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

જો સમીકરણ સંહતિ  \(8 x+y+4 z=-2\)  ; \(x+y+z=0\)  ; \(\lambda x-3 y=\mu\)  ને અનંત ઉકેલ છે તો બિંદુ  \(\left(\lambda, \mu,-\frac{1}{2}\right)\) નું સમતલ  \(8 x+y+4 z+\) \(2=0\) થી અંતર મેળવો.

  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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