268937 Three vectors satisfy the relation A→⋅B→=0 and A→⋅C→=0, then A→ is parallel to
268938 Let F→ be the force acting on a particle having position vector r→ and τ→ be the torque of this foce about the origin. Then (AIEEE-2003)
268939 (A→×B→)+(B→×A→) is equal to
268940 If C→=A→×B→, then C→ is