JEE Main 2022 · 28 June · evening · Single correct · +4 / −1

Let the plane ax + by + cz = d pass through (2, 3, -5) and is perpendicular to the planes
2x + y - 5z = 10 and 3x + 5y - 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :

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

Answer (d) 22

Solution

Maths sir · 4 steps

Perpendicular to both planes means the normal is the cross product of their two normals.

  1. Find the normal. The required plane is perpendicular to both given planes, so its normal is perpendicular to both of their normals, (2, 1, -5) and (3, 5, -7). The cross product gives such a vector.

    |î, ĵ, k̂; 2, 1, -5; 3, 5, -7| = (-7+25)î - (-14+15)ĵ + (10-3)k̂ = 18î - ĵ + 7k̂

  2. Use the point. The plane is 18x - y + 7z = d. Putting in (2, 3, -5) gives 36 - 3 - 35 = -2, so the plane is 18x - y + 7z = -2.
  3. Make d positive. Multiply through by -1: -18x + y - 7z = 2. So a = -18, b = 1, c = -7, d = 2, and gcd(18, 1, 7, 2) = 1 as required.
  4. So. a + 7b + c + 20d = -18 + 7 - 7 + 40 = 22, which is option (d).

The idea to keep. A plane perpendicular to two planes has normal n⃗₁ × n⃗₂. Then fix the sign and scale using the given conditions, here d > 0 and gcd 1, before you compute anything.

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 2022.