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Vector Journeys

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Learn how to find vector components using other vectors, navigate parallel vectors, and discover alternative routes in vector equations. Understand key rules and concepts for efficient vector manipulation.

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Vector Journeys

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  1. Vector Journeys

  2. What is to be learned? • How to get components of a vector using other vectors

  3. Useful (Indeed Vital!) To Know components Parallel vectors with the same magnitude will have the same………………… If vector AB has components ai + bj + ck, then BA will have components……………... -ai – bj – ck

  4. Diversions Sometimes need alternative route! B A C E D AE = AB + BC + CD + DE

  5. Diversions Sometimes need alternative route! B A C E D AE = AB + BC + CD + DE

  6. Diversions Sometimes need alternative route! B A C E D AE = AB + BC + CD + DE

  7. Diversions Sometimes need alternative route! B A C E D AE = AB + BC + CD + DE

  8. Using Components AB = 2i + 4j + 5k and BC = 5i + 8j + 3k AC = 7i + 12j + 8k DE = 6i + 5j + k and FE = -3i + 4j + 5k DF = 9i + j – 4k = AB + BC = 3i – 4j – 5k EF = DE + EF

  9. Wee Diagrams and Components ABCD is a parallelogram C T is mid point of DC T Find AT D ( ) 846 AB = = DC B ( ) 351 BC = A = AD Info Given AT + DT = AD + ½ DC = AD ( ) ( ) ( ) 774 351 423 + =

  10. Wee Diagrams and Letters u D C v -u -v ? AB = 4DC A B 4u Find AD in terms of u and v AD = 4u – v – u = 3u – v

  11. Vector Journeys Find alternative routes using diversions With Letters D A u AD = 3BC B C v Find CD in terms of u and v + AD CD = CB + BA = -v + u + 3v = 2v + u

  12. E with components D C EABCD is a rectangular based pyramid T divides AB in ratio 1:2 1 2 A T B EC = 2i + 4j + 5k 2/3 of CD BC = -3i + 2j – 3k ET = EC + CB + BT ( ) ( ) ( ) negative CD = 3i + 6j + 9k 245 3 -2 3 246 + = + Find ET = 7i + 6j + 14k

  13. Questions Cunningly Acquired!

  14. Hint Vectors Higher VABCD is a pyramid with rectangular base ABCD. The vectors are given by Express in component form. Ttriangle rule  ACV Re-arrange also Triangle rule  ABC Previous Quit Quit Next

  15. U B V PQRSTUVW is a cuboid in which T    W PQ , PS & PW are represented by the vectors A R Q A is 1/3 of the way up ST & B is the midpoint of UV. S P ie SA:AT = 1:2 & VB:BU = 1:1. [ ], [ ]and [ ]resp. 4 -2 0   0 2 4 Find the components of PA & PB and hence the size of angle APB. 0 9 0 VECTORS: Question 3 Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT

  16. U B V PQRSTUVW is a cuboid in which T    W PQ , PS & PW are represented by the vectors A R Q A is 1/3 of the way up ST & B is the midpoint of UV. S P ie SA:AT = 1:2 & VB:BU = 1:1. [ ], [ ]and [ ]resp. -2 0 4   4 2 0 Find the components of PA & PB and hence the size of angle APB. 0 0 9 = 48.1° VECTORS: Question 3  Reveal answer only = 29 |PA|  Go to full solution = 106 |PB| Go to Marker’s Comments APB Go to Vectors Menu EXIT

  17. Question 3 PQRSTUVW is a cuboid in which    PQ , PS & PW are represented by vectors A is 1/3 of the way up ST & B is the midpoint of UV.   + + PA = PB = ie SA:AT = 1:2 & VB:BU = 1:1. [ ]= [ ] [ ] [ ] = [ ] [ ]+ [ ], [ ]resp. [ ] [ ]and -2 -2 0 3 4 -1 4 0 -2 0 4 0 0 2 4 2 4 2 0 4 9 0 0 0 3 3 0 0 9 9       PS + 1/3ST PS + 1/3PW PS + SA =       PQ + PW + 1/2PS PQ + QV + VB = =   Find the components of PA & PB and hence the size of angle APB. = Begin Solution = Continue Solution Markers Comments Vectors Menu Back to Home

  18. Question 3 PQRSTUVW is a cuboid in which    PQ , PS & PW are represented by vectors A ie  A is 1/3 of the way up ST & B is the midpoint of UV. P     . PB = PB = PA = PA . B ie SA:AT = 1:2 & VB:BU = 1:1. [ ] [ ], [ ] [ ]and [ ] [ ] [ ]resp. 3 3 0 -2 -2 4 -2 4 4 2 4 0 4 4 0 0 9 9 9 3 3 (b) Let angle APB =    Find the components of PA & PB and hence the size of angle APB. Begin Solution = (-2 X 3) + (4 X 4) + (3 X 9) Continue Solution = -6 + 16 + 27 Markers Comments = 37 Vectors Menu Back to Home

  19. Question 3 PQRSTUVW is a cuboid in which    PQ , PS & PW are represented by vectors A is 1/3 of the way up ST & B is the midpoint of UV.   PA . PB = ie SA:AT = 1:2 & VB:BU = 1:1. [ ]and [ ]resp. [ ], 4 0 -2   2 4 0 |PB| = (32 + 42 + 92) |PA| = ((-2)2 + 42 + 32)   9 0 0 Given that PA.PB = |PA||PB|cos 37 = then cos = 29 106   PA.PB |PA||PB| 37 = 29 = 106   Find the components of PA & PB and hence the size of angle APB. so  = cos-1(37  29  106) Begin Solution Continue Solution = 48.1° Markers Comments Vectors Menu Back to Home

  20. Ex Suppose that AB =( ) and BC =( ) . 3 5 -1 8 Find the components of AC . ******** ( ) + ( ) =( ) . 3 5 8 AC = AB + BC = -1 8 7 Ex Simplify PQ - RQ ******** PQ - RQ = PQ + QR = PR

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