Mathematics · 3D Geometry

JEE Advanced 2022 — Paper 1 — Question 11

Let P1P_{1} and P2P_{2} be two planes given by P1:10x+15y+12z−60=0,P2:−2x+5y+4z−20=0P_{1}: 10 x+15 y+12 z-60=0, P_{2}:-2 x+5 y+4 z-20=0.

Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on P1P_{1} and P2P_{2} ?

  1. Option A:

    x−10=y−10=z−15\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}

    Correct
  2. Option B:

    x−6−5=y2=z3\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}

    Correct
  3. Option C:

    x−2=y−45=z4\frac{x}{-2}=\frac{y-4}{5}=\frac{z}{4}

  4. Option D:

    x1=y−4−2=z3\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}

    Correct

Answer: A, B, D

Step-by-step solution

The line should be either coincident on P1P_{1} or on P2P_{2} or intersect on P1P_{1} and P2P_{2}

on different points.

(D) x1=y−4−2=z3=λ\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}=\lambda

⇒(λ,−2λ+4,3λ)\Rightarrow(\lambda,-2 \lambda+4,3 \lambda)

lie on P2\mathrm{P}_{2}.

(A) x−10=y−10=z−15\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5} intersects P1P_{1} and P2P_{2}

on different points.

(B) x−65=y2=z3\frac{x-6}{5}=\frac{y}{2}=\frac{z}{3} also intersects P1P_{1} and P2P_{2} on different points.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Intersection of lines, line & plane.
Let P 1 and P 2 be two planes given by P 1 : 10 x+15 y+12 z-60=0, P 2… | JEE Advanced 2022 PYQ with Solution · DhiX AI