Correct Answer - `(2:1)`
OA CB is a parallelogram with O is origin . Let
`vec(OA) = vec(a) , vec(OB) = vec(b) , vec(OC) = vec(a) + vec(b)`
`" and " vec(OD) = (vec(a))/(2)`
`vec(CO) " and " vec(BD) ` meets at P.
`:. vec(OP) = (lambda . 0 + 1(vec(a)+vec(b)))/(lambda+1) " " "[along "vec(OC)"]"`
` rArr vec(OP) = (vec(a) + vec(b))/(lambda+1) " "......(i)`
Again `vec(OP) = (mu((vec(a))/(2))+1(vec(b)))/(mu +1) " ""[along " vec(BD)"]"`
`rArr vec(OP) = (mu vec(a) + 2vec(b))/(2(mu +1)) " ".........(ii)`
From Eqs . (i) and (ii)
` (vec(a) +vec(b))/(lambda+1) = (mu vec(a) +2vec(a))/(2(mu+1))rArr (1)/(lambda+1) =(mu)/(2(mu+1)) " and " (1)/(lambda+1) = (1)/(mu+1)`
On solving we get `mu = lambda =2`
Thus required ratio is 2:1