Vector ` vec a` in the plane of ` vec b=2 hat i+ hat ja n d vec c= hat i- hat j+ hat k` is such that it is equally inclined to ` vec ba n d vec d` where ` vec d= hat j+2 hat kdot` The value of` vec a` is `( hat i+ hat j+ hat k)/(sqrt(2))` b. `( hat i- hat j+ hat k)/(sqrt(3))` c. `(2 hat i+ hat j)/(sqrt(5))` d. `(2 hat i+ hat j)/(sqrt(5))`

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`