If a = 1 10 (3 i + k ), b = 1 7 (2 i +3 j -6 k ) , then the value of ( a -2 ~b ) ( a b ) (2 a + b ) is

Step-by-step solution

Given: a = 1 10 (3 i + k ) and b = 1 7 (2 i + 3 j - 6 k ) Evaluating the magnitudes and dot product: | a |^2 = ( 1 10 )^2(3^2 + 1^2) = 1 10 (9 + 1) = 1| b |^2 = ( 1 7 )^2(2^2 + 3^2 + (-6)^2) = 1 49 (4 + 9 + 36) = 1 a b = 1 7 10 [(3)(2) + (0)(3) + (1)(-6)] = 1 7 10 (6 - 6) = 0 The vector triple product simplifies using the identity ( P Q ) R = ( P R ) Q - ( Q R ) P : ( a b ) (2 a + b ) = (2| a |^2 + a b ) b - (2 a b + | b |^2) a = (2 + 0) b - (0 + 1) a = 2 b - a Substituting into the original expression: E = ( a - 2