Polar Coordinates; Vectors

Transcription

1 10.5 The Dot Product 1. v i, w i+ (a) v w 1(1) + ( 1)(1) (b) cos v w ( 1) º (c) The vectors are orthogonal.. v i +, w i+ (a) v w 1( 1) +1(1) (b) cos v w ( 1) º (c) The vectors are orthogonal. 3. v i +, w i+ (a) v w (1)+1() + 4 (b) cos v w (c) The vectors are neither parallel nor orthogonal º 4. v i +, w i + (a) v w (1)+ () +4 6 (b) (c) cos v w º The vectors are neither parallel nor orthogonal. 5. v 3 i, w i + (a) v w 3(1)+ ( 1)(1) 3 1 (b) cos v w (c) + ( 1) º The vectors are neither parallel nor orthogonal

2 6. v i + 3, w i (a) v w 1(1)+ 3( 1) 1 3 (b) cos v w (c) 1 + ( 1) º The vectors are neither parallel nor orthogonal. 7. v 3i + 4, w 4i+3 (a) v w 3(4) + 4(3) (b) cos v w (c) The vectors are neither parallel nor orthogonal. 8. v 3i 4, w 4i 3 (a) v w 3(4) + ( 4)( 3) (b) (c) º cos v w ( 4) 4 + ( 3) º The vectors are neither parallel nor orthogonal. 9. v 4i, w (a) v w 4(0)+0(1) (b) cos v w º (c) The vectors are orthogonal. 10. v i, w 3 (a) v w 1(0) + 0( 3) (b) cos v w ( 3) º (c) The vectors are orthogonal. 11. v i a, w i+3 Two vectors are orthogonal if the dot product is zero. Solve for a: v w 1() + ( a)(3) 3a 3a 0 3a a 3 1. v i +, w i+b Two vectors are orthogonal if the dot product is zero. Solve for b: v w 1(1) + 1( b) 1 + b 1 + b 0 b

3 Section 10.5 The Dot Product 13. v i 3, w i pro w v v w (1) + ( 3)( 1) w w 1 + ( 1) v v i v 3i +, wi+ 5 i 5 i 5 ( i ) 5 i 5 1 i 1 pro w v v w 3() + (1) w ( i + ) 4 ( w ( + 1 ) 5 i + ) 8 5 i 4 5 v v 3i+ 8 5 i v i, w i+ pro w v v w 1(1) + ( 1)() w w 1 + v v i 1 5 i i i + 1 ( 5 i + ) 1 5 i i v i, w i pro w v v w (1) + ( 1)( ) w w 1 + ( ) v v i 4 5 i 8 5 i 6 5 i v 3i +, w i pro w v v w w w 3( ) + 1( 1) i ( ( ) + ( 1) ) v v ( 3i + ) 14 5 i i v i 3, w 4i pro w v v w 1(4) + ( 3)( 1) w w 4 + ( 1) v v i 3 8 4i 17 i i ( 5 i ) 4 5 i i i 14 5 i i

4 19. Let v a the velocity of the plane in still air. the velocity of the wind. the velocity of the plane relative to the ground. + v a 550( cos( 5º )i + sin( 5º )) 550 i 75 i 75 80i + 75 i i ( )i 75 The speed of the plane relative to the ground is: + ( 75 ) miles per hour To find the direction, find the angle between and a convenient vector such as due south,. ( ) cos v g ( 1) ( 1) º The plane is traveling with a ground speed of about miles per hour in a direction of 38.5 west of south. 0. Let v a the velocity of the plane in still air. the velocity of the wind. the velocity of the plane relative to the ground. + v a 50 cos i + sin 40 i i 40 i i + 50cos i + 50sin + 0 i 0 a i Examining the components: 50sin sin 0 sin º 50 The heading of the plane should be N83.5 E. Examining the i components: 50cos6.5º i+ 0 i ai 76.7 a The speed of the plane relative to the ground is 76.7 miles per hour. 1. Let the positive x-axis point downstream, so that the velocity of the current is v c 3i. Let the velocity of the boat in the water. Let the velocity of the boat relative to the land. 1006

5 Section 10.5 The Dot Product Then The speed of the boat is 0 ; we need to find the direction. Let ai + b so a + b 0 a + b 400. Let k. Since, k ai + b + 3i k (a + 3)i + b a and k b a 3 a + b b 400 b 391 k b i and Find the angle between and : cos v w (1) º The heading of the boat needs to be 8.7 upstream. The velocity of the boat directly across the river is kilometers per hour. The time to cross the river is: t hours or t 1.5 minutes.. Let the positive x-axis point downstream, so that the velocity of the current is v c 5i. Let the velocity of the boat in the water. Let the velocity of the boat relative to the land. Then The speed of the boat is 0 ; we need to find the direction. Let ai + b so a + b 0 a + b 400. Let k. Since, k ai + b + 5i k (a + 5)i + b a and k b a 5 a + b b 400 b 375 k b i and Find the angle between and : cos v w (1) º The heading of the boat needs to be 14.5 upstream. The velocity of the boat directly across the river is kilometers per hour. The time to cross the river is: t hours or t 1.56 minutes. 1007

6 3. Split the force into the components going down the hill and perpendicular to the hill. F d F sin( 8º ) 5300sin( 8º ) Fd 5300(0.139) 738 pounds 8 F F p F cos( 8º ) 5300cos( 8º ) p F 5300(0.9903) 549 pounds The force required to keep the car from rolling down the hill is about 738 pounds. The force perpendicular to the hill is approximately 549 pounds. 4. Split the force into the components going down the hill and perpendicular to the hill. F d F sin( 10º ) 4500sin( 10º ) Fd 4500(0.1736) 781. pounds 10 F F p F cos( 10º ) 4500cos( 10º ) p F 4500(0.9848) pounds The force required to keep the car from rolling down the hill is about 781. pounds. The force perpendicular to the hill is approximately pounds. 5. Let v a the velocity of the plane in still air. the velocity of the wind. the velocity of the plane relative to the ground v a 500 cos( 45º )i + sin( 45º ) 60 cos( 10º )i+ sin( 10º ) 60 1 i i i i The speed of the plane relative to the ground is: + ( ) i + 50 i i kilometers per hour To find the direction, find the angle between and a convenient vector such as due north,. cos v g ( ) 0 + ( )(1) º The plane is traveling with a ground speed of about kilometers per hour in a direction of 38.6 east of north. 1008

7 Section 10.5 The Dot Product 6. Let v a the velocity of the plane in still air. the velocity of the wind. the velocity of the plane relative to the ground. + v a 600( cos( 60º )i sin( 60º )) i 3 300i cos( 45º )i sin( 45º ) i + 300i i i The speed of the plane relative to the ground is: + ( ) i kilometers per hour To find the direction, find the angle between and a convenient vector such as due north,. ( ) cos v g (1) º The plane is traveling with a ground speed of about 639 kilometers per hour in a direction of 31 east of south. 7. Let the positive x-axis point downstream, so that the velocity of the current is v c 3i. Let the velocity of the boat in the water. Let the velocity of the boat relative to the land. Then The speed of the boat is 0 ; its direction is directly across the river, so Let i 3i + 0 Let miles per houṙ Find the angle between and : cos v g (1) º The heading of the boat will be 8.1 downstream. 1009

8 8. Let the positive x-axis point downstream, so that the velocity of the current is v c 4i. Let the velocity of the boat in the water. Let the velocity of the boat relative to the land. Then The speed of the boat is 10 ; its direction is directly across the river, so Let i 4i +10 Let miles per hour. Find the angle between and : cos v g (1) º The heading of the boat will be. downstream. 9. F 3( cos( 60º )i + sin( 60º )) 3 1 i i W F AB 3 i i 3 () foot -pounds F 1cos45º i + sin( 45º ) W F AB i + i + i + 5i (5) foot -pounds F 0 cos( 30º )i + sin( 30º ) W F AB 10 3i W F AB W, AB 4i F cos i sin ( cos i sin ) 4i 3 i i i 10 3(100) foot-pounds 4cos 1 cos 60º 33. Let u a 1, v a i + b, w a 3 i + b 3 u v + w ( a 1 ) ( a i + b + a 3 i + b 3 ) ( a 1 ) ( a i + a 3 i + b + b 3 ) ( a 1 ) ((a + a 3 )i + (b + b 3 )) a 1 (a + a 3 ) (b + b 3 ) a 1 a + a 1 a 3 b b 3 a 1 a b + a 1 a 3 b 3 ( a i + b ) + ( a 1 ) ( a 3 i + b 3 ) u v + u w a

9 Section 10.5 The Dot Product i + 0 and v a i +b. 0 v0 a+0 b0 35. Let v a i + b. Since v is a unit vector, v a + b 1 or a + b 1 If is the angle between v and i, then cos v i or cos v i a +b 1 cos + b 1 b 1 cos b sin b sin Thus, v cos i + sin ( ai +b) i 1 1 a. 36. If v a 1 cos i + sin and w a i + b cos i + sin, then cos( ) v w a 1 a b cos cos + sin sin 37. Let v a i + b. pro i v v i i i ( ai + b ) i i v i a, v b, v ( v i)i + ( v ) a(1)+ b(0) i a i (a) Let u a 1 and v a i + b ( u + v) ( u v) ((a 1 + a )i +(b 1 + b )) (a 1 a )i + (b 1 b ) (b) (a 1 + a )(a 1 a ) + (b 1 + b )(b 1 b ) a 1 a b a ( + b ) a 1 Since the vectors have the same magnitudes and these two quantities represent the squares of the magnitudes of each, the difference is 0 and the vectors are orthogonal. Because the vectors u and v are radii of the circle, we know they have the same magnitude. Since u + v and u v are sides of the angle inscribed in the semicircle and we know that they are orthogonal (part a), then this angle must be a right angle. 39. (v w) w v w w w v w w v w v w w w 0 Therefore the vectors are orthogonal. 40. ( w v + ) ( w v ) w v v w v v w + w v v w v w w w v v v w w w v v w If F is orthogonal to AB, then W F AB ( i + ) ( i ) 1 1+1( 1) 0 4. u + v u v ( u + v) ( u + v) ( u v) ( u v) (u u + u v + v u + v v) (u u u v v u + v v) (u v)+(u v) 4(u v) 1011