Prove that: `[veca+vecb " "vecb+vecc " "vecc+veca]=2[veca" " vecb" " vecc]`

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

For any four vectors, prove that ( veca × vecb )×( vecc × vecd )=[ veca vecc vecd ] vecb −[ vecb vecc vecd ] veca

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)

for any three vectors, veca, vecb and vecc , (veca-vecb) . (vecb -vecc) xx (vecc -veca) = 2 veca.vecb xx vecc .

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

given that veca. vecb = veca.vecc, veca xx vecb= veca xx vecc and veca is not a zero vector. Show that vecb=vecc .

If veca , vecb and vecc are three vectors such that vecaxx vecb =vecc, vecb xx vecc= veca, vecc xx veca =vecb then prove that |veca|= |vecb|=|vecc|

Prove that veca. [(vecb + vecc) xx (veca + 3 vecb + 4 vecc) ]= [{:(veca,vecb,vecc):}]

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 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])