If | a |=2,| b |=3 and a , b are mutually perpendicular vectors, then the area of the triangle whose vertices are 0, a+2 b, a-2 b is

Step-by-step solution

Let position vectors of A , B , C be 0, a+2 b, a-2 b aligned & Area of ABC = 1 2 | AB AC | & = 1 2 |( a +2 ~b )( a -2 ~b )| & = 1 2 | a a - a 2 ~b +2 ~b a =2 ~b b | & = 1 2 |2 ~b a +2 ~b a | & = 1 2 4| ~b a | & =2 2 3 & =12 sq. units. aligned