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

રેખાઓ \(\mathrm{L}_1\) અને \(\mathrm{L}_2\), વચ્ચેનું ન્યુનત્તમ અંતર મેળવો. જ્યાં \(\mathrm{L}_1: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{-3}=\frac{\mathrm{z}+4}{2}\) અને \(\mathrm{L}_2\) એ \(A(-4,4,3), B(-1,6,3)\) માંથી પસાર થાય તથા રેખા \(\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}\) ને લંબ છે.

  1. A \(\frac{121}{\sqrt{221}}\)
  2. B \(\frac{24}{\sqrt{117}}\)
  3. C \(\frac{141}{\sqrt{221}}\)
  4. D  \(\frac{42}{\sqrt{117}}\)
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Answer & Solution

Correct Answer

(C) \(\frac{141}{\sqrt{221}}\)

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Detailed explanation

\begin{aligned} & \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\ & \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \mid \\ 2 & -3 & 2 \\ 3 & 2…

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