JEE Main 2024 Mathematics question
Let a⃗ = 2î + αĵ + k̂, b⃗ = −î + k̂, c⃗ = β ĵ − k̂, where α and β are integers and αβ = −6. Let the values of the ordered pair (α, β) for which the area of the parallelogram of diagonals a⃗ + b⃗ and b⃗ + c⃗ is √21/2, be (α₁, β₁) and (α₂, β₂). Then α₁² + β₁² − α₂β₂ is equal to
Answer: (D)
Solution
a⃗ + b⃗ = î + αĵ + 2k̂ and b⃗ + c⃗ = −î + βĵ. The area of the parallelogram is ½|(a⃗ + b⃗) × (b⃗ + c⃗)| = √21/2, with the cross product (−2β)î − 2ĵ + (β + α)k̂. So 4β² + 4 +
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