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

રેખાઓ \(\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}\) અને \(\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}\) વચ્ચેનું ન્યૂનત્તમ અંતર \(.......\) છે.

  1. A \(6\)
  2. B \(9\)
  3. C \(7\)
  4. D \(8\)
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Answer & Solution

Correct Answer

(B) \(9\)

Step-by-step Solution

Detailed explanation

Given lines \(\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2} \& \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}\) Formula for shortest distance S.D.…
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