If `vecd=vecaxxvecb+vecbxxvecc+vecc+veccxxveca` is a non- zero vector and `|(vecd.vecc)(vecaxxvecb)+(vecd.veca)(vecbxxvecc)+(vecd.vecb) (veccxxveca) =0` then

If vecd=vecaxxvecb+vecbxxvecc+veccxxveca is a on zero vector and |(vecd.vecc)(vecaxxvecb)+(vecd.veca)(vecbxxvecc)+(vecd.vecb)(veccxxveca)|=0 then (A) |veca|+|vecb|+|vecc|=|vecd| (B) |veca|=|vecb|=|vecc| (C) veca,vecb,vecc are coplanar (D) veca+vecc=vec(2b)

Prove that: [(vecaxxvecb)xx(vecaxxvecc)].vecd=[veca vecb vecc](veca.vecd)

If veca, vecb, vecc are three non coplanar, non zero vectors then (veca.veca)(vecbxxvecc)+(veca.vecb)(veccxxveca)+(veca.vecc)(vecaxxvecb) is equal to

If veca, vecb, vecc are non-coplanar non-zero vectors, then (vecaxxvecb)xx(vecaxxvecc)+(vecbxxvecc)xx(vecbxxveca)+(veccxxveca)xx(veccxxvecb) is equal to

If vecA, vecB, vecC are non-coplanar vectors then (vecA.vecBxxvecC)/(vecCxxvecA.vecB)+(vecB.vecAxxvecC)/(vecC.vecAxxvecB)=

Prove that (vecbxxvecc)xx(veccxxveca)=[veca vecb vecc]vecc

Prove that (vecbxxvecc)xx(veccxxveca)=[veca vecb vecc]vecc

If veca, vecba and vecc are non- coplanar vecotrs, then prove that |(veca.vecd)(vecbxxvecc)+(vecb.vecd)(veccxxveca)+(vecc.vecd)(vecaxxvecb) is independent of vecd where vecd is a unit vector.

For vectors veca,vecb,vecc,vecd, vecaxx(vecbxxvecc)=(veca.vecc)vecb-(veca.vecb)vecc and (vecaxxvecb).(veccxxvecd)=(veca.vecc)(vecb.vecd)-(veca.vecd)(vecb.vecc) Now answer the following question: {(vecaxxvecb).xxvecc}.vecd would be equal to (A) veca.(vecbxx(veccxxvecd)) (B) ((vecaxxvecc)xxvecb).vecd (C) (vecaxxvecb).(veccxxvecd) (D) none of these

For vectors veca,vecb,vecc,vecd, vecaxx(vecbxxvecc)=(veca.vecc)vecb-(veca.vecb)vecc and (vecaxxvecb).(veccxxvecd)=(veca.vecc)(vecb.vecd)-(veca.vecd)(vecb.vecc) Now answer the following question: (vecaxxvecb).(veccxxvecd) is equal to (A) veca.(vecbxx(veccxxvecd)) (B) |veca|(vecb.(veccxxvecd)) (C) |vecaxxvecb|.|veccxxvecd| (D) none of these