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

यदि रेखाओं \(\frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1},(\alpha \neq-1)\) तथा \(x+y+z+1=0=2 x-y+z+3\) के बीच की न्यूनतम दूरी \(\frac{1}{\sqrt{3}}\) है, तो \(\alpha\) का एक मान है

  1. A \(-\frac {16}{19}\)
  2. B \(-\frac {19}{16}\)
  3. C \(\frac {32}{19}\)
  4. D \(\frac {19}{32}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac {32}{19}\)

Step-by-step Solution

Detailed explanation

Plane passing through \(x+y+z+1=0\) and \(2 x-y+z+3=0\) is \(x+y+z+1+\lambda\) \((2 x-y+z+3)=0\) \(\Rightarrow(2 \lambda+1) x+(1-\lambda) y+(1+\lambda) z+3 \lambda+1=0\) Parallel to the given line if \(\alpha(2 \lambda+1)-1(1-\lambda)+1(1+\lambda)=0\)…
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