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

રેખાાઓ \(x+1=2 y=-12 z\) ખને \(x=y+2=6 z-6\) વચ્ચેનું ન્યૂનત્તમ અંતર \(............\) છે.

  1. A \(2\)
  2. B \(3\)
  3. C \(\frac{5}{2}\)
  4. D \(\frac{3}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

\(\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}\) and \(\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}\) \(\Rightarrow\) Shortest distance…
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