An equation of a plane parallel to the plane \(x-2y+2z-5=0\) and at a unit distance from the origin is?
Solution and Explanation
The correct answer is \(x-2y+2z-3=0.\)
Let an equation of a plane parallel to the plane \(x−2y+2z−5=0\) be \(x−2y+2z+k=0\ \ \ ...(i)\)
Perpendicular distance from O(0,0,0) to (1) is 1
\(\frac{|k|}{\sqrt{1+4+4}}=1\)
\(⇒|k|=3\)
\(⇒k=+3\ \ or−3\)
\(∴x−2y+2z−3=0\)
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Concepts Used:
Distance of a Point from a Plane
The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
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