` vec a , vec b ,a n d vec c` are unimodular and coplanar. A unit vector ` vec d` is perpendicular to then. If `( vec axx vec b)xx( vec cxx vec d)=1/6 hat i-1/3 hat j+1/3 hat k ,` and the angel between ` vec aa n d vec b` is `30^0,t h e n vec c` is `( hat i-2 hat j+2 hat k)//3` b.`(- hat i+2 hat j-2 hat k)//3` c. `(2 hat i+2 hat j- hat k)//3` d. `(-2 hat i-2 hat j+ hat k)//3`

Options

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

Answer

a,b

Explanation

Given `1/6hati -1/3 hatj + 1/3hatk = (veca xx vecb) xx (veccxxvecd)`
`[ veca vecb vecd]vecc - [veca vecb vecc] vecd`
`[veca vecb vecd]vecc`
`[ veca, vecb and vecc` are coplanar]
`[veca vecb vecd] = (veca xx vecb).vecd`
` |veca xx vecb| |vecd| cos theta`
`(therefore vecd bot veca,vecd,bot vecb, therefore vecd||vecaxxvecb)`
`ab sin 30^(@) .1. (+-1) ( theta = 0 or pi) `
`1.1. 1/2 .1(+-1)= +- 1/2`
form (i) m we have
`vecc= +- (1/3 hati -2/3hatj +2/3hatk) =+- (hati -2hatj + 2hatk)/3`