JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{\mathrm{x}-\lambda}{2}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}-3}{4}\) and \(\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}\) is \(\frac{13}{\sqrt{29}}\), then a value of \(\lambda\) is :
- A \(-\frac{13}{25}\)
- B \(\frac{13}{25}\)
- C \(1\)
- D \(-1\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
Shortest dist. \(=\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}} \)…
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