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

माना रेखा \(\mathrm{x}-2 \mathrm{y}-\mathrm{z}-5=0=\mathrm{x}+\mathrm{y}+3 \mathrm{z}-5\) से होकर जाने वाले तथा रेखा \(\mathrm{x}+\mathrm{y}+2 \mathrm{z}-7=0\) \(=2 x+3 y+z-2\) के समांतर समतल का समीकरण \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=65\) है। तो बिन्दु \((\mathrm{a}, \mathrm{b}, \mathrm{c})\) की समतल \(2 \mathrm{x}+2 \mathrm{y}-\mathrm{z}+16=0\) से दूरी ____________ है।

  1. A \(8\)
  2. B \(9\)
  3. C \(10\)
  4. D \(11\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(9\)

Step-by-step Solution

Detailed explanation

Equation of plane is \((x-2 y-z-5)+b(x+y+3 z-5)=0\) \(\left|\begin{array}{lll}1+b & -2+b & -1+3 b \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|=0\) \(\Rightarrow b =12\) \(\therefore \text { plane is } 13 x+10 y +35 z =65\) Distance from given point to plane \(=9\)
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