Prove that `vecaxx{vecbxx(veccxxvecd)}=(vecb.vecd)(vecaxxvecc)-(vecb.vecc)(vecaxxvecd)`

Prove that: vecaxx[vecbxx(veccxxveca)]=(veca.vecb)(vecaxxvecc)

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.

Prove that: (vecaxxvecb)xx(veccxxvecd)+(vecaxxvecc)xx(vecd xx vecb)+(vecaxxvecd)xx(vecbxxvecc) = -2[vecb vecc vecd] veca

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

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

If [vecbvecc vecd]=24 and (vecaxxvecb)xx(veccxxvecd)+(vecaxxvecc)xx(vecd xx vecb)+(vecaxxvecd)xx(vecbxxvecc)+kveca=0 then k is equal to …………….

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 vecaxx(vecb+vecc)+vecbxx(vecc+veca)+veccxx(veca+vecb)=0

If veca,vecb,vecc and vecd are the position vectors of the vertices of a cycle quadrilateral ABCD, prove that (|vecaxxvecb+vecb xxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))+(|vecbxxvecc+veccxxvecd+vecd+vecd xx vecb|)/((vecb-vecc).(vecd-vecc))