Vectors

1. Block 3
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!
A
B
C
D
E
AE = AB + BC + CD + DE

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

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

7. Diversions
Sometimes need
alternative route!
A
B
C
D
E
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
= DE + EF
= 3i – 4j – 5k
EF

9. Wee Diagrams and Components
A
B
C
D
ABCD is a parallelogram
TT is mid point of DC
AB =
8
4
6( )
BC =
3
5
1
( )
AT
Info Given
= DC
= AD
= AD + DT
= AD + ½ DC
3
5
1( )
4
2
3
( )+
7
7
4
( )=
Find AT

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

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

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

13. Key Question
A
B
C
D
ABCD is a parallelogram
T
T is mid point of BC
AB =
8
4
6( )
BC =
4
6
2
( )
DT
= DC
= DC + CT
= DC + ½ CB
8
4
6
( )
-2
-3
-1( )+
6
1
5( )=
Find DT

14. Questions
Cunningly Acquired!

17. Hint
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VABCD is a pyramid with rectangular base ABCD.
The vectors are given by
Express in component form.
, andAB AD AV
uuur uuur uuur
8 2 2AB = + +
uuur
i j k 2 10 2AD = − + −
uuur
i j k
7 7AV = + +
uuur
i j k
CV
uuur
AC CV AV+ =
uuur uuur uuur
CV AV AC→ = −
uuur uuur uuur
BC AD=
uuur uuur
AB BC AC+ =
uuur uuur uuur
Ttriangle rule ∆ ACV Re-arrange
Triangle rule ∆ ABC also
( )CV AV AB AD→ = − +
uuur uuur uuur uuur 1 8 2
7 2 10
7 2 2
CV
−     
 ÷  ÷  ÷
→ = − − ÷  ÷  ÷
 ÷  ÷  ÷−     
uuur
5
5
7
CV
− 
 ÷
→ = − ÷
 ÷
 
uuur
9 5 7CV = − +→ −
uuur
i j k

18. VECTORS: Question 3
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Reveal answer only
EXIT
P
Q
R
S
T
U V
W
A
B
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by the
vectors
→ → →
[ ],
4
2
0
[ ]and
-2
4
0
[ ]resp.
0
0
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.
→ →

19. VECTORS: Question 3
Go to full solution
Go to Marker’s Comments
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Reveal answer only
EXIT
P
Q
R
S
T
U V
W
A
B
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by the
vectors
→ → →
[ ],
4
2
0
[ ]and
-2
4
0
[ ]resp.
0
0
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.
→ →
|PA|
→
= √29
|PB|
→
= √106
= 48.1°APB

20. Markers Comments
Begin Solution
Continue Solution
Question 3
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PA =
→
PS + SA =
→ →
PS + 1
/3ST
→→
PS + 1
/3PW
→→
=
[ ]
-2
4
0
[ ] =
0
0
3
+ [ ]
-2
4
3
=
PB =
→
PQ + QV + VB
→ → →
PQ + PW + 1
/2PS
→ → →
=
[ ]
4
2
0
[ ]+
0
0
9
+ [ ]=
-1
2
0
[ ]
3
4
9
=
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by
vectors
→ → →
[ ],
4
2
0
[ ]and
-2
4
0
[ ]resp.
0
0
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.
→ →

21. Markers Comments
Begin Solution
Continue Solution
Question 3
Vectors Menu
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PA =
→
[ ]
-2
4
3
PB =
→
[ ]
3
4
9PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by
vectors
→ → →
[ ],
4
2
0
[ ]and
-2
4
0
[ ]resp.
0
0
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.
→ →
(b) Let angle APB = θ
θ
A
P
B
ie
PA .
→
PB =
→
[ ]
-2
4
3
[ ]
3
4
9
.
= (-2 X 3) + (4 X 4) + (3 X 9)
= -6 + 16 + 27
= 37

22. Markers Comments
Begin Solution
Continue Solution
Question 3
Vectors Menu
Back to Home
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by
vectors
→ → →
[ ],
4
2
0
[ ]and
-2
4
0
[ ]resp.
0
0
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.
→ →
PA .
→
PB =
→
37
|PA| = √((-2)2
+ 42
+ 32
)
→
= √29
|PB| = √(32
+ 42
+ 92
)
→
= √106
Given that PA.PB = |PA||PB|cosθ
→ →
then cosθ =
PA.PB
→ →
|PA||PB|
=
37
√29 √106
so θ = cos-1
(37 ÷ √29 ÷ √106)
= 48.1°

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