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JEE Mains · Maths · STD 12 - 11. three dimension geometry

જો રેખાઓ \(\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}\) અને \(\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}\) વચ્ચેનું ન્યુનતમ અંતર \(\frac{13}{\sqrt{29}}\) હોય, તો \(\lambda\) નું એક મૂલ્ય ............ છે.

  1. A  \(-\frac{13}{25}\)
  2. B \(\frac{13}{25}\)
  3. C \(1\)
  4. D \(-1\)
Verified Solution

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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