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
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

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
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
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 + =

10. Wee Diagrams and Lettersu 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
AD =
3BC B C v
Find CD in terms of u and v CD
= CB
+ BA
+ AD
= -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
246 = + +
Find ET
= 7i + 6j + 14k

13. Questions
Cunningly Acquired!

16. 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

17. [ ], [ ]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
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EXIT

18. [ ], [ ]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
= °
APB
Go to Vectors Menu
EXIT

19. [ ], [ ]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
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20. [ ] [ ] [ ], [ ]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 =
Markers Comments
= 37
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21. [ ], [ ]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) ) 
= 29
|PB| = ( ) 
= 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
= °
Markers Comments
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22. ( ) + ( ) 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