JEE Advanced · Mathematics · 31. 3D Geometry
A plane passes through \((1,-2,1)\) and is perpendicular to two planes \(2 x-2 y+z=0\) and \(x-y+2 z=4\), then the distance of the plane from the point \((1,2,2)\) is
- A
0
- B
1
- C
\(\sqrt{2}\)
- D
\(2 \sqrt{2}\)
Answer & Solution
Correct Answer
(D)
\(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Let the equation of plane be,
\[
a(x-1)+b(y+2)+c(z-1)=0
\]
which is perpendicular to \(2 x-2 y+z=0\) and \(x-y+2 z=4\)
\[
\begin{array}{llrl}
\Rightarrow & 2 a-2 b+c & =0 \text { and } a-b+2 c=0 \\
\text { or } & \frac{a}{-2} 1 & =\frac{b}{2}=\frac{c}{-2} \\
\Rightarrow & -12 & 1 & -1 \\
\Rightarrow & \frac{a}{-3} & =\frac{b}{-3}=\frac{c}{0} \\
& \frac{a}{1}=\frac{b}{1}=\frac{c}{0}
\end{array}
\]
So, the equation of plane is,
\(x-1+y+2=0\) or \(x+y+1=0\), its distance from the point \((1,2,2)\) is \(\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}\).
\[
a(x-1)+b(y+2)+c(z-1)=0
\]
which is perpendicular to \(2 x-2 y+z=0\) and \(x-y+2 z=4\)
\[
\begin{array}{llrl}
\Rightarrow & 2 a-2 b+c & =0 \text { and } a-b+2 c=0 \\
\text { or } & \frac{a}{-2} 1 & =\frac{b}{2}=\frac{c}{-2} \\
\Rightarrow & -12 & 1 & -1 \\
\Rightarrow & \frac{a}{-3} & =\frac{b}{-3}=\frac{c}{0} \\
& \frac{a}{1}=\frac{b}{1}=\frac{c}{0}
\end{array}
\]
So, the equation of plane is,
\(x-1+y+2=0\) or \(x+y+1=0\), its distance from the point \((1,2,2)\) is \(\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}\).
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