JEE Main 2021 · 24 February · evening · Single correct · +4 / −1
The vector equation of the plane passing through the intersection
of the planes r⃗ .( î + ĵ + k̂ ) = 1 and r⃗ .( î - 2ĵ ) = - 2, and the point (1, 0, 2) is :
- (a) r⃗ .( î + 7ĵ + 3k̂ ) = 7/3
- (b) r⃗ .( î + 7ĵ + 3k̂ ) = 7
- (c) r⃗ .( 3î + 7ĵ + 3k̂ ) = 7
- (d) r⃗ .( î - 7ĵ + 3k̂ ) = 7/3
Answer and solution
Answer (b) r⃗ .( î + 7ĵ + 3k̂ ) = 7
Solution
Maths sir · 4 steps
Any plane through the line of intersection is plane 1 plus λ times plane 2. Let the given point fix λ.
- Cartesian form. With r⃗ = xî + yĵ + zk̂, the planes are x + y + z - 1 = 0 and x - 2y + 2 = 0. Every plane through their line of intersection has the form below.
(x + y + z - 1) + λ (x - 2y + 2) = 0
- Use the point. Put in (1, 0, 2): (1 + 0 + 2 - 1) + λ(1 - 0 + 2) = 2 + 3λ = 0, so λ = -2/3.
- Clear fractions. Multiply the equation by 3: 3x + 3y + 3z - 3 - 2x + 4y - 4 = 0, which gives x + 7y + 3z = 7. Check with the point: 1 + 0 + 6 = 7.
- So. In vector form the plane is r⃗ · (î + 7ĵ + 3k̂) = 7, which is option (b).
The idea to keep. For a plane through the intersection of P₁ = 0 and P₂ = 0, write P₁ + λ P₂ = 0 and fix λ with the extra condition. Always check the given point in your final answer.
Answer from the NTA final answer key. Solution by PrepWiser.