Final non-C maths revision thread

if theres another maths revision thread its quite far back so I thought id start one.

My question (higher):

OABC is a parallelogram

P is the point on AC such that AP= 2/3 AC

O->A= 6a

O->C= 6c


a) Find vector O->P
give answer in terms of a or c

b) the midpoint of CB is M
prove that OPM is a straight line

Ok, here we go,

If ive drawn the parellelogram right and A ->C is the diagonal of the parallelogram, then:

(Draw the diagram to understand this)
a) To find O ->P we need to first find A->P

A->P is 2/3 A->C [A]
Thus, A->P = 2/3(6a+6c) = 4a+4c

O -> P = 4a+4c-6a = 4c-2a or 2(2c-a)

b) O->P = 2(2c-a)
P ->M = - 4a - 4c +6c + 3a
= 2c - a

Thus we can see that
1) Both vectors are multiples of the vector 2c-a which means they are parellel
2) They both share the point 'P' thus O,P and M must lie on a straight line.

There we go Hope that helps!