Let a → , b → , c → be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ , with the vector a → + b → + c → . Then 36 cos 2 2 θ is equal to

Step-by-step solution

We have, a → = b → = c → a → · b → = b → · c → = c → · a → = 0 Since, vectors are mutually perpendicular and of equally magnitude, hence let a → = i ^ ,   b → = j ^ ,   c → = k ^ Now, a → + b → + c → · a → = i ^ + j ^ + k ^ · i ^ ⇒ i ^ + j ^ + k ^ · i ^ = 3 · 1 · cos θ ⇒ 1 + 0 + 0 = 3 · 1 · cos θ ⇒ 3 cos θ = 1 ⇒ cos θ