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Mathematics · 3D Geometry

JEE Main 2026 — 4 April, Morning Shift — Question 38

The square of the distance of the point (-2, -8, 6) from the line x−11=y−12=z−1\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{-1} along the line x+51=y+5−1=z2\frac{x + 5}{1} = \frac{y + 5}{-1} = \frac{z}{2} is equal to:

  1. Option A:

    3

  2. Option B:

    6

    Correct
  3. Option C:

    8

  4. Option D:

    12

Answer: B

Step-by-step solution

DR or AB is 1,−1,21,-1,2 ∴ Line AB is x+21=y+8−1=z−62=λ\frac{\mathrm{x}+2}{1}=\frac{\mathrm{y}+8}{-1}=\frac{\mathrm{z}-6}{2}=\lambda B is ( λ−2,−λ−8,2λ+6\lambda-2,-\lambda-8,2 \lambda+6 ) B is (μ+1,2μ+1,−μ)(\mu+1,2 \mu+1,-\mu) Solving point B is (−3,−7,4)(-3,-7,4) AB=(−1)2+(1)2+(2)2=6\mathrm{AB}=\sqrt{(-1)^{2}+(1)^{2}+(2)^{2}}=\sqrt{6} AB2=6\mathrm{AB}^{2}=6

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
3D Geometry
Topic
D.C'S ,D.R.'S of angular bisectors

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