Let a⃗=î+ĵ+k̂, b⃗=-î-8 ĵ+2 k̂ and c⃗=4 î+c₂ ĵ+c₃ k̂ be three vectors such that b⃗ × a⃗=c⃗ × a⃗. If the angle between the vector c⃗ and the vector 3 î+4 ĵ+k̂ is θ, then the greatest integer less than or equal to tan ² θ is _______________.

Solution

Maths sir · 4 steps

Equal cross products with a⃗ mean c⃗ - b⃗ is parallel to a⃗.

1. Use the condition. b⃗ × a⃗ = c⃗ × a⃗ gives (c⃗ - b⃗) × a⃗ = 0⃗. So c⃗ - b⃗ is parallel to a⃗, that is c⃗ = b⃗ + λ a⃗ for some scalar λ.
2. Find c. c⃗ = (-1+λ)î + (-8+λ)ĵ + (2+λ)k̂. The î component must be 4, so λ = 5 and c⃗ = 4î - 3ĵ + 7k̂.
3. Find the angle. With d⃗ = 3î + 4ĵ + k̂: c⃗ · d⃗ = 12 - 12 + 7 = 7, |c⃗|² = 16 + 9 + 49 = 74 and |d⃗|² = 9 + 16 + 1 = 26. So cos² θ = 49/(74 × 26) = 49/1924.

   tan² θ = 1/(cos² θ) - 1 = 1924/49 - 1 = 1875/49 ≈ 38.27

4. So. The greatest integer less than or equal to tan² θ is 38.

The common slip: 39. Stopping at sec² θ = 1924/49 ≈ 39.27 and forgetting to subtract 1 gives 39, not 38.

The idea to keep. Whenever b⃗ × a⃗ = c⃗ × a⃗, write c⃗ = b⃗ + λ a⃗. One known component of c⃗ then fixes λ.

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