Question

In a `triangle OAB`,E is the mid point of OB and D is the point on AB such that `AD:DB=2:1` If OD and AE intersect at P then determine the ratio of `OP: PD` using vector methods

Answer 1

Correct Answer - `(3:2)`
Let O be origin and `vec(A) = vec(a) , vec(OB) = vec(b)`
`vec(OE) = (vec(b))/(2)" ""[Since E being mid- point of " vec(OB)"]"`

`vec(OD) =(vec(a)"." 1 + vec(b) "."2)/(1+2)`
(since D divides `vec(AB) ` in the ratio of 2:1)
`rArr` Equation of `vec(OD) " is " vec(r ) = t ((vec(a) + 2vec(b))/(3))`
and equation of `vec(AE) " is " vec(r ) = vec(a ) + s ((vec(b))/(2)-a)`
if `vec(OD) " and " vec(AE)` intersect at P then there must be some `vec(r )` for which they are equal .
`rArr t((vec(a) + 2vec(b))/(3))= vec(a) + s ((vec(b))/(2)-vec(a))`
`rArr (t)/(3)=1 - s "and " (2t)/(3) = (s)/(2)`
` rArr t= (3)/(5) " and " s=(4)/(5)`
`:. " Point P is " (vec(a) + 2 vec(b))/(5)`
Since P divides `vec(OD)` in the ratio of `lambda :1`
`:. (lambda((vec(a) +2vec(b))/(3))+1.0)/(lambda+1) = ((vec(a) + 2vec(b))/(5))`
from Eqs. (i) and (ii)
`(lambda)/(3(lambda+1)) = (1)/(5)`
`rArr 5lambda =3lambda +3`
`rArr lambda = (3)/(2)`
`:. (OP)/(PD) = (3)/(2)`