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Question #41649

Difficulty: 8/10 | 🟢 Free

The Question

Let L1L_1 be the line of intersection of the planes P1:2x+y−2z=5P_1: 2x + y - 2z = 5 and P2:x−2y+2z=6P_2: x - 2y + 2z = 6. Let L2L_2 be the line given by the symmetric form x−20=y+11=z−31\frac{x-2}{0} = \frac{y+1}{1} = \frac{z-3}{1}. A plane Π\Pi contains the line L1L_1 and is parallel to the line L2L_2. The perpendicular distance from the origin to the plane Π\Pi is:

Options

  • A.

    22

  • B.

    53\frac{5}{3}

  • C.

    11

  • D.

    1110\frac{11}{\sqrt{10}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Symmetric vs Non-symmetric form|Line lying in a plane|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

The core trap lies in the standard parameterization of a family of planes.

Students habitually use P1+λP2=0P_1 + \lambda P_2 = 0 to represent any plane passing through the intersection of P1P_1 and P2P_2.

However, this asymmetric form inherently excludes the plane P2P_2 itself (which corresponds to λ→∞\lambda \to \infty).

When students apply the parallelism condition (n⃗1+λn⃗2)⋅d⃗=0(\vec{n}_1 + \lambda \vec{n}_2) \cdot \vec{d} = 0, they obtain a contradictory equation (−1=0-1 = 0), leading them to falsely conclude that no such plane exists or to guess incorrectly.

By using the homogeneous symmetric form αP1+βP2=0\alpha P_1 + \beta P_2 = 0, the linear dependence is properly handled, revealing that the required plane is exactly P2P_2.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptSymmetric vs Non-symmetric form
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree