Mathematics · 3D Geometry

JEE Main 2025 — 23 January, Evening Shift — Question 18

If the square of the shortest distance between the lines x−21=y−12=z+3−3\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3} and x+12=y+34=z+5−5\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5} is mn\frac{\mathrm{m}}{\mathrm{n}}, where m,n\mathrm{m}, \mathrm{n} are coprime numbers, then m+n\mathrm{m}+\mathrm{n} is equal to:

  1. Option A:

    6

  2. Option B:

    9

    Correct
  3. Option C:

    21

  4. Option D:

    14

Answer: B

Step-by-step solution

a⃗=(2,1,−3)\vec{a} = (2, 1, -3) b⃗=(−1,−3,−5)\vec{b} = (-1, -3, -5) p⃗×q⃗=∣i^j^k^12−324−5∣\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}

=2i^−j^=2 \hat{i}-\hat{j}

b→−a→=−3i^−4j^−2k^\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}

Sd=∣(b→−a→)⋅(p→×q→)∣∣p→×q→∣\mathrm{S}_{\mathrm{d}}=\frac{|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}.

=25=\frac{2}{\sqrt{5}} (Sd)2=45\left(\mathrm{S}_{\mathrm{d}}\right)^{2}=\frac{4}{5}

m=4,n=5⇒ m+n=9\mathrm{m}=4, \mathrm{n}=5 \Rightarrow \mathrm{~m}+\mathrm{n}=9

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them