Prove that `[lambda veca +mu vecb" "vecc" "vecd]=lambda [veca" "vecc" "vecd]+mu[vecb" "vecc" "vecd]`.

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

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

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

If [(2veca+4vecb,vecc,vecd)]=lamda[(veca,vecc,vecd)]+mu[(vecb,vecc,vecd)] , then lamda+mu=

If vectors, vecb, vecc and vecd are not coplanar, the prove that vector (veca xx vecb) xx (vecc xx vecd) + ( veca xx vecc) xx (vecd xx vecb) + (veca xx vecd) xx (vecb xx vecc) is parallel to veca .

Let veca , vecb and vecc be three non-zero vectors such that veca + vecb + vecc = vec0 and lambda vecb xx veca + vecbxxvecc + vecc xx veca = vec0 then find the value of lambda .

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

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

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])

Show that the points whose position vectors are veca,vecb,vecc,vecd will be coplanar if [veca vecb vecc]-[veca vecb vecd]+[veca vecc vecd]-[vecb vecc vecd]=0