Mathematics · 3D Geometry

JEE Main 2024 — 29 January, Shift 1 — Question 15

Let PQRP Q R be a triangle with R(−1,4,2)R(-1,4,2). Suppose M(2,1,2)\mathrm{M}(2,1,2) is the mid point of PQ . The distance of the centroid of △PQR\triangle P Q R from the point of intersection of the line x−20=y2=z+3−1\frac{\mathrm{x}-2}{0}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}+3}{-1} and x−11=y+3−3=z+11\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3}{-3}=\frac{\mathrm{z}+1}{1} is

  1. Option A:

    69

  2. Option B:

    9

  3. Option C:

    69\sqrt{69}

    Correct
  4. Option D:

    99\sqrt{99}

Answer: C

Step-by-step solution

Centroid G divides MR in 1:21: 2

G(1,2,2)\mathrm{G}(1,2,2)

Point of intersection A of given lines is (2,−6,0)(2,-6,0)

AG=69\mathrm{AG}=\sqrt{69}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
3D Geometry
Topic
Intersection of lines, line & plane.
Let P Q R be a triangle with R(-1,4,2) . Suppose M (2,1,2) is the mid… | JEE Main 2024 PYQ with Solution · DhiX AI