Let a⃗=2 î+α ĵ+k̂, b⃗=-î+k̂, c⃗=β ĵ-k̂, where α and β are integers and α β=-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-α₂ β₂ is equal to

Solution

Maths sir · 4 steps

With diagonals given, the area is half the cross product of the diagonals, so do not halve a side product.

1. Write the diagonals. d₁⃗ = a⃗+b⃗ = î+αĵ+2k̂ and d₂⃗ = b⃗+c⃗ = -î+βĵ. For a parallelogram with diagonals d₁⃗ and d₂⃗, the area is 1/2|d₁⃗×d₂⃗|.

   d₁⃗×d₂⃗ = |î, ĵ, k̂; 1, α, 2; -1, β, 0| = -2β î - 2 ĵ + (α+β) k̂

2. Use the area. 1/2|d₁⃗×d₂⃗| = √21/2 gives 4β² + 4 + (α+β)² = 21. So 4β² + (α+β)² = 17.
3. Try integer pairs. With αβ = -6, the term 4β² ≤ 17 allows only β = ±1 or ±2. For β = ±1 you need (α+β)² = 13, which no integer gives. For β = ±2 you need (α+β)² = 1. This works for (α,β) = (-3, 2) and (3, -2).
4. So. Both pairs have α²+β² = 13 and αβ = -6, so either labelling gives 13 - (-6) = 19, option (c).

The idea to keep. Area is |p⃗×q⃗| for adjacent sides but 1/2|d₁⃗×d₂⃗| for diagonals. With an integer condition like αβ = -6, bound one variable first and test the few cases left.

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