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

उस समतल, जो \((1,2,-3)\) से होकर जाता है तथा समतलों, \(3 x + y -2 z =5\) तथा \(2 x -5 y - z =7\) के लम्बवत् है, का समीकरण है

  1. A \(3 x-10 y-2 z+11=0\)
  2. B \(6 x-5 y-2 z-2=0\)
  3. C \(11 x+y+17 z+38=0\)
  4. D \(6 x-5 y+2 z+10=0\)
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Answer & Solution

Correct Answer

(C) \(11 x+y+17 z+38=0\)

Step-by-step Solution

Detailed explanation

Normal vector : \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1\end{array}\right|=-11 \hat{i}-\hat{j}+17 \hat{k}\) So drs of normal to the required plane is \(<11,1,17>\) plane passes through \((1,2,-3)\) So eq \(^{n}\) of plane :…
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