JEE Main 2021 · 27 August · evening · Single correct · +4 / −1
The equation of the plane passing through the line of intersection of the planes r⃗ .( î + ĵ + k̂ ) = 1 and r⃗ .( 2î + 3ĵ - k̂ ) + 4 = 0 and parallel to the x-axis is :
- (a) r⃗ .( ĵ - 3k̂ ) + 6 = 0
- (b) r⃗ .( î + 3k̂ ) + 6 = 0
- (c) r⃗ .( î - 3k̂ ) + 6 = 0
- (d) r⃗ .( ĵ - 3k̂ ) - 6 = 0
Answer and solution
Answer (a) r⃗ .( ĵ - 3k̂ ) + 6 = 0
Solution
Maths sir · 4 steps
Take the family P₁ + λ P₂ = 0, then kill the x-coefficient so the plane runs parallel to the x-axis.
- Cartesian form. The two planes are x + y + z - 1 = 0 and 2x + 3y - z + 4 = 0. Every plane through their line of intersection has the form below, for some number λ.
(x + y + z - 1) + λ(2x + 3y - z + 4) = 0
- Parallel to x-axis. A plane is parallel to the x-axis when its normal is perpendicular to î, that is, when the coefficient of x is zero. Here that coefficient is 1 + 2λ, so λ = -1/2.
- Substitute back. With λ = -1/2 the equation becomes -1/2y + 3/2z - 3 = 0. Multiplying by -2 gives y - 3z + 6 = 0.
- So. In vector form the plane is r⃗ · (ĵ - 3k̂) + 6 = 0, which is option (a).
The idea to keep. Parallel to an axis means that variable drops out of the plane's equation. Use P₁ + λ P₂ = 0 and choose λ to make that coefficient zero.
Answer from the NTA final answer key. Solution by PrepWiser.