Let O be an interior point of triangle A B C, such that 2 O A → + 3 O B → + 4 O C → = 0, then the ratio of the area of ∆ A B C to the area of ∆ A O C is

Step-by-step solution

Let position vectors of O , A , B , C are O → , a → , b → , c → respectively. 2 O A → + 3 O B → + 4 O C → = 0 ⇒ 2 a → + 3 b → + 4 c → = 0 Area of ∆ A O C = 1 2 O A → × O C → = 1 2 a → × c → Area of ∆ A B C = 1 2 B A → × B C → = 1 2 a → × b → + b → × c → + c → × a → = 1 2 a → × - 4 c → - 2 a → 3 + - 4 c → - 2 a → 3 × c → + c → × a → = 1 2 4 3 c → × a → + 2 3 c → × a → + c → × a → = 3 2 | c → × a → | A r e a o f ∆ A B C A r e a o f ∆ A O C = 3 1