JEE Main 2024 Mathematics question
Let L₁ : r⃗ = (î − ĵ + 2k̂) + λ(î − ĵ + 2k̂), λ ∈ ℝ, L₂ : r⃗ = (ĵ − k̂) + μ(3î + ĵ + pk̂), μ ∈ ℝ, and L₃ : r⃗ = δ(ℓî + mĵ + nk̂), δ ∈ ℝ be three lines such that L₁ is perpendicular to L₂ and L₃ is perpendicular to both L₁ and L₂. Then the point which lies on L₃ is
Answer: (A)
Solution
L₁ ⊥ L₂: 3 − 1 + 2p = 0, so p = −1. L₃ ⊥ L₁: ℓ − m + 2n = 0, and L₃ ⊥ L₂: 3ℓ + m − n = 0. Adding, 4ℓ + n = 0, so n = −4ℓ and m = ℓ
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