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

एक समतल \(E\) दो समतलों \(2 x -2 y + z =0\) तथा \(x - y +2 z =4\), के लंबवत है, तथा बिंदु \(P (1,-1\), 1) से होकर जाता है। समतल \(E\) की बिंदु \(Q ( a , a\), 2) से दूरी \(3 \sqrt{2}\) है, तो \(( PQ )^2\) बराबर है

  1. A \(9\)
  2. B \(12\)
  3. C \(21\)
  4. D \(33\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(21\)

Step-by-step Solution

Detailed explanation

First plane, \(P_{1}=2 x-2 y+z=0\), normal vector \(\equiv \bar{n}_{1}=(2,-2,1)\) Second plane, \(P_{2} \equiv x-y+2 z=4\), normal vector \(\equiv \bar{n}_{2}=(1,-1,2)\) Plane perpendicular to \(P_{1}\) and \(P_{2}\) will have normal vector \(\bar{n}_{3}\) Where…
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