Let a⃗ and b⃗ be the vectors along the diagonals of a parallelogram having area 2√2. Let the angle between a⃗ and b⃗ be acute, |a⃗ | = 1, and |a⃗ . b⃗ | = |a⃗ × b⃗ |. If c⃗ = 2√2 ( a⃗ × b⃗ ) - 2b⃗ , then an angle between b⃗ and c⃗ is :

Solution

Maths sir · 4 steps

Diagonals halve the area formula, and a⃗ × b⃗ is perpendicular to b⃗: use both.

1. Area from diagonals. For a parallelogram with diagonals a⃗ and b⃗, the area is 1/2|a⃗ × b⃗|. So 1/2|a⃗ × b⃗| = 2√2, which gives |a⃗ × b⃗| = 4√2.
2. Find the angle. |a⃗ · b⃗| = |a⃗ × b⃗| means |cos θ| = sin θ. With θ acute, θ = π/4. Then 1 × |b⃗| × 1/√2 = 4√2, so |b⃗| = 8.
3. Dot and length. Since a⃗ × b⃗ is perpendicular to b⃗, b⃗ · c⃗ = -2|b⃗|² = -128. The same perpendicularity removes the cross term in |c⃗|²: |c⃗|² = 8|a⃗ × b⃗|² + 4|b⃗|² = 256 + 256 = 512, so |c⃗| = 16√2.

   cos φ = (b⃗ · c⃗)/(|b⃗| |c⃗|) = -128/(8 × 16√2) = -1/√2

4. So. The angle between b⃗ and c⃗ is φ = 3π/4, option (d).

The idea to keep. a⃗ × b⃗ is perpendicular to both a⃗ and b⃗, so it vanishes from any dot product with them. And when the given vectors are diagonals, the area is half of |a⃗ × b⃗|.

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