Let S be the set of all a ∈ R for which the angle between the vectors u⃗=a(log ₑ b) î-6 ĵ+3 k̂ and v⃗=(log ₑ b) î+2 ĵ+2 a(log ₑ b) k̂, (b>1) is acute. Then S is equal to :

Solution

Maths sir · 4 steps

Put t = ln b > 0: near b = 1 the dot product sits near -12, so no value of a keeps it positive.

1. Acute means positive. The angle between two vectors is acute exactly when their dot product is positive. Since b > 1, write t = ln b. Then t can be any positive number.

   u⃗·v⃗ = a t² - 12 + 6at = at(t+6) - 12

2. For every b. For a to belong to S, we need at(t+6) - 12 > 0 for every t > 0, because b can be any number greater than 1.
3. Test small t. If a ≤ 0, then at(t+6) ≤ 0, so the dot product is at most -12. If a > 0, take b very close to 1, so t is close to 0. Then at(t+6) is close to 0 and the dot product is close to -12, which is negative.
4. So. No real a makes the dot product positive for every b > 1. So S = Φ, option (b).

The idea to keep. When a condition must hold for every value of a parameter, check the edge values first. Here b → 1⁺ sends ln b → 0 and rules out every a at once.

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