The angle between vectors `(AxxB)and(BxxA)` is

Sets A and B have n elements in common. How many elements will (AxxB) and (BxxA) have in common?

If a,b and c are three mutually perpendicular vectors, then the projection of the vectors l(a)/(|a|)+m(b)/(|b|)+n((axxb))/(|axxb|) along the angle bisector of the vectors a and b is

If a,b and c are three mutually perpendicular vectors, then the projection of the vectors l(a)/(|a|)+m(b)/(|b|)+n((axxb))/(|axxb|) along the angle bisector of the vectors a and b is

If a,b and c are three mutually perpendicular vectors, then the projection of the vectors l(a)/(|a|)+m(b)/(|b|)+n((axxb))/(|axxb|) along the angle bisector of the vectors a and b is

If A=(1,2) and B=(2,3), then find the number of elements in (AxxB)cap(BxxA) . The following are the steps involved in solving the above problem. Arrange them in sequential order. (A) (AxxB)cap(BxxA)=(2,2) (B) Given A=(1,2) and B=(2,3) (C) n[)AxxB)cap(BxxA)]=1 (D) AxxB={(1,2)(1,3)(2,2)(2,3)} and B xxA={(2,1),(2,2),(#,1),(3,2)}