Triangle AOB

VECTORS
Vectors is taught both in form two and three. It is one of the key topics that appear each year in KCSE examinations.
In understanding this topic it must be noted that
i) A vector quantity is defined by both direction and magnitude (size)
ii) A change of direction implies a change of sign e.g OA= a but AO = -a
iii) Fractions and ratios are related in that if AR:RB = 1:2 then AR = 1/3 AB

In the triangle OAB below, M is the mid-point of AB and R divides OB in the ratio 1:2

Triangle AOB
a) Given that OA=a and OB=b express the following vectors in terms of a and b
i) AB = AO + OB
= a+b
ii) OM = OA AM (or OB+BM)
= a+ ½ AB (a midpoint M divides AB into two halves)
= (AO+OB)
= a+ ½ (-a+b) nb. AB= -a+b
= a- ½a + ½b
= ½a + ½b

iii) AR = AO+OR
=-a+ 1/3OB (OR:RB = 1:2)
= -a+1/3b

b) Further OX=hOM and AX = tAR where h and t are constants. Express OX in TWO ways hence determine the values of h and t. To do this, we must use the constants h and t. Therefore we have

OX = hOM
OX = OA+AX
= OA+tAR (rmb. AX = tAR
OX = h(1/2a + ½b)
= ½ha + ½hb

OX = OA +tAR
= a+t(-a+1/3b)
= a-ta+1/3tb
= (1-t)a + 1/3tb

The two expressions are
OX= ½ha + ½hb
OX = (1-t)a+1/3tb
Forming simultaneous equations
½ha = 1-ta
½h = 1-t (cancel a) – I
½hb = 1/3tb
½h = 1/3t (cancel b) – II
Making t the subject in equation I
t = (1- ½h)
Substituting t in equation II
½h = 1/3(1- ½h)
½h = 1/3-1/6h
½h + 1/6h = 1/3
2/3h = 1/3
h= 1/3×3/2
= ½
therefore t= 1- ½h
= 1- ½ ( ½)
=1- ¼
= ¾

c) Hence express OX in terms of a and b
Rmb OX= ½ha + ½hb
but h = ½
OX = ½ (½ )a + ½ (½ )b
= ¼ b + ¼b