Let a and b be two non-collinear vectors. The position vectors of four points A, B, C and D are a + b , 4 a + b , 5 a +4 b and 2 a +5 b respectively. If the area of the quadrilateral ABCD is 66 sq. units, then the area (in sq. units) of the parallelogram whose adjacent sides are represented by the vectors u =2 a + b and v = a -3 b is equal to :

Step-by-step solution

The diagonal vectors of the quadrilateral ABCD are : AC = (5 a +4 b ) - ( a + b ) = 4 a +3 b BD = (2 a +5 b ) - (4 a + b ) = -2 a +4 b The area of the quadrilateral ABCD is given by 1 2 | AC BD | . AC BD = (4 a +3 b ) (-2 a +4 b ) = -8( a a ) + 16( a b ) - 6( b a ) + 12( b b ) = 16( a b ) + 6( a b ) = 22( a b ) Area of ABCD = 1 2 |22( a b )| = 11| a b | Given that the area is 66 , we have : 11| a b | = 66 | a b | = 6 Now, the adjacent sides of the parallelogram are u =2 a + b and v = a -3 b . The area of this paral