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

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