Prove that `veca\'xxvecb\'+vecb\'xxvecc\'+vecc\'xxveca\'=(veca+vecb+vecc)/([veca vecb vecc])`

If veca,vecb, vecc and veca',vecb',vecc' are reciprocal system of vectors, then prove that veca'xxvecb'+vecb'xxvecc'+vecc'xxveca'=(veca+vecb+vecc)/([vecavecbvecc])

If veca, vecb and vecc be any three non coplanar vectors. Then the system of vectors veca\',vecb\' and vecc\' which satisfies veca.veca\'=vecb.vecb\'=vecc.vecc\'=1 veca.vecb\'=veca.veca\'=vecb.veca\'=vecb.vecc\'=vecc.veca\'=vecc.vecb\'=0 is called the reciprocal system to the vectors veca,vecb, and vecc . [veca,vecb,vecc]-(veca\'xxvecb\')+(vecb\'xxvec\')+(vecc\'xxveca\')= (A) veca+vecb+vecc (B) veca+vecb-vecc (C) 2(veca+vecb+vecc) (D) 3(veca\'+vecb\'+vecc\')

Prove that veca\'.(vecb+vecc)+vecb\'.(vecc+veca)+vecc\'.(veca+vecb)=0

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

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

If veca, vecb and vecc are three non-coplanar non-zero vectors, then prove that (veca.veca) vecb xx vecc + (veca.vecb) vecc xx veca + (veca.vecc)veca xx vecb = [vecb vecc veca] veca

If veca, vecb and vecc are three non-coplanar non-zero vectors, then prove that (veca.veca) vecb xx vecc + (veca.vecb) vecc xx veca + (veca.vecc)veca xx vecb = [vecb vecc veca] veca

If veca,vecb,vecc are unity vectors such that vecd=lamdaveca+muvecb+gammavecc then gamma 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 and vecc be any three non coplanar vectors. Then the system of vectors veca\',vecb\' and vecc\' which satisfies veca.veca\'=vecb.vecb\'=vecc.vecc\'=1 veca.vecb\'=veca.veca\'=vecb.veca\'=vecb.vecc\'=vecc.veca\'=vecc.vecb\'=0 is called the reciprocal system to the vectors veca,vecb, and vecc . The value of [veca\' vecb\' vecc\']^-1 is (A) 2[veca vecb vecc] (B) [veca,vecb,vecc] (C) 3[veca vecb vecc] (D) 0

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