Let a and b be two vectors such that | a | = 1 , | b | = 2 and a b > 0 . If the area of the parallelogram adjacent sides of which are represented by the vectors p = 2 a + b and q = a - 2 b is 5 3 , and the vectors u = a + b and v = 2 a - b are perpendicular to each other, then the value of ^2 + 2 is :

Step-by-step solution

The area of the parallelogram is given by | p q | . p q = (2 a + b ) ( a - 2 b ) = 2( a a ) - 4( a b ) + ( b a ) - 2( b b ) = 0 - 4( a b ) - ( a b ) - 0 = -5( a b ) Given the area is 5 3 : |-5( a b )| = 5 3 5| a || b | = 5 3 5(1)(2) = 5 3 = 3 2 Since a b > 0 , > 0 , so = 3 . Thus, a b = | a || b | = (1)(2) ( 1 2 ) = 1 . Now, u and v are perpendicular, so u v = 0 : ( a + b ) (2 a - b ) = 0 2 | a |^2 - ^2( a b ) + 2( a b ) - | b |^2 = 0 Substitute the known values: 2 (1)^2 - ^2(1) + 2(1) - (2)^2 = 0 2 - ^2 + 2 - 4 =