If `vecA=(vecbxxvecc)/([vecb vecc veca]), vecB=(veccxxveca)/([vecc veca vecb]), vecC=(vecaxxvecb)/([veca vecb vecc])` find `[vecA vecB vecC]`

Let veca,vecb,vecc be non coplanar vectors and vecp= (vecbxxvecc)/([veca vecb vecc]), vecq= (veccxxveca)/([veca vecb vecc]), vecr= (vecaxxvecb)/([veca vecb vecc]) . What is the vaue of (veca-vecb-vecc).vecp+(vecb-vecc-veca).vecq+(vecc-veca-vecb).vecr? (A) 0 (B) -3 (C) 3 (D) -9

If vecP = (vecbxxvecc)/([vecavecbvecc]).vecq=(veccxxveca)/([veca vecb vecc])and vecr = (vecaxxvecb)/([veca vecbvecc]), " where " veca,vecb and vecc are three non- coplanar vectors then the value of the expression (veca + vecb + vecc ). (vecq+ vecq+vecr) is

If vecp=(vecbxxvecc)/([(veca,vecb,vecc)]),vecq=(veccxxveca)/([(veca,vecb,vecc)]),vecr=(vecaxxvecb)/([(veca,vecb,vecb)]) where veca,vecb,vecc are three non-coplanar vectors, then the value of the expression (veca+vecb+vecc).(vecp+vecq+vecr) is

If is given that vecx= (vecbxxvecc)/([veca,vecb,vecc]), vecy=(veccxxveca)/[(veca,vecb,vecc)], vecz=(vecaxxvecb)/[(veca,vecb,vecc)] where veca,vecb,vecc are non coplanar vectors. Find the value of vecx.(veca+vecb)+vecy.(vecc+vecb)+vecz(vecc+veca)

If veca,vecb,vecc are unity vectors such that vecd=lamdaveca+muvecb+gammavecc then lambda is equal to (A) ([veca vecb vecc])/([vecb veca vecc]) (B) ([vecb vecc vecd])/([vecb vecc veca]) (C) ([vecb vecd vecc])/([veca vecb vecc]) (D) ([vecc vecb vecd])/([veca vecb vecc])

If veca, vecb, vecc are non-coplanar vectors, then (veca.(vecb xx vecc))/(vecb.(vecc xx veca)) + (vecb.(vecc xx veca))/(vecc.(veca xx vecb)) +(vecc.(vecb xx veca))/(veca. (vecb xx vecc)) is equal to:

Let veca,vecb and vecc be a set of non- coplanar vectors and veca'vecb' and vecc' be its reciprocal set. prove that veca=(vecb'xxvecc')/([veca'vecb'vecc']),vecb=(vecc'xxveca')/([veca'vecb'vecc'])andvecc=(veca'xxvecb')/([veca'vecb'vecc'])

Prove that [veca+vecb vecb+vecc vecc+veca]=2[veca vecb vecc]

If veca, vecb and vecc are three non-coplanar vectors, then find the value of (veca.(vecbxxvecc))/(vecb.(veccxxveca))+(vecb.(veccxxveca))/(vecc.(vecaxxvecb))+(vecc.(vecbxxveca))/(veca.(vecbxxvecc))

[ veca + vecb vecb + vecc vecc + veca ]=[ veca vecb vecc ] , then