Mathematics · 3D Geometry

JEE Main 2024 — 8 April, Shift 1 — Question 9

If the shortest distance between the lines

L1:r→=(2+λ)i^+(1−3λ)j^+(3+4λ)k^,λ∈R\mathrm{L}_{1}: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R}

L2:r→=2(1+μ)i^+3(1+μ)j^+(5+μ)k^,μ∈R\mathrm{L}_{2}: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{\mathrm{k}}, \mu \in \mathbb{R} is mn\frac{\mathrm{m}}{\sqrt{\mathrm{n}}},

where gcd⁡(m,n)=1\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1, then the value of m+n\mathrm{m}+\mathrm{n} equals.

  1. Option A:

    384

  2. Option B:

    387

    Correct
  3. Option C:

    377

  4. Option D:

    390

Answer: B

Step-by-step solution

(CD)=∣AB‾⋅p→×q→∣p→×q→∣∣(C D)=\left|\frac{\overline{\mathrm{AB}} \cdot \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|

=∣(0i^+2j^+2k^)⋅(−15i^+7j^+9k^)355∣=\left|\frac{(0 \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-15 \hat{i}+7 \hat{j}+9 \hat{k})}{\sqrt{355}}\right|

=0+14+18355=32355=\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}}

∴m+n=32+355=387\therefore \mathrm{m}+\mathrm{n}=32+355=387

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them