In a parallelogram OABC, vectors `veca, vecb, vecc` are, respectively, tehe position vectors of vertices A, B, C with reference to O as origin. A point E is taken on the side BC which divides it in the ratio `2 : 1`. Also, the line segment AE intersects the line bisecting the angle `angle AOC` internally at point P. If CP when extended meets AB in point F, then
The ratio in which F divides AB is

Options

A. `(2|veca|)/(||veca - 3|vecc||)`
B. `(|veca|)/(||veca | - 3|vecc||)`
C. `(3|veca|)/(||veca| - 3 |vecc||)`
D. `(3|vecc|)/(3|vecc| - |veca|)`

Answer

D

Explanation

Let the position vector of A and C be `veca and vecc` respectively. Therefore,
Position vector of B `= vecb = vec a + vec c" " `(i)
Also Position vector of E `=( vecb + 2vecc)/( 3) = ( veca + 3vecc)/( 3)" " ` (ii)
Now point P lies on angle bisector of `angle AOC`. Thus,
Position vector of point P `= lamda((veca)/(|veca|) + ( vecb)/(|vecb|))" "` (iii)
Also let P divides EA in ration `mu : 1`. Therefore,
Position vector of P
`" "=(mu veca + (veca + 3 vecc)/(3)) /(mu+1) = ((3 mu +1) veca + 3 vecc)/( 3 ( mu +1))" "` (iv)
Comparing (iii) and (iv), we get
`lamda ((veca )/(|veca|) + (vecc)/( |vecc|) ) = (( 3mu + 1) veca + 3vecc)/( 3(mu+1))`
`rArr (lamda ) /(|veca|) = ( 3mu +1)/( 3(mu +1)) and(lamda ) /(|vecc|) = (1)/( mu+1)`
`rArr (3|vecc|- |veca|)/(3|veca|)=mu`
`rArr (lamda)/(|vecc|) = (1)/((3|vecc| - |veca|)/(3|veca|)+1)`
`rArr lamda =( 3|veca||vecc|)/(3|vecc|+ 2|veca|)`
Hence, positive vector of P is `(3 |veca ||vecc|)/(3|vecc|+ 2|veca|)((veca)/(|veca|) + (vecc)/(|vecc|))`
Let F divides AB in ratio `t : 1`, then position vector of F is `(tvecb + veca)/(t+1)`
Now points C, P, F are collinear. Then,
`vec(CF) = mvec(CP)`
`rArr =(t(veca+ vecc))/(t+1) - vecc= m{(3|veca||vecc|)/(3|vecc|+2|veca|)((veca)/(|veca|)+ (vecc)/(|vecc|))- vecc}`
Comparing coefficients we get
`(t)/(t+1) = m(3|vecc|)/(3|vecc| + 2|veca|) and (-1)/(t+1) = m (|veca|- 3|vecc|)/( 3 |vecc| +2|veca|)`
`t = ( 3|vecc|)/( 3|vecc|-|veca|)`