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

मूलबिंदु से \(\sqrt{\frac{2}{21}}\) की दूरी पर एक समतल, जिसमें समतलों \(x - y - z -1=0\) तथा \(2 x + y -3 z +4=0\) की प्रतिच्छेदन रेखा स्थित है, का समीकरण है

  1. A \(3 x-y-5 z+2=0\)
  2. B \(3 x-4 z+3=0\)
  3. C \(-x+2 y+2 z-3=0\)
  4. D \(4 x-y-5 z+2=0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(4 x-y-5 z+2=0\)

Step-by-step Solution

Detailed explanation

Required equation of plane \(\mathrm{P}_{1}+\lambda \mathrm{P}_{2}=0\) \((\mathrm{x}-\mathrm{y}-\mathrm{z}-1)+\lambda(2 \mathrm{x}+\mathrm{y}-3 \mathrm{z}+4)=0\) Given that its dist. From origin is \(\frac{2}{\sqrt{21}}\) Thus…
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