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Mathematics · 3D Geometry

JEE Main 2024 — 5 April, Shift 1 — Question 20

If the line 2−x3=3y−24λ+1=4−z\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z makes a right angle with the line x+33μ=1−2y6=5−z7\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}, then 4λ+9μ4 \lambda+9 \mu is equal to :

  1. Option A:

    13

  2. Option B:

    4

  3. Option C:

    5

  4. Option D:

    6

    Correct

Answer: D

Step-by-step solution

2−x3=3y−24λ+1=4−z\frac{2-\mathrm{x}}{3}=\frac{3 y-2}{4 \lambda+1}=4-z

x−2(−3)=y−23(4λ+13)=z−4(−1)\frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\frac{z-4}{(-1)}

x+33μ=1−2y6=5−z7\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}

x+33μ=y−12(−3)=z−5(−7)\frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{(-3)}=\frac{z-5}{(-7)}

Right angle ⇒(−3)(3μ)+(4λ+13)(−3)+(−1)(−7)=0\Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 −9μ−4λ−1+7=0-9 \mu-4 \lambda-1+7=0

4λ+9μ=64 \lambda+9 \mu=6

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Angle between planes, plane & a line.

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