Let a⃗ and b⃗ be two unit vectors such that the angle between them is π/3. If λ a⃗+2 b⃗ and 3 a⃗-λ b⃗ are perpendicular to each other, then the number of values of λ in [-1,3] is :

Solution

Maths sir · 4 steps

Expand the dot product, get a quadratic in λ, then check honestly where its roots land.

1. Use perpendicularity. Perpendicular vectors have zero dot product, so (λ a⃗+2b⃗)·(3a⃗-λb⃗)=0. Here |a⃗|=|b⃗|=1 and a⃗·b⃗=cos π/3=1/2.

   3λ|a⃗|²-λ²(a⃗·b⃗)+6(a⃗·b⃗)-2λ|b⃗|²=3λ-λ²/2+3-2λ=0

2. Solve the quadratic. Multiply by -2: λ²-2λ-6=0. So λ=(2±√(4+24))/2=1±√7.
3. Check the interval. Since √7≈ 2.65, the roots are λ≈ 3.65 and λ≈ -1.65. The first is above 3 and the second is below -1.
4. So. Neither value lies in [-1,3], so the number of values of λ is 0, option (d).

The idea to keep. Perpendicular means dot product zero. Expand term by term using |a⃗|² and a⃗·b⃗, and always test the roots against the given interval before counting them.

Answer from the NTA final answer key. Solution by PrepWiser.