For p > 0 , a vector v → 2 = 2 i ^ + p + 1 j ^ is obtained by rotating the vector v → 1 = 3 p i ^ + j ^ by an angle θ about origin in counter clockwise direction. If tan θ = α 3 - 2 4 3 + 3 , then the value of α is equal to

Step-by-step solution

We have, v → 1 = 3 p i ^ + j ^ v → 2 = 2 i ^ + p + 1 j ^ tan θ = α 3 - 2 4 3 + 3 v → 1 = v → 2 ⇒ 3 p 2 + 1 = 4 + ( p + 1 ) 2 ⇒ 2 p 2 - 2 p - 4 = 0 ⇒ p 2 - p - 2 = 0 ⇒ p = 2 ,   - 1 (rejected) Now, cos θ = v → 1 · v → 2 v → 1 · v → 2 = 2 3 p + ( p + 1 ) ( p + 1 ) 2 + 4 3 p 2 + 1 ⇒ cos θ = 4 3 + 3 13 13 = 4 3 + 3 13 ⇒ tan θ = 112 - 24 3 4 3 + 3 = 6 3 - 2 2 4 3 + 3 ⇒ tan θ = 6 3 - 2 4