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In the DeltaOAB,M is the mid-point of AB...

In the `DeltaOAB,M` is the mid-point of AB,C is a point on OM, such that 2OC=CM. X is a point on the side OB such that OX=2XB. The line XC is produced to meet OA in Y. then, `(OY)/(YA)` is equal to

A

`1/3`

B

`2/7`

C

`3/2`

D

`2/5`

Text Solution

Verified by Experts

The correct Answer is:
B

`bar(OA)=bara,bar(OB)=barb`
`therefore vec(OM)=(veca+vecb)/(2)`
`therefore vec(OA) = (veca+vecb)/(6)`
`vec(OX)=2/3vecb`
Let `(vec(OY))/(vec(YA))=lambda`
`therefore vec(OY) = lambda/(lambda+1)veca`
Now points Y,C an X are collinear
`therefore vec(YC) = mvec(CX)`
`therefore (veca+vecb)/(6) -lambda/(lambda+1)veca=m(2vecb)/(3) -m(veca+vecb)/(6)`
Comparing coefficients of `veca` and `vecb`
`therefore 1/6-lambda/(lambda+1)=-m/6` and `1/6=(2m)/(3)-m/6`
`therefore m=1/3` and `lambda=2/7`
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