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

જો સમતલ \(2x -y + 2z + 3 = 0\) નું સમતલો \(4x -2y + 4z + \lambda = 0\) અને \(2x -y + 2z + \mu = 0\) થી અંતર અનુક્રમે \(\frac {1}{3}\) અને \(\frac {2}{3}\) હોય તો \(\lambda + \mu \) ની મહતમ કિમંત મેળવો.

  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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