`veca ,vecb and vecc` are unimodular and coplanar. A unit vector `vecd` is perpendicualt to them , `(veca xx vecb) xx (vecc xx vecd) = 1/6 hati - 1/3 hatj + 1/3 hatk` , and the angle between `veca and vecb is 30^(@)` then `vecc` is

Options

A. `1/3(-2hati-2hatjhatk)`
B. `+-1/3(-hati-2hatj+2hatk)`
C. `1/3(2hati+hatj-hatk)`
D. `+-1/3(-hati+2hatj-2hatk)`

Answer

D

Explanation

Cleary `(vecaxxvecb)xx(veccxxvecd)`
`={(vecaxxvecb).vecd}vecc-{(vecaxxvecb).vecc}vecd`
`=[(veca, vecb, vecd)]vecc-[(veca, vecb, vecc)]vecd`
`={(vecaxxvecb).vecd}vecc-0vecd`
`=|vecaxxvecb||vecd|vecc [ :' vecaxxvecb||vecd, :. (vecaxxvecb).vecd=+-|vecaxxvecb||vecd|]`
`(+-|veca||vecb|sin 30^(@)) [ :' |vecd|=1]`
`=+-1/2vecc`
`:.1/6hati-1/3hatj+1/3hatk=+-1/2veccimpliesvecc=+-1/3(hati-2hatj+2hatk)`