JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \( \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} \) \( \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}\) is \(\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}\), where \(\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\), then the value of \(\mathrm{m}+\mathrm{n}\) equals:
- A \(384\)
- B \(387\)
- C \(377\)
- D \(390\)
Answer & Solution
Correct Answer
(B) \(387\)
Step-by-step Solution
Detailed explanation
Shortes distance \((\mathrm{CD})=\left|\frac{\overline{\mathrm{AB}} \cdot \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|\)…
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