The position verctors of the points A and B with respect of O are `2hati+2hatj+hatk and 2hati+4hatj+4hatk`, the length of the internal bisector of `angleBOA` of `DeltaAOB` is

Options

A. `(sqrt(136))/(9)`
B. `(sqrt(136))/(3)`
C. `(20)/(3)`
D. `(sqrt(217))/(9)`

Answer

B

Explanation

`|OA|=sqrt(4+4+1)=3`
and `|OB|=sqrt(4+16+16)=6`
`therefore` Required vector`=lamda(OA+OB)`
`=lamda[(1)/(3)(2hati+2hatj+hatk)+(1)/(6)(2hati+4hatj+4hatk)]`
`(lamda)/(3)(3hati+4hatj+3hatk)`
`therefore` Length of vector
`=(lamda)/(3)sqrt(9+16+9)+(lamda)/(3)sqrt(34)`
Take `lamda=2`
Required length of a vector is `(sqrt(136))/(3)`