Show that: `(veca-vecd)xx(vecb-vecc)+(vecb-vecd)xx(vecc-veca)+(vecc-vecd)xx(veca-vecb)` is independent of `vecd`.

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.

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(vecaxxvecc)].vecd=[veca vecb vecc](veca.vecd)

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

If vecaxxvecb=veccxxvecd and vecaxxvecc=vecbxxvecd then (A) (veca-vecd)=lamda(vecb-vecc) (B) veca+vecd=lamda(vecb+vecc) (C) (veca-vecb)=lamda(vecc+vecd) (D) none of these

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))

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))

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).(vecxxvecd) is equal to (A) (vecaxxvecd).(vecbxxvecc) (B) (vecbxxveca).(veccxxvecd) (C) (vecdxxvecc).(vecbxxveca0 (D) none of these

If veca,vecb,vecd,vecd be vectors such that [vecavecbvecc]=2 and (vecaxxvecb) xx (vecc xx vecd)+(vecb xx vecc) xx (vecc xx vecd) + (vecc xx veca) xx (vecb xx vecd)=-muvecd Then the value of mu is

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.(vecxx(veccxxvecd)) (B) ((vecaxxvecc)xxvecb).vecd (C) (vecaxxvecb).(vecdxxvecc) (D) none of these