The condition for equations `vecrxxveca = vecb and vecr xx vecc = vecd` to be consistent is

Solve the vector equation vecr xx vecb = veca xx vecb, vecr.vecc = 0 provided that vecc is not perpendicular to vecb

If veca, vecb, vecc and vecd ar distinct vectors such that veca xx vecc = vecb xx vecd and veca xx vecb = vecc xx vecd . Prove that (veca-vecd).(vecc-vecb)ne 0, i.e., veca.vecb + vecd.vecc nevecd.vecb + veca.vecc.

If veca,vecb,vecc,vecd are four distinct vectors satisfying the conditions vecaxxvecb=veccxxvecd and vecaxxvecc=vecbxxd then prove that veca.vecb+vecc.vecd!=veca.vecc+vecb.vecd

[vecaxx vecb " " vecc xx vecd " " vecexx vecf] is equal to

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|

If vecb is not perpendicular to vecc . Then find the vector vecr satisfying the equation vecr xx vecb = veca xx vecb and vecr. vecc=0

If vecb is not perpendicular to vecc . Then find the vector vecr satisfying the equation vecr xx vecb = veca xx vecb and vecr. vecc=0

veca,vecb,vecc are three non-coplanar such that veca + vecb + vecc = alpha vecd and vecb + vecc + vecd = beta veca , then veca + vecb + vecc + vecd is equal to:

If vecA xx vecB = vecC+ vecD , them select the correct alternative:

The vectors veca and vecb are not perpendicular and vecc and vecd are two vectors satisfying : vecbxxvecc=vecbxxvecd and veca.vecd=0. Then the vecd is equal to (A) vecc+(veca.vecc)/(veca.vecb)vecb (B) vecb+(vecb.vecc)/(veca.vecb)vecc (C) vecc-(veca.vecc)/(veca.vecb)vecb (D) vecb-(vecb.vecc)/(veca.vecb)vecc