JEE Main 2023 · 31 January · evening · Single correct · +4 / −1

Let the plane P: 8 x+α₁ y+α₂ z+12=0 be parallel to
the line L: (x+2)/2=(y-3)/3=(z+4)/5. If the intercept of P
on the y-axis is 1 , then the distance between P and L is :

  1. (a) 6/√14
  2. (b) √14
  3. (c) √(2/7)
  4. (d) √(7/2)
Answer and solution

Answer (b) √14

Solution

Maths sir · 4 steps

Find the two unknowns from the intercept and the parallel condition, then any point of L gives the distance.

  1. Use the intercept. On the y-axis, x = 0 and z = 0, so α₁ y + 12 = 0 gives y = -12/α₁. Setting this equal to 1 gives α₁ = -12.
  2. Use parallel condition. A line is parallel to a plane when its direction (2, 3, 5) is perpendicular to the plane's normal (8, α₁, α₂). So 16 + 3α₁ + 5α₂ = 0, that is 16 - 36 + 5α₂ = 0, which gives α₂ = 4.
  3. Tidy the plane. P is 8x - 12y + 4z + 12 = 0. Dividing by 4 gives 2x - 3y + z + 3 = 0. Since L is parallel to P, every point of L is the same distance from P, so take the point (-2, 3, -4) on L.

    d = (|2(-2) - 3(3) + (-4) + 3|)/√(2² + 3² + 1²) = 14/√14 = √14

  4. So. The distance between P and L is √14, which is option (b).

The idea to keep. For a line parallel to a plane, normal dot direction is 0. The distance is then the distance from any one point of the line to the plane.

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.