Let `veca = 2i + j+k, vecb = i+ 2j -k and a` unit vector `vecc` be coplanar. If `vecc` is pependicular to `veca`. Then `vecc` is

Options

A. `1/sqrt2(-j+k)`
B. `1/sqrt3(i-j-k)`
C. `1/sqrt5(i-2j)`
D. `1/sqrt3(i-j-k)`

Answer

a

Explanation

As `vecc` is coplanar with `veca and vecb` we take
`vecc = alpha veca + betavecb`
where `alpha and beta` are scalars.
As `vecc` is perpendicular to `veca` , using (i), we get
`0 = alphaveca.veca alpha +betavecb.veca`
`or 0 =alpha (6) +beta(2+2-1) =3 (2alpha+beta) `
`or beta = -2alpha`
Thus `vecc=alpha(veca -2vecb)=alpha(-3j+3k)=3alpha(-j+k)`
`or |vecc|^(2)=18alpha^(2)`
`or 1=18alpha^(2)`
`or alpha= +- 1/(3sqrt2)`
`vecc =+- 1/sqrt2 (-j+k)`