Chapter 7 Introduction to vectors
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1 Introduction to ectors MC Qld-7 Chapter 7 Introduction to ectors Eercise 7A Vectors and scalars a i r + s ii r s iii s r b i r + s Same as a i ecept scaled by a factor of. ii r s Same as a ii ecept scaled by a factor of. iii s r a A to D s + t b A to B s + t + u+ c D to A s t d B to E u t e C to A u t s Displacement ( north km north The answer is C. a u + A to C b u D to B c u B to D d u + u u + A to C u + w w u u + u u u u The answer is D. 6 a CH CG + GH r + s uur uur b CJ CG + GJ s + t c GD GH + HD r s uur uur d FI FE + EI r + s uur uur e HE HI + IE t s uur uur f DJ DH + HG + GJ s r + t uur uur g CI CD + DE + EI r + t + s h JC uur uur JG + GC t s 7 a and b 8 c R km d tan θ θ tan ( θ. clockwise from north. EF TE 00 cos 00 0 km Total distance east of the starting point is km
2 MC Qld-7 Introduction to ectors Resultant bearing: 0 tan θ ( θ. Resultant bearing is clockwise from north 9 θ tan 0.8 Bearing is True R km tan θ θ tan ( θ 6. Resultant bearing a a + b 8 east and 8 north b a + b 8 east and north c a b west and north d b a east and south e b a east and south f 0.a +.b east and 0 north g a.b 9. west and. south h a east and 0 north i.a.b 8 north b.a. west and 9. south 6 7 AC a b BD a + b AC + BD a b + a + b a ( u + u + ( a + b + c a + ( b + c r s ( s r The horizontal component of w is The ertical component of w is The horizontal component of is The ertical component of is w+ (6, 8 AB w (,
3 Introduction to ectors MC Qld-7 8 a b w (6, 0 (, 6 9 One can deduce that and y components can be added, subtracted and multiplied separately. 0 OD a + b a + b The answer is B. EO a + b a b The answer is D. Net displacement ector is O Displacement, elocity, force Speed, time, length magnitude and angles (N S and (E W. Eercise 7B Position ectors in two and three dimensions a i + k, y, z b 6i k 6, y 0, z c.i + + k., y z a i ii θ tan 6 ( 6 tan b i w ( ii tan 7 ( tan ( θ (second quadrant c i a (. + ( ii tan. (. tan (.09.8 θ (third quadrant d i b 0 + ( ( 0.6 ii tan ( 0 0 tan ( or θ (fourth quadrant a 0 T b T c T d Τ θ 90 + ( w cos 0 00 cos 0 0 y w sin 0 00 sin 0 00 sin 60 0 w 0i 0 i 0 cos 0, y 0 sin 0 The answer is C.
4 MC Qld-7 6 Introduction to ectors 6 7 θ cos ( y 7 sin ( 7 8. u 8.9 i 8. 0 ( 0 0 ( ( 0 i ( 0 a+ b i + + i + + i ( ( a+ b θ tan. Take 6 steps in a direction. south of east. 8 θ 90 + (60 9 cos y sin b 60.6 i a 0cos i 0sin 0 i 0 o o b 00cos60 i 00sin 60 00i 00 a+ b ( 0 i 0 + ( 00i 00 ( i ( a+ b ( ( θ tan The resultant displacement is 6 km at 9.8 south of east. 9 a 0cos i + 0sin 0 i + 0 b 0cos i 0sin i a cos0 i + sin 0 i + b sin 0 i + cos 0 a b i + ( sin 0 i + cos0 + sin 0 i + cos i.69 a b The scouts are 0.8 km apart. a a i + a + â i + b d i d + ( ˆd i c b i + b + ˆb i + d e i + e ( + ê i + e c i + c + ( ĉ i +
5 Introduction to ectors MC Qld-7 7 f f.i +.7 f (. + ( ˆf i i i + ( ˆ i The answer is B. 0.i ˆ i i (0.i + 0. w 0.i 0.0 w ( 0. + ( ŵ i i 0.0 w 0i 0 w ( 0 + ( ŵ i i 6 a AB ( 0 i + ( i + b BA (0 i + ( i 7 a i AB ( 0 i + ( i 7 ii AB + ( 7 6 b i AB ( i + ( i + ii AB + 0 c i AB (0 i + ( i + 7 ii AB ( d i AB ( i + ( i ii AB ( + ( 0 e i AB ( i + (7 7 i ii AB f i AB ( 7 i + ( i ii AB ( 8 a BA AB i + 7 b BA i c BA i 7 d BA i + e BA i f BA i 9 a AB i 7 AB 6 ˆ AB ( i 7 6 i b AB i + AB 0 ˆ AB i c AB i + 7 AB 6 ˆ 7 AB i d AB i AB 0 ˆ AB i 0 0 e AB i AB ˆ AB i i f AB i AB
6 MC Qld-7 8 Introduction to ectors ˆ AB i i 0 u i and e i + a i u + ( 9 ii e ( + iii û i 9 9 i ê i + u + e i i + i + i u + e + 0 b u + e 9 + > u + e 0 Therefore, reect the statement as the magnitudes are different. u i + and e i a i u ( + ii e + ( 6 iii û i + i ê i 6 6 u + e ( + i + ( i + i u + e + b u + e + 6 > u+ e Therefore, reect the statement as the magnitudes are different. a a b ( i + ( i a b + b a b ( ( i + i 7 a b a r i b b a c i + d tanθ θ tan 9.0 True bearing e + km/h θ tan.0 True bearing The boat should trael on a bearing of 9.0 to arrie at the opposite bank due north of the starting position. b c d e f + + ( 0 ( + ( ( +.6 ( + ( + ( 7 + ( + ( or a a i + k (i + k i + k a + ( + b b i + + k ( i + k i + k b + c c i + k ( i+ + k i k c + ( + ( 6 d d 8i+ + k ( i + k 7i + + k
7 Introduction to ectors MC Qld-7 9 d C (, 6, 0, D (,,, E (, 8, 0 and F (, 6, 6 CD i k (i + 6 i 7 k uur EF i k ( i k i k ( i 7 k CD uur EF CD 8 a tan y θ [] ( θ tan 90 ( θ ( θ sin 90 cos 90 y y cosθ y sinθ sinθ cosθ y tan θ [] y Equating [] and [] y y yy 0 + y y 0 a a 0i, b a b 0 i + b a b 0 + c tan θ 0 θ tan ( True bearing u i+ tan θ θ tan θ. i tan θ θ tan ( θ 6.9 Difference That is, the two ectors are perpendicular to each other. + 0 To confirm that this is a pattern for all perpendicular ectors, let u i + y and let u make an angle of θ to the positie direction of the -ais. Let i + y be at right angles to ector u. As the ectors are perpendicular, will be at 90 θ to the negatie direction of the ais, i + b tan θ θ 8. True bearing c time distance speed 0. km km / h hours Time taken is. minutes. Eercise 7C Multiplying two ectors the dot product If u i + and 6i + u tan β 6 tan α β 8. α θ α β cos θ 0.89 u
8 MC Qld-7 0 Introduction to ectors u i + and 6 i+ u This is more accurate as no angle is required. a u ( ( 6 i + + k i + + k b u ( ( i + k i + k c u ( i+ k ( i 7 + k d u ( 9 ( i + i e u ( i+ ( + k f u (0 i ( i 0 0 g u ( + k ( i h u (6 i + k ( i k u ( i + k ( i + 6 k The answer is E. u u cos θ 6 cos. The answer is C. 6 u cos θ u 7, 8 and θ 80 0 u 7 8 cos u u ( i+ y ( i + y + y y + y ( i + k ( i + k u 9 w ( u (i [i + ( i ] (i (i wu w (i (i + (i ( i + ( + ( + 9 w ( u w u w 0 w ( u+ (i (i+ + i (i ( i w u + w (i (i + + (i ( i w ( u+ w u + w Two ectors are perpendicular if their dot product is 0. A: (i + + k ( i k B: (i + + k (i+ + k C: (i + + k ( i D: (i + + k ( i + k + 0 The answer is D. ( u ( u + 0 u u 0 u 0 u The answer is B. ( u ( u + u u or u The answer is D. a (i k (7 + k b ( i + k ( 9i+ k
9 Introduction to ectors MC Qld-7 c (8i+ (i + k d (i + k (i + k a u + (, 6 cos θ u u θ 07 b u 98 cos θ 98 c u + + (, ( ( θ , cos θ 7 9 d u 6 u 0. θ 8 + ( +, 7 7 cos θ 7 7 u θ 09 ( i + ( i 9 +, cos θ θ.6 The answer is D. 7 u ( i ( i u + (, ( ( + ( cos θ ( + 6 θ 80 The answer is E. 8 u ( ai + (6 i 0 6a 6 0 6a 6 a 9 u ( ai a k ( i k + + a + 6a a a 6 6 a 0 a 0 u i + Let k(i + ki + k u (i + (ki + k 0 k + 6 k 0 0 k 0 k i + 8 u i Let k(i ki k u (i (ki k 80 6 k + 9 k 80 k 80 k i 0 Eercise 7D Resoling ectors scalar and ector resolutes a u i+ and a i + i u + û ( i + a ( i+ ( i ii a + â ( i + â u ( i + ( i +
10 MC Qld-7 Introduction to ectors 8 + b u i and a i i u + ( 9 û ( i 9 a ( i ( i ii a ( 0 â ( i 0 aˆ. u ( i ( i c u i + 6 and a i i u ( û ( i a ( i + 6 ( i ii a ( 7 â ( 7 i â u ( ( 6 7 i i d u i and a i i u + ( û ( i a ( i ( i ii a 6 ( + ( â ( i â u ( i (i e u 8i 6 and a i + i u 8 + ( 6 0 û (8i 6 0 a (8i 6 ( i ii a 6 ( + â ( i+ 6 â u ( i + (8 i
11 Introduction to ectors MC Qld-7 a u i and i + i u + ( 0 û ( i 0 ( i ( i ( uˆ uˆ ( i 0 0 ii ( i 0 i 0 0 iii i + i i+ 0 0 b u i + and 8 i 0 i u + û ( i+ ( i+ (8 i ii ( uˆ uˆ (i + 8i + 0 iii 8i + 0 (8i c u i + and i + i u + û (i + (i + ( i ii ( uˆ uˆ 0 (i+ 0 iii i + 0 i + d u i+ + k and i+ k i u + + û ( i k ( ( i k i + ii ( uˆ uˆ ( i k + + iii i k + i + + k + + e u i + + k and i k i u û ( i+ + k 9 ( i+ + k ( i k ii iii 9 ( uˆ uˆ ( i + + k i k i k + 9 i k 00 i k ( i 6 8 k 9
12 MC Qld-7 Introduction to ectors f u i + k and k i u + + ( û ( i + k ( i + k ( k + ii ( uˆ uˆ ( i+ k + iii k i + k i + i+ i+ ( i+ ( + 0. km Let the yacht s position be denoted by. The path of the rescue boat be denoted by u. i u k( i represents the closest distance the rescue boat gets to the yacht. u k + ( a Let the inured bushwalker s position be denoted by. Let the path of the searcher be denoted by u. i + u k(i + u k + k û k(i k + i + i + (i ( uˆ uˆ 8 i + 7 i The searcher is.6 km from the camp when closest to the bushwalker. b represents the minimum distance between the searcher and the bushwalker. k 0 û ( 0 k i k ( i 0 ( i ( i ( uˆ uˆ 7 ( i ( i 0 7 i i i 0 0 i 0 0 ( i 0 ( + ( km or 0.6 km
13 Introduction to ectors MC Qld-7 Eercise 7E Time-arying ectors a u ti t t or t y t y b u ( t i t t t + y t ( + c u ( t + i + t t + t y t y ( d u ti + t t t y t y y 8 u ( t + i + t when t 0, u i + 0 The answer is B. t a i + ( t + t, t 0 t, 0 or t y t + t ( + y + b When 0, y 0, y + 6 -intercepts: + 0 ( ( 0 or a u cos (t i + sin (t, t 0 cos t or cos t y sin t or y sin t + y cos t + sin t + y b Circle with centre (0, 0 and radius. When t 0, u i. π When t, u. So the particle is moing from (, 0 in an anticlockwise direction. c Period π π 6 a u cos (t i + sin (t, t 0 cos t or 9 cos t y sin t or y 9 sin t + y 9 cos t + 9 sin t + y 9 b Circle with centre (0, 0 and radius. When t 0, u i. π When t, u. The particle is moing from (, 0 in an anticlockwise direction. a ( t + i + ( t t, t 0 t +, (since t 0 or t y t t ( ( + + y + b y +, ( + ( Turning point is (, c Period π π 7 a u ( + cos t i + ( + sin t, t 0 + cos t or cos t cos t ( y + sin t or sin t y + sin t (y + ( + (y + cos t + sin t ( + (y +
14 MC Qld-7 6 Introduction to ectors b Circle with centre (, and radius. When t 0, u i. π When t, u i. So the particle is moing from (, in an anticlockwise direction. c Period π π 8 u cos t i + sin t, cos t or cos t cos t 9 y sin t or sin t y 9 + y cos t + sin t t 0 + y 9 Graph is an ellipse with centre (0, 0 with a and b. When t 0, u i When t π, u The particle is moing in an anticlockwise direction from (, 0. 9 u cos t i sin t, cos t or cos t cos t y sin t y or sin t sin y t 6 y + cos t + sin t 6 t 0 y + 6 Graph is an ellipse with centre (0, 0 with a and b. When t 0, u. i π When t, u. The particle is moing in a clockwise direction from (, 0. t + t 0 a u i + ( t + ( t + t + ( t + t y ( t + t + + t + y ( t + ( t + ( t + + y or y When t 0, u ( 0 i + 0 i y t + t When ( t + ( t + or t + t t t + 0 (t 0 t So y when t. and u ( + i i + or ( i+ b u ( t + i + ( t + t +, or t y (t + y ( + y (, c u ( cos t + i + ( sin t cos t + cos t or cos t y sin t y + sin t y + or sin t y + + cos t + sin t ( ( y Ship A s position ector is u (t + i + (t Ship B s position ector is (t + i + (t +
15 Introduction to ectors MC Qld-7 7 Ship A: t + t or t y t y y 0 [] Ship B: t + t t y t [] Equate [] and [] to determine where the ships path cross Substitute 9 into [] y i k i 0 +. k i +. k The answer is A. i + y + z k Where 00 sin y 00 cos z 0 00 i k A: + + B: C: D: + ( E: The answer is B. PQ i + k ( i + k i + k PQ ( + ( + 8 The answer is A. 6 7 The ships paths cross at 9 7 and y 7. u ti + t t y t or y t t ei+ e e t y e t ( e t or y To coincide: (-components t e t and (y-components t t e or t e t Paths coincide if t e t. But e t > t for all t 0. Therefore the paths can neer coincide. Furthermore, is always ahead of u (Since e t > t. Chapter reiew u i + 0. k and i + k u. ( i + 0. k.( i + k a HX i + 0 cos i + 0 sin + i + i + + ( + i + ( + b HX ( km 6 u i + k and i + + k u ( i + k ( i + + k The answer is D. 7 u i and 6 i + u cos θ u θ cos (0 90 The answer is E.
16 MC Qld-7 8 Introduction to ectors 8 A: Magnitude + Not a unit ector. B: Magnitude 0 0 Not a unit ector. C: Magnitude A unit ector. and (i (0.8i So the ectors are perpendicular. The answer is C. 9 u i + a and ai a ( i + a ( ai a 0 6a a 0 a (6 a 0 a 0 or a 6 Reect a 0 as 0 in this case. The answer is D. (i + k (i 0 cos θ + ( + + ( θ cos ( 6 The answer is E. u i + and i + u cos θ u (i + ( i + + ( θ cos.909 u i and i + a u + i + i + i b u i ( i + 7i 6 c u (i ( i u d û u i + ( ( i i u e cos θ u 7 + ( ( θ i + k + ( + 9 -ais: cos θ 9 θ 6. y-ais: cos θ 9 θ.8 z-ais: cos θ 9 θ.0 [ pi + ( p ] (pi + 0 p + 6( p 0 p + 6 8p 0 p 9p ± ( 9 (( p 9 ± 69 a i and b i + b + ˆb ( i + ˆb a (i + ( i ( 6 The answer is B. 6 a + ( â ( i
17 Introduction to ectors MC Qld-7 9 ( aˆ b aˆ ( ( ( i i + i ( 6 ( i ( i The answer is B. 7 a u i + k and y The path is hyperbolic. Asymptotes: 0, y i + k b û ( u u (i + k + + ( ( i + k i + k ( uˆ û ( i + k (i + k ( + + ( i + k ( i + k i + k ( i + k k ( + k 8 u sin ti + cos t sin t sin t sin t 9 y cos t y cos t cos t y y + sin t + cos t 9 The path is an ellipse. The answer is D. 9 u i + ( t t t, t > 0 > 0 or t y ( t ( i + k Modelling and problem soling a 8 i + k ( i k b ˆ 6 i + k 6i + k 6 + ( + 6i + k (6 i 8 + k ( i + k c The fighter starts at i k and then continues along the path m ˆ. The equation of motion is therefore m i + 8+ k + mˆ i + 8+ k + ( i + k d ˆ ˆ ( i + k u ( ˆ u ˆ ( ( 8 i k i k + ( i + k ( + 6 ( i 9 + k u 9 ( i + k ( i + k from fighter s starting position. u + u i k + ( i + k 0 i k u (0 i k e u The distance is 7.8 km. f Speed km h 6 + ( + from station.
18 MC Qld-7 0 Introduction to ectors The speed of the fighter is 080 km/h. g Vertical speed of the fighter 70 km/h. a C will hae the same coordinate as D, the same y coordinate as E and the z coordinate will be 0. C is (,., 0 b CE OE OC. + k ( i i + k c Other diagonal i +. + k. i + k ( i + k (i + k cos θ ( θ 7.7 d V. 66 cm in litres V litres. e d i +. + k f d cm g Other longest diagonal (connecting D to the y ais is gien by: b i +. k b. 7. d b cos θ d b ( θ 8. a c OY 7i + 7 0i+ 0 7i+ 7 ZX i + 7 i i + 7 ( i+ 7 (7i + 7 d cos θ ( e tan θ 7. θ f The ector resolute of OX on the -ais is the -component of X. i g Let P be the point such that OP i + 7 (P lies on XY PZ i ( i + 7 ( i 7 OY 7i + 7 PZ OY [( i 7 ] (7i ( OP i + 7 Coordinates of P are (, 7. h Area bh where b, h 7 Area 7 a square units. b ZY 7i + 7 i i + 7 YX i + 7 (7i + 7 i b Speed +.8 m/s c tan θ.6667 θ 9.0 The bearing is 09 Τ. d Distance speed time where speed m/s
19 Introduction to ectors MC Qld-7 time minutes or 0 seconds. Distance 0 60 The width of the rier is 60 metres. e Distance The swimmer is carried 600 metres downstream. f If the swimmer started on the north bank: Speed.8 m/s (same as before Rier width 60 m Distance downstream 600 m (same as before Bearing: tan θ θ 9.0 Bearing is 80 9 T The results are the same, ecept the bearing is now T. a Assuming that Sally is at the origin, the ector would be: sin i + cos + 8 k c Total height is m 9.7 θ tan. The angle of eleation is.. 6 a Let represent the skydier s path. i + k (6i + + k i 9 8k b ( + ( 9 + ( 8 ˆ ( i 9 8 k c i + k Distance + ( + 0 km d Speed 60(km/h km/h. i k b Distance m
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