If `vecxXvecb=veccxxvecb and vecx_|_veca then vecx` is equal to (A) `((vecbxxvecc)xxveca)/(vecb.veca)` (B) `((vecbxx(vecaxxvecc))/(vecb.vecc))` (C) `((vecaxx(veccxxvecb))/(veca.vecb))` (D) none of these

The vectors veca and vecb are not perpendicular and vecac 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

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

Assertion: If vecx xx vecb=veccxxvecb and vecx_|_veca then vecx=((vecbxxvecc)xxveca)/(veca.vecb) , Reason: vecaxx(vecbxxvecc)=(veca.vecc)vecb-(veca.vecb)vecc (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

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

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

If veca,vecb,vecc are unity vectors such that vecd=lamdaveca+muvecb+gammavecc then lambda is equal to (A) ([veca vecb vecc])/([vecb veca vecc]) (B) ([vecb vecc vecd])/([vecb vecc veca]) (C) ([vecb vecd vecc])/([veca vecb vecc]) (D) ([vecc vecb vecd])/([veca vecb vecc])

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 vecaxx(vecaxxvecb)=vecbxx(vecbxxvecc) and veca.vecb!=0 , then [(veca,vecb,vecc)]=

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

If veca is parallel to vecb xx vecc, then (veca xx vecb) .(veca xx vecc) is equal to (a) |veca|^(2)(vecb.vecc) (b) |vecb|^(2)(veca .vecc) (c) |vecc|^(2)(veca.vecb) (d) none of these