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Mathematics · 3D Geometry

JEE Main 2025 — 3 April, Morning Shift — Question 39

Line L1L_{1} passes through the point (1,2,3)(1,2,3) and is parallel to z-axis. Line L2\mathrm{L}_{2} passes through the point (λ,5,6)(\lambda, 5,6)

and is parallel to yy-axis. Let for λ=λ1,λ2,λ2<λ1\lambda=\lambda_{1}, \lambda_{2}, \lambda_{2}<\lambda_{1}, the shortest distance between the two lines be 3 . Then the square of the distance of the point (λ1,λ2,7)\left(\lambda_{1}, \lambda_{2}, 7\right) from the line L1L_{1} is

  1. Option A:

    40

  2. Option B:

    32

  3. Option C:

    25

    Correct
  4. Option D:

    37

Answer: C

Step-by-step solution

L1≡x−10=y−20=z−31\mathrm{L}_{1} \equiv \frac{\mathrm{x}-1}{0}=\frac{\mathrm{y}-2}{0}=\frac{\mathrm{z}-3}{1}

L2≡x−λ0=y−51=z−60\mathrm{L}_{2} \equiv \frac{\mathrm{x}-\lambda}{0}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-6}{0}

SD=∣λ−133001010∣∣i^j^k^001010∣\mathrm{SD}=\frac{\left|\begin{array}{ccc}\lambda-1 & 3 & 3\\ 0 & 0 & 1\\ 0 & 1 & 0\end{array}\right|}{\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ 0 & 0 & 1\\ 0 & 1 & 0\end{array}\right|}

=∣λ−1∣=3=|\lambda-1|=3

λ=4,−2\lambda=4,-2

λ1=4\lambda_{1}=4

λ2=−2\lambda_{2}=-2

Let foot of perpendicular from P(4,−2,7)\mathrm{P}(4,-2,7) is Q(1,2,t+3)\mathrm{Q}(1,2, \mathrm{t}+3)

So (3,−4,4−t).(0,0,1)=0(3,-4,4-\mathrm{t}) .(0,0,1)=0

t=4\mathrm{t}=4

So Q(1,2,7)\mathrm{Q}(1,2,7)

PQ2=9+16\mathrm{PQ}^{2}=9+16

PQ2=25\mathrm{PQ}^{2}=25

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them

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