If `vecaxxvecb=veccxxvecd and vecaxxvecc=vecbxxvecd` show that `(veca-vecd)` is parallel to `(vecb-vecc)`. It is given that `veca!=vecd` and `vecb!=vecc`.

STATEMENT-1 : If vecaxxvecb=veccxxvecd and vecaxxvecc=vecbxxvecd , then veca-vecd is perpendicular to vecb-vecc . And STATEMENT-2 : If vecP and vecQ are perpendicular then vecP.vecQ=0 .

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 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=veca.vecc, vecaxxvecb=vecaxxvecc and veca!=vec0, then prove that vecb=vecc.

If vecaxxvecb=vecc and vecbxxvecc=veca , show that veca,vecb,vecc are orthogonal in pairs. Also show that |vecc|=|veca| and |vecb|=1

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

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

For any three vectors veca,vecb,vecc their product would be a vector if one cross product is folowed by other cross product i.e (vecaxxvecb)xxvecc or (vecbxxvecc)xxveca etc. For any four vectors veca,vecb,vecc,vecd the product would be a vector with the help of sequential cross product or by cross product of two vectors obtained by corss product of two pair i.e. (vecaxx(vecbxxvecc))xxvecd or (vecaxxvecb)xx(veccxxvecd). (vecaxxvecb)xx(veccxxvecd0 is a vector (A) along the line off intersection of two planes containing veca,vecb and vecc,vecd (B) perpendicular to plane containing veca,vecb and vecc,vecd (C) parallel to the plane containing veca,vecb and vecc,vecd (D) none of these

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

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