1. Addition of vectors
(i) Triangle Rule [For vectors with a common point] C B A

2. (ii) Parallelogram Rule [for vectors with same initial point]D C B A

3. (iii) Extensions follow to three or more vectors
p+q+r q p

4. First we need to understand what is meant by the vector – a
Subtraction
First we need to understand what is meant by the vector – a a
– a
a and – a are vectors of the same magnitude, are parallel, but act in opposite senses.

5. A few examples
b – a  a a b b

6. Which vector is represented by p – q ?

7. B
CB = CA + AB
= - AC + AB
= AB – AC C A

8. OP is a position vector of a point P. We usually associate p with OP
Position Vectors
Relative to a fixed point O [origin] the position of a Point P in space is uniquely determined by OP P p
OP is a position vector of a point P. We usually associate p with OP O

9. A very Important result!B
AB = b - a b A a O

10. The Midpoint of AB A M
OM = ½(b + a) a B b O

11. An Important technique
To establish or express the co-linearity of three points [Lie in a straight line]
Choose any two line segments, AB, AC or BC.
For the points to be co-linear AB, AC or BC must lie in the same direction
Example
Given OA = p, OC = q and OB = 2p – q , show that A, B and C are co-linear.
AB = AO + OB
= – p + (2p – q)
= p – q B A
BC = BO + OC
= – (2p – q) + q
= 2q – 2p
= –2(p – q ) = –2AB C
Hence A , B and C are co-linear. O
AB &BC are parallel (even though in opposite directions) and have a common point B

12. Example M, N, P and Q are the mid-points of OA, OB, AC and BC.
OA = a, OB = b, OC = c
(a) Find, in terms of a, b and c expressions for
(i) BC (ii) NQ (iii) MP
(b) What can you deduce about the quadrilateral MNQP? a)
BC = BO + OC
= c – b
(ii) NQ = NB + BQ b
=  c a
(ii) MP = MA + AP c
=  c
MNPQ is a parallelogram as NQ and MP are equal and parallel.

13. The diagram shows quadrilateral OABC. OA = a, OC = c and OB = 2a + c
(a) Find expressions, in terms of a and c, for
(i) AB (ii) CB
(iii) What kind of quadrilateral is OABC?
Give a reason for your answer.
(b) Point P lies on AC and AP =  AC.
(i) Find an expression for OP in terms of a and c.
Write your answer in its simplest form.
(ii) Describe, as fully as possible, the position of P. a)
AB = AO + OB
= a + c
(ii) CB = CO + OB P
= 2a
Trapezium :
CB is parallel to OA. b)
OP = OA + AP
=  a +  c
(ii) OB = 3 x (OP)
They are parallel and have a common point, hence O, P & B are co-linear.

14. Example
OACB is a parallelogram with OA = a and OB = b
M is the midpoint of AC
P is the intersection of OM with AB
Obtain the position vector of M
Given that AP = kAB use the triangle OAP to obtain an expression for OP in terms of a, b and k.
Deduce the position vector P.

15. A C a M P O B b OM = OB + BM = b +  a AP = kAB OP = OA + AP = a + kAB
Example
OACB is a parallelogram with OA = a and OB = b
M is the midpoint of AC
P is the intersection of OM with AB
Obtain the position vector of M
Given that AP = kAB use the triangle OAP to obtain an expression for OP in terms of a, b and k.
Deduce the position vector P.
OM =
OB + BM
= b +  a
AP = kAB
OP = OA + AP
= a + kAB
= a + k(b – a)
= (1 – k)a +kb O C A b a B
(iii) OP = hOM
OP = h(b +  a) M P
a 1 – k =  h
b k = h
Hence 1 =  h h = 
OP =  a +  b