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Mathematics · 3D Geometry

JEE Main 2026 — 6 April, Evening Shift — Question 33

The shortest distance between the lines x−41=y−32=z−2−3\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3} and x+22=y−64=z−5−5\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5} is:

  1. Option A:

    566\frac{5 \sqrt{6}}{6}

  2. Option B:

    252 \sqrt{5}

  3. Option C:

    353 \sqrt{5}

    Correct
  4. Option D:

    454 \sqrt{5}

Answer: C

Step-by-step solution

SD =∣4−(−2)3−62−512−324−5∣∣i^j^k^12−324−5∣=\frac{\left|\begin{array}{ccc}4-(-2) & 3-6 & 2-5\\ 1 & 2 & -3\\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & -3\\ 2 & 4 & -5\end{array}\right|} SD=∣6(−10+12)−(−3)(−5+6)−3(4−4)∣∣2i^−j^∣S D=\frac{|6(-10+12)-(-3)(-5+6)-3(4-4)|}{|2 \hat{i}-\hat{j}|} SD=∣12+35∣\mathrm{SD}=\left|\frac{12+3}{\sqrt{5}}\right| =155=\frac{15}{\sqrt{5}} SD=35\mathrm{SD}=3 \sqrt{5}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them

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