If a , b and c are non-coplanar unit vectors such that the angle between any two of them is 60^ , and the vector d = x a + y b + z c is perpendicular to both a and b , then the value of (x+y) z is

Step-by-step solution

Given | a | = | b | = | c | = 1 and the angle between any two vectors is 60^ . a b = b c = c a = 1 1 60^ = 1 2 Since d is perpendicular to a and b , we have d a = 0 and d b = 0 . Substituting d = x a + y b + z c into d a = 0 : (x a + y b + z c ) a = 0 x| a |^2 + y( b a ) + z( c a ) = 0 x(1) + y ( 1 2 ) + z ( 1 2 ) = 0 2x + y + z = 0 Similarly, substituting d into d b = 0 : (x a + y b + z c ) b = 0 x( a b ) + y| b |^2 + z( c b ) = 0 x ( 1 2 ) + y(1) + z ( 1 2 ) = 0 x + 2y + z = 0 Adding the two equations, we get: (2