JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
A vector a⃗ is parallel to the line of intersection of the plane determined by the vectors î, î+ĵ and the plane determined by the vectors î-ĵ, î+k̂. The obtuse angle between a⃗ and the vector b⃗=î-2 ĵ+2 k̂ is :
- (a) (3 π)/4
- (b) (2 π)/3
- (c) (4 π)/5
- (d) (5 π)/6
Answer and solution
Answer (a) (3 π)/4
Solution
Maths sir · 4 steps
A line lying in both planes is perpendicular to both normals, so cross the two normals.
- Normals of planes. The normal to the first plane is î × (î+ĵ) = k̂. The normal to the second plane is (î-ĵ) × (î+k̂) = -î-ĵ+k̂.
- Line of intersection. The line lies in both planes, so it is perpendicular to both normals. Its direction is the cross product of the two normals.
k̂ × (-î-ĵ+k̂) = -ĵ + î = î-ĵ
- Angle with b. Take a⃗ = λ(î-ĵ). Then a⃗·b⃗ = λ(1+2) = 3λ, |a⃗| = √2 |λ| and |b⃗| = 3. So cos θ = 3λ/(3√2 |λ|) = ±1/√2.
- So. The two possible angles are π/4 and 3π/4, because a⃗ can point either way along the line. The obtuse one is 3π/4, option (a).
The idea to keep. The line where two planes meet is parallel to n⃗₁ × n⃗₂. A line has two directions, so you get an angle and its supplement; pick the one the question asks for.
Answer from the NTA final answer key. Solution by PrepWiser.