If a → , b → and c → are three non-zero vectors and a → + b → + c → = 0 → , then prove that: a → × b → = b → × c → = c → × a → .

Given: a → + b → + c → = 0 → Multiplying both sides by cross product a → , we get: a → × a → + b → + c → = a → × 0 → ⇒ a → × a → + a → × b → + a → × c → = 0 → ⇒ 0 → + a → × b → - c → × a → = 0 → [Using: a → × a → = b → × b → = c → × c → = 0 → ; a → × c → = - c → × a → ] ⇒ a → × b → = c → × a → → 1 Similarly, Multiplying both sides by cross product b → , we get: b → × a → + b → + c → = b → × 0 → ⇒ b → × a → + b → × b → + b → × c → = 0 → ⇒ - a → × b → + 0 → + b → × c → = 0 → ⇒ b → × c → = a → × b → → 2 Thus, 1 and 2 , we get: a → × b → = b → × c → = c → × a → Hence, proved.

Prove by vector method, that in any △ A B C a sin ⁡ A = b sin ⁡ B = c sin ⁡ C

We have from triangle law, a → + b → + c → = 0 Taking cross product with b → throughout a → × b → + b → × b → + c → × b → = 0 or a → × b → + c → × b → = 0 ∵ b → × b → = 0 or a → × b → = - c → × b → = b → × c → Thus, a → × b → = b → × c → or a b sin π - C = b c sin π - A or a b sin C = b c sin A ∵ sin π - z = sin z or a sin A = c sin C Similarly, it can shown that b sin B = c sin C ∴ a sin ⁡ A = b sin ⁡ B = c sin ⁡ C Hence, proved.