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

The distance of the point  \((1, -2, 4)\)  from the plane passing through the point \((1, 2, 2 )\)  and perpendicular to the planes  \(x - y + 2 z = 3\)  and \(2x - 2y+ z+ 12=0,\)  is

  1. A \(2\)
  2. B \(\sqrt 2\)
  3. C \(2\sqrt 2\)
  4. D \(\frac {1}{\sqrt 2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\sqrt 2\)

Step-by-step Solution

Detailed explanation

Let equation of plane be \(a(x-1)+b(y-2)+c(z-2)=0\) .....\((1)\) \((1)\) is perpendicular to given planes then \(a-b+2 c=0\) \(2 a-2 b+c=0\) Solving above equation \(c=0\) and \(a=b\) equation of plane \(( 1)\) can be \(x+y-3=0\) distance from \((1,-2,4)\) will be…
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