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

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

Let veca,vecb, vecc be any three vectors, Statement 1: [(veca+vecb, vecb+vecc,vecc+veca)]=2[(veca, vecb, vecc)] Statement 2: [(vecaxxvecb, vecbxxvecc, veccxxveca)]=[(veca, vecb, vecc)]^(2)

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])

Show that [veca vecb vecc]\^2=|(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc)|

If the vectors veca, vecb, and vecc are coplanar show that |(veca,vecb,vecc),(veca.veca, veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc)|=0

If vector veca,vecb,vecc are coplanar show that |(veca,vecb,vecc),(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc)|

If veca,vecb,vecc are coplanar vectors , then show that |{:(veca,vecb,vecc),(veca*veca,veca*vecb,veca*vecc),(vecb*veca,vecb*vecb,vecb*vecc):}|=vec0

veca,vecb,vecc are three vectors such that vecaxxvecb=vecc, vecbxxvecc=veca. Prove that veca,vecb,vecc are mutually at righat angles and |vecb|=1, |vecc|=|veca| .

If [veca xx vecb vecb xx vecc vecc xx veca]=lambda[veca vecb vecc^(2)] , then lambda is equal to