MHT CET · Maths · Three Dimensional Geometry
A plane passes through \((1,-2,1)\) and is perpendicular to the planes \(2 x-2 \mathrm{y}+\mathrm{z}=0\) and \(x-\mathrm{y}+2 \mathrm{z}=4\). The distance of the point \((1,2,2)\) from this plane is _______ units.
- A \(1\)
- B \(\sqrt{2}\)
- C \(2 \sqrt{2}\)
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\mathbf{n} = \langle 2,-2,1 \rangle \times \langle 1,-1,2 \rangle = \langle -3,-3,0 \rangle\) Use normal vector \(\mathbf{n}' = \langle 1,1,0 \rangle\).
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