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JEE Advanced Mathematics Three Dimensional Geometry 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Mathematics Question (2023) — Solution

Question

Let ℓ 1  and ℓ 2  be the lines r → 1 = λ i ^ + j ^ + k ^ and r → 2 = j ^ - k ^ + μ i ^ + k ^ , respectively, Let X be the set of all the planes H that contain the line ℓ 1 . For a plane H , let d H denote the smallest possible distance between the points of ℓ 2  and H . Let H 0  be a plane in X for which d H 0 is the maximum value of d H as H varies over all planes in X . Match each entry in List-I to the correct entries in List-II.   List-I   List-II P The value of d H 0 is 1 3 Q The distance of the point 0 ,   1 ,   2 from H 0 is 2 1 3 R The distance of origin from H 0  is 3 0 S The distance of origin from the point of intersection of planes y = z ,   x = 1 and H 0  is 4 2     5 1 2 The correct option is

Options

  1. A. P → 2   Q → 4   R → 5   S → 1
  2. B. P → 5   Q → 4   R → 3   S → 1
  3. C. P → 2   Q → 1   R → 3   S → 2
  4. D. P → 5   Q → 1   R → 4   S → 2

Answer

B. P → 5   Q → 4   R → 3   S → 1

Step-by-step solution

Given, H 0  will be the plane containing the line ℓ 1 and parallel to ℓ 2 . So, the normal vector of plane parallel to  ℓ 1 and ℓ 2 is given by, i ^ j ^ k ^ 1 1 1 1 0 1 = j ^ 1 - j ^ 1 - 1 + k ^ - 1 = i ^ - k ^ Hence, the equation of plane  H 0    will be, H 0   :   x - z = C  which passes through origin, So, C = 0 ∴   H 0   :   x - z = 0 Now solving, P d H 0 = 1  distance of point 0 ,   1 ,   - 1 from H . d = 0 - - 1 2 = 1 2   ∴   P → 5 Q   d = 0 - 2 2 = 2   ∴   Q → 4 R   d = 0 2 = 0   ∴   R → 3 S  Point of intersection will be of given planes y = z ,   x = 1   &   x - z = 0   will be,  1 ,   1 ,   1   Hence, distance  d = 1 + 1 + 1 = 3   ∴   S → 1 ∴ Option (B) is correct.

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