Let OABCD be a pentagon in which the sides OA and CB are parallel and the sides OD and AB are parallel as shown in figure. Also, OA:CB=2:1 and OD:AB=1:3. if the diagonals OC and AD meet at x, find OX:OC.

Options

A. `5//2`
B. `6`
C. `7//3`
D. `4`

Answer

B

Explanation

Let the position vectors of A, B, C and D be `veca, vecb, vecc and vecd`, respectively. Then,
`" "OA: CB = 2 : 1`
`rArr vec(OA) = 2vec(CB)`
`rArr veca = 2 (vecb - vecc)" "` (i)
and `OD : AB = 1: 3`
`" " 3 vec(OD) = vec(AB)`
`rArr 3 vecd = ( vecb - veca) = vecb - 2 ( vecb - vecc)" "`[ using(i)]
`" " = - vecb + 2vecc" "` (ii)
Let `OX : XC = lamda : 1 and AX : XD= mu : 1`
Now, X divides OC in the ratio `lamda : 1`. Therefore,
`P.V of X = (lamda vecc)/(lamda +1)" "` (iii)
X also divides AD in the ratio `mu: 1`. Therefore,
P.V. of X = `(muvecd + veca)/(mu +1)" "` (iv)
From (iii) and (iv), we get
`(lamda vecc)/(lamda + 1) = ( mu vecd + veca)/( mu + 1)`
or `((lamda)/(lamda + 1)) vecc = ((mu)/(mu+ 1))vecd + ((1)/(mu+1))veca`
or `((lamda )/(lamda+1))vecc = ((mu)/(mu+1))((-vecb + 2 vecc)/(3)) + ((1)/(mu+1)) 2(vecb - vecc)`
`" "` [using (i) and (ii)]
or `" "((lamda )/(lamda + 1))vecc = ((6-mu)/( 3(mu+1))) vecb + (( 2 mu )/(3(mu+1)) - (2)/(mu +1))vecc`
or `((lamda)/(lamda + 1)) vecc = ((6-mu)/(3(mu+1)))vecb + (( 2mu -6)/(3(mu+1)))vecc`
or ` ((6 -mu)/(3(mu+1)))vecb+ ((2mu-6)/(3(mu+1))- (lamda )/(lamda +1)) vecc = vec0`
or `" "( 6-mu)/(3(mu+1))=0 and ( 2mu-6)/( 3( mu +1)) - (lamda) /(lamda +1)=0`
`" "`( as `vecb and vecc` are non-collinear )
or `" " mu = 6, lamda = (2)/(5)`
Hence, `OX : XC = 2 : 5`