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

रेखाओं \(\mathrm{x}+1=2 \mathrm{y}=-12 \mathrm{z}\) तथा \(\mathrm{x}=\mathrm{y}+2=6 \mathrm{z}-6\) के बीच न्यूनतम दूरी है -

  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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