7. Applying to Vectors
A (2,5,3) B (5,8,9)
AP = PB2
use position
vector rule
p – a = 2(b – p)
p – a = 2b – 2p
3p = 2b + a
=2 +
( )
=
( )12
21
21
3p
p =
( )4
7
7P = (4,7,7)
( ) 2
5
3
+
5
8
9
→ →
8. position vectorsp – a = 2(b – p)
p – a = 2b – 2p
( )
3p = 2b + a
= +( )9
2
8
=
( )
21
0
18
3p p =
( )
7
0
6P = (7,0,6)
3
-4
2
AP = PB2
→ →
A BP2 1
A (3,-4,2) B (9,2,8)
P divides AB in ratio 2:1
2
9. Vectors, Ratios
and Unknown Points
• Form a wee equation
(diagram v helpful)
• Use position vector rule
• Make required vector the subject
• Remember to change back to coordinates!
10. A (2,5,3) B (6,5,7)
P divides AB in ratio 1:3
A BP1 3
AP = PB3
→ →
11. position vectors3(p – a) = b – p
3p – 3a = b – p
( )
4p = b + 3a
= +3( )6
5
7
=
( )
12
20
16
4p p =
( )3
5
4
P = (3,5,4)
2
5
3
AP = PB3
→ →
12. position vectors3(p – a) = 2(b – p)
3p – 3a = 2b – 2p
( )
5p = 2b + 3a
= +3( )68
9
=
( )
15
25
30
5p p =
( )
3
5
6P = (3,5,6)
1
3
4
AP = PB2
→ →
A BP2
A (1,3,4) B (6,8,9)
P divides AB in ratio 2:3
2
3
3
Key
Question