A unit vector `veca` in the plane of `vecb=2hati+hatj` and `vecc=hati-hatj+hatk` is such that angle between `veca` and `vecd` where `vecd=vecj+2veck` is

Options

A. `(hati+hatj+hatk)/sqrt3`
B. `(hati-hatj+hatk)/sqrt3`
C. `( 2 hati + hatj)/sqrt5`
D. `( 2 hati + hatj)/sqrt5`

Answer

b

Explanation

Let ` veca = lambdavecb + mu vecc`
`veca` is equally inclined to `vecb and vecd`, where,
`vecd = hatj + 2hatk`
` Rightarrow (veca.vecb)/(ab) = (veca.vecb)/(ab)`
`((lambdavecb+muvecc).vecb)/b= ((lambdavecb+muvecc).vecd)/d`
`([lamda(2hati+hatj)+mu(hati - hatj + hatk)].(hatj +2hatk))/sqrt5`
` lambda ( 4+1) = mu ( 2-1) = lambda(1) + mu(-1+2)`
`or 4 lamda = 0 .,i.e lamda =0`
` hata (hati -hatj +hatk)/sqrt3`