Let a , b and c be three coplanar unit vectors. A unit vector d is perpendicular to them. If ( a b ) ( c d ) = 3 26 i - 2 13 j + 6 13 k and the angle between a and b is 30^ , then c is equal to...

Step-by-step solution

Using the vector quadruple product expansion: ( a b ) ( c d ) = [ a b d ] c - [ a b c ] d Since a , b , c are coplanar, their scalar triple product [ a b c ] = 0 . Thus, the expression simplifies to [ a b d ] c . The scalar triple product is [ a b d ] = ( a b ) d . Since d is perpendicular to the coplanar vectors a and b , the vector a b is parallel to d . The magnitude of a b is given by | a | | b | 30^ = 1 1 1 2 = 1 2 . Since d is a unit vector, ( a b ) d = | a b | | d | = 1 2 . Therefore, 1 2 c = 3 26 i - 2 13 j