JEE Main 2023 · 8 April · morning · Single correct · +4 / −1

If the equation of the plane containing the line
x+2 y+3 z-4=0=2 x+y-z+5 and perpendicular to the plane
r⃗=(î-ĵ)+λ(î+ĵ+k̂)+μ(î-2 ĵ+3 k̂)
is a x+b y+c z=4, then (a-b+c) is equal to :

  1. (a) 18
  2. (b) 22
  3. (c) 20
  4. (d) 24
Answer and solution

Answer (b) 22

Solution

Maths sir · 4 steps

Write every plane through the line as one family with a parameter k, then fix k using perpendicularity.

  1. Family of planes. Any plane through the line of intersection is (x+2y+3z-4) + k(2x+y-z+5) = 0. Its normal is (1+2k, 2+k, 3-k).
  2. Normal of given plane. The given plane contains the directions î+ĵ+k̂ and î-2ĵ+3k̂. Their cross product is its normal: (1,1,1)×(1,-2,3) = (5, -2, -3).
  3. Perpendicular planes. Two planes are perpendicular when their normals have zero dot product. So 5(1+2k) - 2(2+k) - 3(3-k) = 0, which gives 11k - 8 = 0 and k = 8/11.

    11(x+2y+3z-4) + 8(2x+y-z+5) = 0 ⇒ 27x + 30y + 25z = 4

  4. So. The right side is already 4, so a = 27, b = 30, c = 25. Then a - b + c = 27 - 30 + 25 = 22, option (b).

The idea to keep. Plane through the intersection of two planes: always use P₁ + kP₂ = 0. One extra condition, here perpendicularity through a dot product of normals, fixes k.

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

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Asked about 300 times in JEE Main since 2020. This question is from 2023.