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

यदि समतल \(2 x - y +2 z +3=0\) की समतलों \(4 x -2 y +4 z +\lambda=0\) तथा \(2 x - y +2 z +\mu=0\) से दूरियाँ क्रमश : \(\frac{1}{3}\) तथा \(\frac{2}{3}\) इकाईयाँ है, तो \(\lambda+\mu\) का अधिकतम मान है

  1. A \(15\)
  2. B \(13\)
  3. C \(5\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(13\)

Step-by-step Solution

Detailed explanation

\(4 x-2 y+4 z+6=0\) \(\frac{|\lambda-6|}{\sqrt{16+4+16}}=\left|\frac{\lambda-6}{6}\right|=\frac{1}{3}\) \(|\lambda-6|=2\) \(\lambda=8,4\) \(\frac{|\mu-3|}{\sqrt{4+4+1}}=\frac{2}{3}\) \(|\mu-3|=2\) \(\mu=5,1\) \(\therefore \) Maximum value of \((\mu+\lambda)=13.\)
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