MHT CET · Maths · Three Dimensional Geometry
A plane which is perpendicular to two planes \(2 x-2 y+z=0\) and \(x-y+2 z=4\), passes through \((1,-2,1)\). The distance of the plane from the point \((1,2,2)\) is
- A 0 units
- B 1 units
- C \(\sqrt{2}\) units
- D \(2 \sqrt{2}\) units
Answer & Solution
Correct Answer
(D) \(2 \sqrt{2}\) units
Step-by-step Solution
Detailed explanation
The equation of a plane passing through \((1,-2,1)\) is
\(\mathrm{a}(x-1)+\mathrm{b}(y+2)+\mathrm{c}(\mathrm{z}-1)=0...(i)\)
\(\begin{aligned}
& \text { Plane (i) is perpendicular to planes } \\
& 2 x-2 y+z=0 \text { and } x-y+2 z=4 . \\
& \therefore \quad \begin{array}{l}
2 a-2 b+c=0, \text { and } ...(i)\\
a-b+2 c=0 ...(ii)\\
\text { Solving (ii) and (iii), we get } \\
a=-3, b=-3, c=0
\end{array}
\end{aligned}\)
Substituting the values of \(a, b, c\) in equation (i), we get
\(x+y+1=0\)
\(\therefore \quad\) The distance of this plane from \((1,2,2)\) is
\(d=\left|\frac{1+2+1}{\sqrt{1+1}}\right|=2 \sqrt{2}\)
\(\mathrm{a}(x-1)+\mathrm{b}(y+2)+\mathrm{c}(\mathrm{z}-1)=0...(i)\)
\(\begin{aligned}
& \text { Plane (i) is perpendicular to planes } \\
& 2 x-2 y+z=0 \text { and } x-y+2 z=4 . \\
& \therefore \quad \begin{array}{l}
2 a-2 b+c=0, \text { and } ...(i)\\
a-b+2 c=0 ...(ii)\\
\text { Solving (ii) and (iii), we get } \\
a=-3, b=-3, c=0
\end{array}
\end{aligned}\)
Substituting the values of \(a, b, c\) in equation (i), we get
\(x+y+1=0\)
\(\therefore \quad\) The distance of this plane from \((1,2,2)\) is
\(d=\left|\frac{1+2+1}{\sqrt{1+1}}\right|=2 \sqrt{2}\)
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