In a triangle `A B C ,Da n dE` are points on `B Ca n dA C ,` respectivley, such that `B D=2D Ca n dA E=3E Cdot` Let `P` be the point of intersection of `A Da n dB Edot` Find `B P//P E` using the vector method.

Answer

`8:3`

Explanation

Let the vertices of the triangle be `A (vec0), B(vecb) and C(vecc)`.
Given that D divides BC in the ratio `2:1`.
Therefore, position vector of D is `(vecb + 2 vecc)/(3)`.

E divides AC in the ratio `3:1`.
Therefore, position vector of E is `(vec0+ 3 vecc)/(4) = (3vecc)/(4)`.
Let point of intersection P of AD and BE divide BE in the ratio `1:k` and AD in the ratio `1:m`. Then position vectors of P in these two cases are `(kvecb+ 1(3vecc//4))/(k+1) and (mvec0 + m ((vecb+ 2 vecc)//3))/(m+1)`, respectively.
Equating the position vectors of P in these two cases, we get
`" "(kvecb)/(k+1) + ( 3vecc)/(4(k+1))= (mvecb)/(3(m+1)) + (2mvecc)/( 3(m+1))`
`rArr (k)/(k+1) = (m)/(3(m+1)) and (3)/(4(k+1)) = (2m)/(3(m+1))`
Dividing, we have `(4k)/(3) = (1)/(2) or k = (3)/(8)`
Required ratio is `8 : 3`.