Vectors `vecx,vecy,vecz` each of magnitude `sqrt(2)` make angles of `60^0` with each other. If `vecx xx(vecy xx vecz) = veca, vecy xx( vecz xx vecx)=vecb` and `vecx xx vecy=vecc,` find `vecx, vecy, vecz` in terms of `veca, vecb` and `vecc`.

Options

A. `gamma/(|veca xxvecb|^(2))[veca+vecbxx(vecaxxvecb)]`
B. `gamma/(|veca xxvecb|^(2))[veca+vecb-vecaxx(vecaxxvecb)]`
C. `gamma/(|veca xxvecb|^(2))[veca+vecb+vecaxx(vecaxxvecb)]`
D. none of these

Answer

c

Explanation

Given
`vecx xx vecy=veca` (i)
`vecy xx vecz=vecb` (ii)
`vecx.vecb =gamma` (iii)
`vecx.vecy=1` (iv)
`vecy .vecz=1` (v)
from (ii) , `vecx.(vecy xx vecz) =vecx .vecb=gamma Rightarrow [vecx vecy vecz] = gamma`
From (i) and (ii) , `(vecx xx vecy) xx (vecy xx vecz) =veca xx vecb`
` [vecx vecy vecz]-[vecy vecy vecz] = veca xx vecb`
` or vecy = (veca xx vecb)/ gamma`
Also from (i), we get ` (vecx xx vecy) xx vecy =veca xx vecy `
` or (vecx.vecy) vecy- (vecy.vecy)vecx =veca xx vecy`
`or vecx = (1//|vecy|^(2))(vecy - veca xx vecy)`
`gamma^(2)/(|vecaxx vecb|^(2))[(vecaxxvecb)/gamma-(vecaxx(vecaxxvecb))/gamma]`
`gamma/(|veca xx vecb|^(2))[veca xx vecb-vecaxx (veca xx vecb)]`
Also from (ii) , `(vecy xx vecz)xxvecy = vecb xx vecy`
`or |vecy|^(2)vecz-(vecz.vecy)vecy= vecbxxvecy`
`or vecz= 1/(|vecy|^(2))[vecy+vecb xx vecy]`
`gamma/(|vecaxxvecb|^(2))[vecaxxvecb +vecb xx (veca xx vecb)]`