Given that A → + B → + C → = 0 . Out of three vectors, two are equal in magnitude and the magnitude of third vector is 2 times that of either of the two having equal magnitude. Then the angle between vectors are given by

Step-by-step solution

Let, A → = B → = x Then C → = 2 x Now A → + B → = - C → A → + B → · A → + B → = - C → · - C → 2 A.B + A 2 + B 2 = C 2 2 ABcos θ + x 2 + x 2 = 2 x 2 cos θ = 0 θ = 9 0 o ⇒ Angle between A and B is 90 0 Again A → + C → = - B → A → + C → · A → + C → = - B → - B → A 2 + C 2 + 2 Acos θ = B 2 3 x 2 + 2 . x 2 x cos θ = x 2 cos θ = - 2 2 2 = - 1 2 θ = 1 3 5 o i.e Angle between A and C is 135 o Similarly, angle between B and C is θ 2 B → + C → = - A → B 2 + C 2 + 2 Bcos θ = A 2 cos θ = - 2 2 2 = - 1 2 θ 2 = 1 3 5 o