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

यदि समतल \(23 x-10 y-2 z+48=0\) तथा रेखाओं \(\frac{ x +1}{2}=\frac{ y -3}{4}=\frac{ z +1}{3}\) और \(\frac{ x +3}{2}=\frac{ y +2}{6}=\frac{ z -1}{\lambda}(\lambda \in R )\) को अंतर्विष्ट करने वाले समतल के बीच की दूरी \(\frac{ k }{\sqrt{633}}\) है, तो \(k\) बराबर है

  1. A \(2\)
  2. B \(3\)
  3. C \(6\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

If \(\lambda=-7,\) then planes will be parallel \& distance between them will be \(\frac{3}{\sqrt{633}} \Rightarrow \mathrm{k}=3\) But if \(\lambda \neq-7,\) then planes will be intersecting and distance between them will be \(0\)
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