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

यदि रैखिक समीकरण निकाय \(8 x + y +4 z =-2\) \(x + y + z =0\) \(\lambda x -3 y =\mu\) के अनंत हल हैं, तो समतल \(8 x + y +4 z +2=0\) से बिंदु \(\left(\lambda, \mu,-\frac{1}{2}\right)\) की दूरी है :

  1. A \(3 \sqrt{5}\)
  2. B \(4\)
  3. C \(\frac{26}{9}\)
  4. D \(\frac{10}{3}\)
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Answer & Solution

Correct Answer

(D) \(\frac{10}{3}\)

Step-by-step Solution

Detailed explanation

\(D =\left|\begin{array}{ccc}8 & 1 & 4 \\1 & 1 & 1 \\\lambda & -3 & 0\end{array}\right|=0 \Rightarrow \lambda=4\) Also \(D _{1}= D _{2}= D _{3}=0\) So \( \mu=-2\) Point \(\left(4,-2,-\frac{1}{2}\right)\) Distance from plane \(=\frac{10}{3}\)
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