Let a → and b → be two unit vectors such that a → + b → = 3 . If c → = a → + 2 b → + ( a → × b → ) , then 2 c → is equal to:

Step-by-step solution

As a → + b → = 3 Squaring both the sides a → 2 + b → 2 + 2 a → ⋅ b → = 3 1 + 1 + 2 ⋅ 1 ⋅ 1 ⋅ cos ⁡ θ = 3 2 c o s θ = 1 c o s θ = 1 2 θ = 60° ∴ Angle between a → a n d b → i s 60 o Now, | c → | = | a → + 2 b → + 3 ( a → × b → ) | Squaring both the sides | c → | 2 = | a → | 2 + 4 | b → | + 9 | a → × b → | 2 + 4 a → . ( b → ) + 3 a → . ( a → × b → ) + 6 b → . ( a → × b → ) | c → | 2 = 1 + 4 + 9 sin 2 θ + 4 cos θ + 0 + 0 | c → | 2 = 5 + 9. 3 4 + 4. 1 2 = 55 4 ∴ 2 | c → | = 55