Let a⃗ be a non-zero vector parallel to the line of intersection of the two planes described by î+ĵ, î+k̂ and î-ĵ, ĵ-k̂. If θ is the angle between the vector a⃗ and the vector b⃗=2 î-2 ĵ+k̂ and a⃗ · b⃗=6, then the ordered pair (θ,|a⃗ × b⃗|) is equal to :

Solution

Maths sir · 4 steps

Each plane's normal is the cross product of its two vectors, and the line runs along normal × normal.

1. Find both normals. A normal to a plane is perpendicular to both vectors lying in it. So n⃗₁ = (î+ĵ) × (î+k̂) = î-ĵ-k̂ and n⃗₂ = (î-ĵ) × (ĵ-k̂) = î+ĵ+k̂.
2. Direction of the line. The line of intersection lies in both planes, so it is perpendicular to both normals. That makes it parallel to n⃗₁ × n⃗₂ = -2ĵ+2k̂, so a⃗ = λ(-ĵ+k̂).
3. Fix the length. a⃗ · b⃗ = λ(0 + 2 + 1) = 3λ = 6, so λ = 2 and a⃗ = -2ĵ+2k̂. Then |a⃗| = 2√2 and |b⃗| = √(4+4+1) = 3.

   cos θ = (a⃗·b⃗)/|a⃗||b⃗| = 6/(2√2 × 3) = 1/√2

4. So. θ = π/4 and |a⃗ × b⃗| = |a⃗||b⃗|sin θ = 2√2 × 3 × 1/√2 = 6. The pair is (π/4, 6), option (d).

The idea to keep. Line of intersection of two planes is parallel to n⃗₁ × n⃗₂. And once you know a⃗·b⃗, |a⃗| and |b⃗|, the cross product's size follows from sin θ.

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