Mathematics · 3D Geometry

JEE Main 2026 — 4 April, Morning Shift — Question 37

A line with direction ratios 1, -1, 2 intersects the lines x2=y3=z+13\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3} and x+1−1=y−21=z4\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4} at the points P and Q, respectively. If the length of the line segment PQ is α, then 225α² is equal to:

  1. Option A:

    1024

  2. Option B:

    1014

    Correct
  3. Option C:

    1104

  4. Option D:

    1204

Answer: B

Step-by-step solution

P(2λ,3λ,3λ−1)\mathrm{P}(2 \lambda, 3 \lambda, 3 \lambda-1) and Q(−μ−1,μ+2,4μ)\mathrm{Q}(-\mu-1, \mu+2,4 \mu) D.R of PQ 2λ+μ+1,3λ−μ−2,3λ−4μ−12 \lambda+\mu+1,3 \lambda-\mu-2,3 \lambda-4 \mu-1, 2λ+μ+11=3λ−μ−2−1=3λ−4μ−12\frac{2 \lambda+\mu+1}{1}=\frac{3 \lambda-\mu-2}{-1}=\frac{3 \lambda-4 \mu-1}{2} Solving λ=15\lambda=\frac{1}{5} and μ=−815\mu=\frac{-8}{15} P(25,35,−25)&Q(−715,2215,−3215)\mathrm{P}\left(\frac{2}{5}, \frac{3}{5}, \frac{-2}{5}\right) \& \mathrm{Q}\left(\frac{-7}{15}, \frac{22}{15}, \frac{-32}{15}\right) PQ2=(1315)2+(1315)2+(2615)2\mathrm{PQ}^{2}=\left(\frac{13}{15}\right)^{2}+\left(\frac{13}{15}\right)^{2}+\left(\frac{26}{15}\right)^{2} α2=1014225\alpha^{2}=\frac{1014}{225}

∴225α2=1014\therefore 225 \alpha^{2}=1014

Solution figure

Answer key and solution verified before publishing.

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