Let P Q R be a triangle. The points A , B and C are on the sides Q R , R P and P Q respectively such that Q A A R = R B B P = P C C Q = 1 2 . Then Area ∆ P Q R Area ∆ A B C is equal to

Step-by-step solution

Let the position vector of P is O → , Q is q → and R is r → . Now, position vector of A is 2 q → + r → 3 Position vector of B is 2 r → 3 . Position vector of C is q → 3 . A B → = r → - 2 q → 3 A C → = - r → - q → 3 So, A B → × A C → = 1 3 r → - 2 q → × - r → - q → ⇒ A B → × A C → = 1 3 - r → × q → + 2 q → × r → ⇒ A B → ×