Vector Journeys

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

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

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

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

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

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

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

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

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

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

( ) ( ) ( ) 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

Questions Cunningly Acquired!

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

[ ], [ ]and [ ]resp. VECTORS: Question 3 PQRSTUVW is a cuboid in which PQ , PS & PW are represented by the vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. Find the components of PA & PB and hence the size of angle APB.  Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT

[ ], [ ]and [ ]resp. VECTORS: Question 3 PQRSTUVW is a cuboid in which PQ , PS & PW are represented by the vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. Find the components of PA & PB and hence the size of angle APB.   Reveal answer only |PA| = 29 Go to full solution  |PB| = 106 Go to Marker’s Comments = 48.1° APB Go to Vectors Menu EXIT

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

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

[ ], [ ]and [ ]resp. Question 3 PA .  PB = 37 PQRSTUVW is a cuboid in which PQ , PS & PW are represented by vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 |PA| = ((-2)2 + 42 + 32)  = 29 |PB| = (32 + 42 + 92)  = 106 Given that PA.PB = |PA||PB|cos  A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. then cos = 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

( ) + ( ) 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