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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Three Dimensional Geometry

If the shortest distance between the lines array ll L_1: r =(2+λ) i...

If the shortest distance between the lines L1:r⃗=(2+λ)i^+(1−3λ)j^+(3+4λ)k^,λ∈RL2:r⃗=2(1+μ)i^+3(1+μ)j^+(5+μ)k^,μ∈R\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \quad \mu \in \mathbb{R} \end{array} is mn\frac{m}{\sqrt{n}}, where gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1, then the value of m+nm+n equalsIf the shortest distance between the lines array ll L_1: r =(2+λ) i +(1-3 λ) j +(3+4 λ) k , & λ ∈ ℝ L_2: r =2(1+μ) i

Options

  1. A390
  2. B384
  3. C377
  4. D387

Correct answer

D. 387

Step-by-step solution

If the shortest distance between the lines array ll L_1: r =(2+λ) i +(1-3 λ) j +(3+4 λ) k , & λ ∈ ℝ L_2: r =2(1+μ) i
Shortes distance (CD) =∣AB‾⋅p→×q→∣p→×q→∣∣=\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∣=0+14+18355=32355∴m+n=32+355=387\begin{aligned} & =\left|\frac{(0 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(-15 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})}{\sqrt{355}}\right| \\ & =\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\ & \therefore \mathrm{m}+\mathrm{n}=32+355=387\end{aligned}$$
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