 = 18.87 1 16 = 122 (18.9, 122 ) 10 f 8 15 = 17 1 15 = 61.9 (17, 6.9 ) 8 g (12, 5) = 13 1 5 = 22.6 (13, 2.6 ) 12 h 9 12 = 15 1 12 = 233.1 (15, 2.1 ) 9 i ( 7, 12) = 193 = 13.89 1 12 = 210.3 (13.9, 210.3 ) 7 j 8 8 = 128 = 11.31 1 8 = 315 (11.3, 315 ) 8 4 a (9, 12) (15, 53.13 ) b (11, 17) (20.25, 57.09 ) c ( 9, 15) (17.49, 120.96 ) d (18, 13) (22.20, 324.16 ) e ( 20, 17) (26.25, 220.37 ) f (16.4, 8.7) (18.56, 27.95 ) g (6.37, 12.8) (14.30, 296.46 ) h ( 3.91, 11.62) (12.26, 108.60 ) Cengage Learning Australia 2014 ISBN 9780170251501 16

5 a (5, 30 ) (4.33, 2.5) b (10, 300 )) (5, 8.66) c (24, 90 ) (0, 24) d (16, 135 ) ( 11.31, 11.31) e (28, 120 ) ( 14, 24.25) f (70, 270 ) (0, 70) g (35, 0 ) (35, 0) h (22, 180 ) ( 22, 0) 6 a (6, 60 ) (3, 5.20) b (12, 120 ) ( 6, 10.39) c (17, 52 ) (10.47, 13.40) d (23, 166 ) ( 22.32, 5.56) e (14, 132 ) ( 9.37, 10.40) f (15, 287 ) (4.39, 14.34) g (24, 221 ) ( 18.11, 15.75) h (43, 17 ) (41.12, 12.57) i (4, 200 ) ( 3.76, 1.37) j (8, 60 ) (4, 6.93) 7 a (1, 5) = 26 (2, 2) = 8 so (1, 5) b (4, 7) = 65 (3, 9) = 90 so (3, 9) c ( 2, 6) = 40 (3, 8) = 73 so (3, 8) d (7, 7) = 98 (4, 8) = 80 so (7, 7) e ( 3, 10) = 109 ( 6, 8) = 100 so ( 3, 10) f (10, 4) = 116 (6, 11) = 157 so (6, 11) Cengage Learning Australia 2014 ISBN 9780170251501 17

Reasoning and communication 8 Given A(5, 6) and B( 2, 2) then AB = ( 7, 8) ( 7, 8) (10.63, 311.2 ) Cengage Learning Australia 2014 ISBN 9780170251501 18

Exercise 1.04: Multiplication by scalars Concepts and techniques 1 a = (4, 2) and b = (10, 5) B (4, 2) = 2 (10, 5) 5 2 a = ( 6, 4) and b = (12, 8) B 2 ( 6, 4) = (12, 8) 3 Given b = (6, 2) 4 Given d = a 3d = 18 12, 54 36 b 2d = 36 24 c 1 d = 6 3 4 d 6.4d = 115.2 76.8 e 1 d = 9 2 6 Cengage Learning Australia 2014 ISBN 9780170251501 19

45 f 2.5d = 30 g 2 d = 12 3 8 h 4 d = 5 If v = (6, 47 ), 72 48 a 3v = (18, 47 ) b 4v = (24, 227 ) c 2.5v = (15, 47 ) d v = (6, 227 ) e 5v = (30, 227 ) f 10v = (60, 47 ) g 7v = (42, 227 ) 1 h 2 v = (3, 47 ) 6 Given a = (2, 7), b = ( 4, 9) and c = (3, 12) a 4a = (8, 28) b 3c = ( 9, 36) c 3.5b = ( 14, 31.5) d 9.7a = (19.4, 67.9) 1 e 3 b = ( 4, 3) 3 f 1 2 c = (1 1 2, 6) g 3 4 b = ( 3, 6 3 4 ) h 2 a = ( 4, 14) Cengage Learning Australia 2014 ISBN 9780170251501 20

7 If a = (4, 195 ), b = (5, 69 ) and c = (2, 304 ), a 3a = (12, 195 ) b 2c = (4, 124 ) c 1.5b = (7.5, 69 ) d 4a = (16, 15 ) 1 e 2 b = (2.5, 249 ) f 1 3 c = ( 2 3, 304 ) g 4b = (20, 249 ) h 5a = (20, 15 ) Cengage Learning Australia 2014 ISBN 9780170251501 21

Exercise 1.05: Unit vectors Concepts and techniques 1 The magnitude is 1. B 2 The magnitude is 2. B 1 2 3 v =, =, = 13 13 13 3 a ( 23) 2 13 3 13, 13 13 1 4 5 v =, =, = 41 41 41 b ( 4 5) 4 41 5 41, 41 41 v 1 2 6,, 1, 3 = = = 40 40 40 10 10 = c ( 2 6) 10 3 10, 10 10 d 1 v = 37., 82. = 041., 091. 2 2 37. + 82. ( ) ( ) ( ) e f g 3 1 3 10 v = = 10 1 = 1 10 5 1 5 13 v = 13 = 12 12 13 24 1 24 25 v = 25 = 7 7 25 3 10 10 10 10 h 1 8. 22 0. 87 v = 2 2 8. 22 4. 69 4. 69 = 0. 50 + i (10, 86 ) has magnitude 10. v = (1, 86 ) j (12, 165 ) has magnitude 12. v = (1, 165 ) Cengage Learning Australia 2014 ISBN 9780170251501 22

k v 2π = 1, 3 l 7π v = 1, 5 4 a (5, 3) = 5i 3j b c d ( 6, 4) = 6i + 4j ( 4, 7) = 4i 7j (6, 5) = 6i + 5j e ( 3.2, 9.4) = 3.2i 9.4j f 1 4 = i 4j g h 2 8 3 5 = 2i 8j = 3i + 5j i j 7.59 3.68 0.07 0.19 = 7.59i + 3.68j = 0.07i + 0.19j k (7, 74 ) (1.93, 6.73) = 1.93i 6.73j l (9, 125 ) ( 5.16, 7.37) = 5.61i + 7.37j m n 11π 11, (9.53, 5.5) = 9.53i 5.5j 6 4π 16, ( 8, 13.86) = 8i 13.86j 3 o (4.8, 119 ) ( 2.33, 4.20) = 2.33i 4.20j Cengage Learning Australia 2014 ISBN 9780170251501 23

Reasoning and communication 5 Given m = 3i 4j then m = 3i + 4j. m = 5 3 4 m = i + j. 5 5 6 Given g = 4i 7j then 4 7 g = i j 65 65 Therefore the required vector of magnitude 6 is 6 g = 24 42 i j= 2. 98i 5. 21j 65 65 Cengage Learning Australia 2014 ISBN 9780170251501 24

Exercise 1.06: Using components Concepts and techniques 1 a ( 4, 3) + ( 9, 10) = ( 13, 13) b (7, 9) + (11, 5) = (18, 14) c ( 13, 6) + (8, 7) = ( 5, 1) d (3.8, 4.5) + (6.2, 7.3) = (10, 11.8) e ( 1.07, 0.35) + (5.24, 4.57) = (4.17, 4.22) 3 2 3 13 (3, 2 ) + ( 1,5 ) = (1, 2 ) 1 4 f 2 5 4 3 4 15 2 a b c d e 14 11 + 96 = 23 17 5 9 + 7 7 = 12 2 13 16 + 7 8 = 6 8 3.6 8.7 + 6.9 4.6 = 3.3 13.3 2.07 4.66 + 0.97 1.78 = 1.1 6.44 f 1 2 2 4 3 9 + 3 3 4 2 1 3 = 1 6 4 7 5 9 3 a 5i 6j + 2i 7j = 7i 13j b c d 8i + 7j + 9i + 3j = 17i + 10j 11i 4j + 6i + 9j = 5i + 5j 5.8i 6.7j + ( 9.2i 5.3j) = 3.4i 12j e 2.14i + 1.79j + ( 6.09i + 3.36j) = 8.23i + 5.15j 2 3 7 1 1 5 f 4 5 i + 3 8 j + ( 5 10 i + 6 4 j) = i 9 8 j 10 10 Cengage Learning Australia 2014 ISBN 9780170251501 25

4 Given a = (1, 4), b = ( 7, 8), c = (2, 4), d = (5, 2) and e = ( 6, 1) a c + e = ( 4, 5) b b + c + d = (0, 10) c 5a = (5, 20) d 2e = (12, 2) e 3a + 2e = ( 9, 14) f 4b 2a = ( 30, 24) g a c e = (5, 1) h 7c 7a + d + 2e = (0, 0) i 7a + 5b + e = ( 34, 69) j 2d 7c = ( 4, 32) 5 a (6, 20 ) + (9, 55 ) (5.64, 2.05) + (5.16, 7.37) = (10.80, 9.42) (14.33, 41.11 ) b (25, 120 ) + (16, 80 ) ( 9.72, 37.41) (38.65, 104.57 ) c (7, 30 ) (9, 100 ) (7.63, 5.36) (9.32, 324.9 ) d (105, 300 ) (95, 60 ) (5, 173.21) (173.28, 271.7 ) e (6, 70 ) (6, 10 ) ( 3.86, 4.60) (6, 130 ) Reasoning and communication 6 a AB = (5 3, 11 4) = (2,7) b RB = [ 7 ( 2), 2 1] = ( 5, 3) c MB = [ 9 ( 3), 1 ( 8)] = ( 6, 9) d QP = (1 1, 10 8) = (0, 18) e BX = (8, 14) f FG = (8, 11) Cengage Learning Australia 2014 ISBN 9780170251501 26

7 Given a = (x 1, y 1 ) and c < 0, prove that ca = c a. ca = ( cx ) + ( cy ) 2 2 1 1 = c ( x + y ) 2 2 2 1 1 = c x + y 2 2 2 1 1 = c a Given that c < 0, then c > 0 so c = c. ca = c a. 8 Given a = (x 1, y 1 ), b = (x 2, y 2 ) and c is a scalar, prove that c(a + b) = ca + cb. c(a + b) = c(x 1 + x 2, y 1 + y 2 ) = [(cx 1 + cx 2 ), (cy 1 + cy 2 )] = [(cx 1, cy 1) + (cx 2, cy 2 )] = c(x 1, y 1 ) + c(x 2, y 2 ) = ca + cb Cengage Learning Australia 2014 ISBN 9780170251501 27

Exercise 1.07: Vector properties Concepts and techniques 1 Otherwise 2d = 2c so d = c and this may not necessarily be true. D 2 rp is meaningless. B 3 The additive inverse of a b c m is (5, 8) as ( 5, 8) + (5, 8) = O d is (11, 4) as ( 11, 4) + (11, 4) = O r is (0, 6) as (0, 6) + (0, 6) = O d q is 3 1 as 3 1 3 + = 1 O e g is 14 3 14 14 as 3 + = 3 O f h is 12 11 as 12 12 + = 11 11 O g a is 4i + 9j as 4i 9j + 4i + 9j = O h p is 2 i 3 4 5 j as 2 3 i + 4 5 j + 2 i 3 4 5 j = O i n is 0.24i + 1.06j as 0.24i 1.06j + 0.24i + 1.06j = O 4 a m + p = ( 3, 5) + (2, 9) = ( 3 + 2, 5 + 9) = (2 + 3, 9 + 5) = (2, 9) + ( 3, 5) = p + m Cengage Learning Australia 2014 ISBN 9780170251501 28

b q + d = (10, 7) + ( 8, 4) = (10 + 8, 7 + 4) = ( 8 + 10, 4 + 7) = ( 8, 4) + (10, 7) = d + q c f + w = 3 12 + 2 5 = = 3+ 2 12 + 5 2+ 3 5 + 12 2 = 5 + 3 12 = w + f d u + a = 15 6 + 7 11 = 15 + 7 6 + 11 = 7 + 15 11+ 5 7 = 11 + 15 6 = a + u Cengage Learning Australia 2014 ISBN 9780170251501 29

e b + n = (4i j) + ( 5i + 7j) = (4 5)i + ( 1 + 7)j = ( 5 + 4)i + (7 + 1)j = ( 5i + 7j) + (4i j) = n + b f e + n = ( 11i 6j) + (13i + 8j) = 11i + 13i 6j + 8j = 13i 11i + 8j 6j = (13i + 8j) + ( 11i 6j = n + e 5 a Given a = ( 2, 6), b = (1, 7) and c = ( 8, 0) Show (a + b) + c = a + (b + c) (a + b) + c = [( 2, 6) + (1, 7)] + ( 8, 0) = [ 2 + 1, 6 + ( 7)] + ( 8, 0) = [ 2 + 1 8, 6 + ( 7) + 0] = [ 2 + (1 8), 6 + ( 7 + 0)] = ( 2, 6) + [(1 8), ( 7 + 0)] = ( 2, 6) + [(1, 7) + ( 8, 0)] = a + (b + c) Cengage Learning Australia 2014 ISBN 9780170251501 30

b Given p = ( 1, 9), q = ( 4, 6) and r = (5, 3) Show (p + q) + r = p + (q + r) (p + q) + r = [( 1, 9) + ( 4, 6)] + (5, 3) = ( 1 4, 9 6) + (5, 3) = ( 1 4 + 5, 9 6 3) = [ 1 + ( 4 + 5), 9 + ( 6 3)] = ( 1, 9) + [( 4 + 5), ( 6 3)] = ( 1, 9) + [( 4, 6) + (5, 3)] = p + (q + r) c Given t = 7 2, u = 8 10 and v = 6 3 Show (t + u) + v = t + (u + v) (t + u) + v = [(7, 2) + ( 8, 10)] + (6, 3) = [7 + ( 8), 2 + 10] + (6, 3) = [7 + ( 8) + 6, 2 + 10 + 3] = [ 7 + ( 8 + 6), 2 + (10 + 3)] = (7, 2) + [( 8 + 6), (10 + 3)] = (7, 2) + [( 8, 10) + (6, 3)] = t + (u + v) Cengage Learning Australia 2014 ISBN 9780170251501 31

5 d k = 0, m = 11 2 and n = 5 4 Show (k + m) + n = k + (m + n) (k + m) + n = [( 5, 0) + (11, 2)] + ( 5, 4) = ( 5 + 11, 0 2) + ( 5, 4) = ( 5 + 11 5, 0 2 4) = ( 5 + (11 5), 0 + ( 2 4)) = ( 5, 0) + ((11 5), ( 2 4)) = ( 5, 0) + ((11, 2) + ( 5, 4)) = k + (m + n) e w = i 3j, r = 2i + 2j and s = 9i 6j Show (w + r) + c = w + (r + c) (w + r) + c = [( 1, 3) + (2, 2)] + (9, 6) = ( 1 + 2, 3 + 2) + (9, 6) = ( 1 + 2 + 9, 3 + 2 + 6) = [ 1 + (2 + 9), 3 + (2 + 6)] = ( 1, 3) + {(2 + 9), [2 + ( 6)]} = ( 1, 3) + {(2, 2) + [9, ( 6)]} = w + (r + c) Cengage Learning Australia 2014 ISBN 9780170251501 32

f e = 4i + 11j, d = i 3j and f = 8i 4j Show (e + d) + c = e + (d + c) (e + d) + c = (( 4, 11) + ( 1, 3)) + (8, 4) = ( 4 1, 11 3) + (8, 4) = ( 4 1 + 8, 11 3 + 4) = { 4 + ( 1 + 8), 11 + [ 3 + ( 4)]} = ( 4, 3) + {( 1 + 8), [ 3 + ( 4)]} = ( 4, 3) + [( 1, 3) + (8, 4)] = e + (d + c) 6 a h = (1, 2), w = 3 and g = 2 Show (w + g) h = wh + gh (w + g)h = (3 + 2) (i 2j) = (3 + 2) i + (3 + 2) ( 2j) = 3i + 2 i 6j 4j = 3i 6j + 2 i 4j = 3(i 2j) + 2(i 2j) = 3 h + 2h = wh + gh Cengage Learning Australia 2014 ISBN 9780170251501 33

b Given d = ( 12, 18), a = and b = Show (a + b) d = ad + bd 1 2 2 3 1 2 1 2 = + i + + ( 2j) 2 3 2 3 1 2 1 2 = i + i + j + 2 3 2 3 j 1 1 2 2 = i + j + i + 2 2 3 3 j 1 2 = ( i 2j ) + ( i 2j) 2 3 = ad + bd ( a + b) d = + ( i 2j) ( 2 ) ( 2 ) ( 2 ) ( 2 ) c Given a = 4 2, m = 4 and n = 5 Show (m + n) a = ma + na (n + n)a = (4 + 5) 4 2 = = = = ( 4+ 54 ) ( 4+ 5)( 2) 4 4 + 5 4 4 ( 2) + 5 ( 2) 4 4 5 4 + 4 ( 2) 5 ( 2) 4 4 4 + 5 2 2 = ma + na Cengage Learning Australia 2014 ISBN 9780170251501 34

16 d b = 8, r = 0.5 and s = 0.25 Show (r + s) b = rb + sb (r + s)b = (0.5 + 0.25) 16 8 = = = = ( 0.5 + 0.25)( 16) ( 0.5 0.25)( 8) + 0.5 ( 16) + 0.25 ( 16) 0.5 ( 8) + 0.25 ( 8) ( ) ( ) ( ) ( ) 0.5 16 0.25 16 + 0.5 8 0.25 8 16 16 0.5 + 0.25 8 8 = rb + sb e Given q = 5i j, c = 2 and d = 6 Show (c + d) q = cq + dq (c + d)q = (2 + 6) (5i j) = (2 + 6) 5i + (2 + 6) ( j) = 2 5i + 6 5 i 2j 6j = 10i 2j + 30i 6j = 2(5i j) + 6(5 i j) = 2 q + 6q = cq + dq Cengage Learning Australia 2014 ISBN 9780170251501 35

f p = 12i 36j, k = 1 3 and v = 3 4 Show (k + v) p = kp + vp 1 3 ( k + v) p = + ( 12i 36j) 3 4 = 1 3 1 3 + ( 12i ) + + 3 4 3 4 ( 36 j) = 1 3 1 3 ( 12) i + ( 12 ) i+ ( 36) j + ( 36 ) j 3 4 3 4 = 1 1 3 3 ( 12 ) i+ ( 36) j+ ( 12) i + ( 36 ) j 3 3 4 4 = 1 3 ( 12i 36 j) + ( 12i 36 j) 3 4 = 1 3 p + p 3 4 = kp+ vp Cengage Learning Australia 2014 ISBN 9780170251501 36

Reasoning and communication 7 Prove that vector addition is associative i.e. that (a + b) + c = a + (b + c) Let a = (x 1, y 1 ), b = (x 2, y 2 ) and c = (x 3, y 3 ) (a + b) + c = [(x 1, y 1 ) + (x 2, y 2 )] + (x 3, y 3 ) = [(x 1 + x 2 ), (y 1 + y 2 )] + (x 3, y 3 ) = (x 1 + x 2 + x 3 ), (y 1 + y 2 + y 3 ) = [x 1 + (x 2 + x 3 ), y 1 + (y 2 + y 3 )] = (x 1, y 1 ) + (x 2 + x 3, y 2 + y 3 ) = a + (b + c) 8 Prove that vector addition is commutative i.e. that a + b = b + a. Let a = (x 1, y 1 ) and b = (x 2, y 2 ). a + b = (x 1, y 1 ) + (x 2, y 2 ) = [(x 1 + x 2 ), (y 1 + y 2 )] = [(x 2 + x 1 ), (y 2 + y 1 )] = (x 2, y 2 ) + (x 1, y 1 ) = b + a Cengage Learning Australia 2014 ISBN 9780170251501 37

9 Given that p and q are vectors and k is a scalar quantity, prove that k(p + q) = kp + kq. Let p = (x 1, y 1 ) and q = (x 2, y 2 ). k(p + q) = k [(x 1, y 1 ) + (x 2, y 2 )] = k [(x 1 + x 2 ), (y 1 + y 2 )] = [k (x 1 + x 2 ), k (y 1 + y 2 )] = [(k x 1 + k x 2 ), (k y 1 + k y 2 )] = [(k x 1, k y 1 ) + (k x 2, k y 2 )] = k (x 1, y 1 ) + k (x 2, y 2 ) = kp + kq. Cengage Learning Australia 2014 ISBN 9780170251501 38

Exercise 1.08: Application of vectors Reasoning and communication 1 Add the two vectors 6000 N 32 39 9000 N r = [6000 cos (58 ), 6000 sin (58 )] + [ 9000 cos (51 ), 9000 sin (51 )] = 2 484.37i + 12 082.6j Total force on the car is r = ( ) 2 2 2484.37 + 12 082.6 = 12 335.4 N The total force on the car is 12 340 N. Cengage Learning Australia 2014 ISBN 9780170251501 39

Direction? (x) = x = 11.6 2 484.37 12 082.6 The direction of the force is 11.6 towards the passenger side 2 125 o 55 o 5 km A 5 km 55 o 100 o 205 o 3 km B 3 km C ABC = 90 55 + 270 205 = 100 Using the cosine law, AC 2 = 5 2 + 3 2 2 5 3 cos (100 ) AC = 6.26 Using the sine law sin (A) sin (100 ) = 3 6. 26 o BAC = 28.15 The bearing of her final position is 125 + 28.15 = 153.2 The result disce is 6.26 km Cengage Learning Australia 2014 ISBN 9780170251501 40

3 400 sin (30 ) = F sin (45 ) The components perpendicular to the banks that have to be equal to prevent the boat moving to one side or the other. F = 282.8 N 400 N 30 o o 45 F F 283 N Cengage Learning Australia 2014 ISBN 9780170251501 41

4 1200 130 600 r = + + 80 300 290 200.38 65 205.21 + + 1181.77 112.58 563.82 478.59 = 505.37 In polar coordinates, r 696.02 o 46.56 The direction and magnitude of the result force on the boat are 696 N at an angle of 46.6 to the shoreline. Cengage Learning Australia 2014 ISBN 9780170251501 42

5 20 m/s 18 m/s Change in velocity v 2 v 1 = 18i ( 20j) = 18i + 20j 18i + 20j = 26.91 1 18 = 42 20 The change in velocity is 26.9 m/s and the direction changes 42. 6 60 km h -1 60 km h -1 Change in velocity v 2 v 1 = 60 0 60 = 0 60 60 60 2 2 = ( 60) + ( 60) = 84.85 60 The change in velocity is 84.85 km h 1 at an angle of 045 to where the police car was going. Cengage Learning Australia 2014 ISBN 9780170251501 43

7 N 197 o N 140 16 o 140 Change in velocity = v 2 v 1 = 140 cos (26 ) 140 cos (73 ) 166.76 = 140 sin (26 ) 140 sin (73 ) 72. 51 v 2 v 1 = 166.76 = 181.85 72.51 1 72.51 = 23.5 166.76 23.5 o 166.76 72.51 Change in velocity is about 182 knots at an angle of 66.5 (from north), Cengage Learning Australia 2014 ISBN 9780170251501 44

8 12 m/sec 15 m/sec Change in velocity = v 2 v 1 = 0 15 cos (45 ) 10.61 = 12 15 cos (45 ) 1.39 v 2 v 1 = 10.61 = 10.70 1.39 1 1.39 = 7.46 10.61 The change in velocity is about 10.7 m sec 1 and the angle is 82.5 (from north) 9 Change in velocity = v 2 v 1 = v v cos ( θ) v v cos ( θ) = 0 v sin ( θ) v sin ( θ) v 2 v 1 = v v cos ( θ) v sin ( θ) [ v v cos ( θ) ] [ v sin ( θ )] = + 2 2 = v = v 2 2 1 2cos ( θ) + cos ( θ ) + sin ( θ) 2 2cos ( θ) Cengage Learning Australia 2014 ISBN 9780170251501 45

Chapter 1 Review Multiple choice 1 D 2 A 3 11 + 7 = 4 B 4 D The result vector is v + w using tip to tail which is the same as w v. 5 4 10 = 2(2i 5j) so it is in the opposite direction to b. B Cengage Learning Australia 2014 ISBN 9780170251501 46

6 A 7 D 8 (1, 1) = 2 A 9 a = ( 4, 2) and b = (12, 6) = 3a i.e. a = 1 b C 3 Short answer 10 a and b Cengage Learning Australia 2014 ISBN 9780170251501 47

11 a DE = 2i + 6j b c d e TS = 8i + 5j GJ = 4i + 4j RK = 2i + 4j YZ = 7i 12j 12 c = (6, 3) and d = (3, 7) a c + d = (6, 3) + (3, 7) = (9, 4) b d + c = (3, 7) + (6, 3) = (9, 4) c d + ( d) = (0, 0) d c + ( d) = (6, 3) (3, 7) = (3, 10) e c + c + d = (6, 3) + (6, 3) + (3, 7) = (15, 1) Cengage Learning Australia 2014 ISBN 9780170251501 48

13 a (5, 8) (9.434, 58 ) (using the calculator) If not using the calculator you would use (θ) = y x and r = 2 2 x + y. Check the quadrant. Quadrant 1. (θ) = y 8 x = 5 so θ = 58 and r = 2 2 5 + 8 = 9.434 (5, 8) (9.434, 58 ) b ( 3, 4) (5, 126.9 ) c ( 6, 10) (11.662, 239 ) d (9, 3) (9.487, 341.6 ) e ( 4, 0) (4, 180 ) Cengage Learning Australia 2014 ISBN 9780170251501 49

14 a (3, 30 ) 3 3 3, 2 2 (using the calculator) If not using the calculator you would use x = 3 cos (30 ) and y = 3 sin 30 x = 3 cos (30 ) = 3 3 3 3 = 2 2 y = 3 sin (30 ) = 1 3 3 = 2 2 Hence (3, 30 ) 3 3 3, 2 2 b (8, 150 ) ( 4 3, 4) c (4, 300 ) (2, 2 3) d (7, 45 ) 7 2 7 2, 2 2 e (9, 230 ) ( 5.785, 6.894) Cengage Learning Australia 2014 ISBN 9780170251501 50

15 a 3 5 (5.831, 59 ) b c 2 6 1 4 (6.325, 251.6 ) (4.123, 284 ) d 0 5 (5, 90 ) e 2 8 (8.246, 104 ) 16 a (24, 10) = 2 2 24 + 10 = 26 b ( 3, 4) = 5 c 8 15 = 17 d 5 9 = 106 10.296 e (10, 12) = 244 15.621 17 18 If a = 12 9, then 6a = 6 12 9 = 72 54 Cengage Learning Australia 2014 ISBN 9780170251501 51

19 m = (7, 24) m 1 1 ˆm = ( 7, 24) ( 7, 24) m = (7, 24) = 25 = 7 24, 25 25 20 a (1, 3) = i 3j b c (4, 5) = 4i + 5j ( 2, 5) = 2i + 5j d e 4 3 = 4i 3j = 2i 5j f (6, 56 ) (3.36, 4.97) = 3.36i + 4.97j (from calculator) g (8, 200 ) 7.52i 2.74j h (3, 180 ) ( 3, 0) = 3i i (10, 142 ) 7.88i + 6.16j j (5, 120 ) 2.5i + 5 3 2 j 21 a 4i 7j + 8i + 2j = 12i 5j b (3, 8) + ( 9, 6) = ( 6, 2) c 1 10 + 66 = 5 4 22 Given a = (3, 5), b = ( 2, 3) and c = ( 1, 4), a a + c = (2, 9) b b a = ( 5, 8) c 3b + 2c = ( 8, 1) d 5a 3b = (21, 34) e a + b + c = (0, 6) Cengage Learning Australia 2014 ISBN 9780170251501 52

2 23 Given x = 3, y = 41 and z = 1 3,: a x y = 6 4 b 2x + 3y = 8 3 c y + 3z = d z 2x = 7 8 5 3 e x + z = 1 6 24 Given p = (5, 85 ), q = (7, 146 ) and r = (6, 212 ) using the calculator a p + q = (10.39, 121.1 ) b q + r = (10.92, 176.1 ) c r + p = (4.99, 158.8 ) d p r = (9.85, 55.9 ) e r q = (7.13, 275.8 ) Cengage Learning Australia 2014 ISBN 9780170251501 53

25 Given f = (5, 150 ), g = (8, 60 ) and h = (9, 225 ), a f g = (5, 150 ) (8, 60 ) ( 33.33, 47.73) (58.22, 124.94 ) = (58.22, 235.06 ) Without using the calculator quick method f g = (5, 150 ) (8, 60 ) [5 cos (150 ), 5 sin (150 )] [8 cos (60 ), 8 sin (60 )] ( 8.33, 4.43) Using (θ) = y x and r = 2 2 x + y we get (θ) = 4.43 8.33 2 2 and r = ( 8.33) + ( 4.43) θ = 28.00 r = 9.43 but quadrant 3 so θ = 208 f g = (9.43, 208 ) b g h (10.36, 13.29) (16.86, 52.1 ) c g + h ( 2.36, 0.56) (2.43, 166.6 ) d f + g ( 0.34, 9.43) (9.43, 92 ) e h + g + f ( 6.69, 3.06) (7.36, 155.4 ) Cengage Learning Australia 2014 ISBN 9780170251501 54

26 Show that vector addition is associative for r = (2, 5), g = ( 2, 7) and k = (3, 2). Let r = (2, 5), b = ( 2, 7) and k = (3, 2) (r + b) + k = ((2, 5) + ( 2, 7)) + (3, 2) = ((2 + 2), ( 5 + 7)) + (3, 2) = (2 + 2 + 3), ( 5 + 7 + 2) = (2 + ( 2 + 3), 5 + ( 7 + 2)) = (2, 5) + ( 2 + 3, 7 + 2) = r + (b + k) 27 If m = (x 1, y 1 ) and n = (x 2, y 2 ), prove that m + n = n + m. Let m = (x 1, y 1 ) and n = (x 2, y 2 ). m + n = (x 1, y 1 ) + (x 2, y 2 ) = [(x 1 + x 2 ), (y 1 + y 2 )] = [(x 2 + x 1 ), (y 2 + y 1 )] = (x 2, y 2 ) + (x 1, y 1 ) = n + m Cengage Learning Australia 2014 ISBN 9780170251501 55

Application 28 Express each of the following in terms of a and c. a BC = AO = a b c AC = AO + OC = a + c OB = OA + AB = a + c d OM = 1 2 OB = 1 (a + c) 2 e AM = AO + OM = a + 1 2 (a + c) = 1 ( a + c) 2 f MC = MO + OC = OM + OC = 1 1 (a + c) + c = ( a + c) 2 2 29 (24,0) + [18 cos (25 ), 18 sin (25 )] = (40.31, 7.61) (40.31, 7.61) = 41.02 (θ) = 7.61 40.31 so θ = 10.69 Approximately 41 N at 10.7 to the horizontal. Cengage Learning Australia 2014 ISBN 9780170251501 56

30 AC = (0, 900) + (600, 0) = (600, 900) 2 AC = ( ) 2 600 + 900 = 1081.7 θ = 1 900 600 = 56.31 The bearing is 90 + 56.31 = 146.3 The displacement 1082 m on a bearing of 146.3. θ Cengage Learning Australia 2014 ISBN 9780170251501 57

31 The result force on the ship is r = [35 000 cos (46 ), 35 000 sin (46 )] + [30 000 cos (23 ), 30 000 sin (23 )] r = ( 13 454.96, 51 928.19) r = 53643 i.e. approximately 53 600 N θ = 1 51928.19 13454.96 = 104.5 The result force on the ship is 53 600 N in the direction N14.5 W. Cengage Learning Australia 2014 ISBN 9780170251501 58

32 Change in velocity v 2 v 1 = 0.2[cos (135 ), sin (135 )] 0.3[cos (240 ), sin (240 )] = ( 0.0086, 0.4012) ( 0.0086, 0.4012) = 0.4013 θ = 1 0.4012 0.0086 = 88.8 o 88.8 o 60 Angle = (90 88.8 ) + 90 + 60 = 151.2 The change in velocity is 0.4013 m/sec at an angle of 151.2 to the original direction. Cengage Learning Australia 2014 ISBN 9780170251501 59