If a , b and c are three vectors such that | a + b + c | = 1 , c = ( a b ) and | a | = 1 3 , | b | = 1 2 , | c | = 1 6 , then the angle between a and b is

Step-by-step solution

Given | a + b + c | = 1 , squaring both sides gives: | a |^2 + | b |^2 + | c |^2 + 2( a b + b c + c a ) = 1 Since c = ( a b ) , the vector c is perpendicular to both a and b . a c = 0 and b c = 0 Substituting the given magnitudes | a | = 1 3 , | b | = 1 2 , and | c | = 1 6 : ( 1 3 )^2 + ( 1 2 )^2 + ( 1 6 )^2 + 2( a b + 0 + 0) = 1 1 3 + 1 2 + 1 6 + 2( a b ) = 1 2 + 3 + 1 6 + 2( a b ) = 1 1 + 2( a b ) = 1 2( a b ) = 0 a b = 0 | a | | b | = 0 Since | a | 0 and | b | 0 , we have = 0 . = ^c 2 Answer: ^c 2