### Vectors II (Q 4,5) solved

```Addition of vectors
(i) Triangle Rule [For vectors with a common point]
C
AB  BC  AC
B
A
(ii) Parallelogram Rule [for vectors with same initial point]
D
C
AB  AC  AD
B
A
(iii) Extensions follow to three or more vectors
r
p+q+r
q
p
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.
A few examples

a
b
b–a
a
b
q
p
p–q
–q
Which vector is represented
by p – q ?
B
CB = CA + AB
= - AC + AB
= AB – AC
C
A
Position Vectors
Relative to a fixed point O [origin] the position of a Point P in space is
uniquely determined by OP
P
p
O
OP is a position vector of a point P.
We usually associate p with OP
A very Important result!
B
AB = b - a
b
A
a
O
The Midpoint of AB
A
M
OM = ½(b + a)
a
B
b
O
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.
A
AB = AO + OB
BC = BO + OC
= – p + (2p – q)
= – (2p – q) + q
=p–q
=
B
C
2q – 2p
= –2(p – q ) = –2AB
O
AB &BC are parallel (even though in opposite directions)
and have a common point B
Hence A , B and C are co-linear.
Example
M, N, P and Q are the mid-points of OA, OB, AC and BC.
B
OA = a, OB = b, OC = c
a)
(i) BC = BO + OC
= c–b
A
(ii) NQ = NB + BQ
N
M
a
Q
b
= c
P
(ii) MP = MA + AP
O
c
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?
= c
MNPQ is a parallelogram as NQ and MP are equal and parallel.
The diagram shows quadrilateral OABC.
OA = a, OC = c and OB = 2a + c
B
C
Not drawn accurately
a)
(i) AB = AO + OB
= a+c
c
2a + c
(ii) CB = CO + OB
P
= 2a
O
a
A
b)
(a) Find expressions, in terms of a and c, for
(i) AB (ii) CB
(iii) What kind of quadrilateral is OABC?
(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.
(iii) Trapezium :
CB is parallel to OA.
(i) OP = OA + AP
=a +c
(ii) OB = 3 x (OP)
They are parallel and
have a common point,
hence O, P & B are colinear.
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
(i) Obtain the position vector of M
(ii) Given that AP = kAB use the triangle OAP to obtain an
expression for OP in terms of a, b and k.
(iii) Deduce the position vector P.
Example
(i) OM = OB + BM
OACB is a parallelogram with OA = a and OB = b
= b +a
M is the midpoint of AC
P is the intersection of OM with AB
(i)
Obtain the position vector of M
(ii) Given that AP = kAB use the triangle OAP to obtain
an expression for OP in terms of a, b and k.
(iii) Deduce the position vector P.
A
C
(ii) AP = kAB
OP = OA + AP
= a + kAB
= a + k(b – a)
= (1 – k)a +kb
(iii) OP = hOM
OP = h(b +  a)
a
O
M
P
b
B
a
1–k=h
b
k=h
Hence 1 =  h h = 
OP =  a +  b
```