JEE Mains · Maths · STD 12 - 11. three dimension geometry
A plane \(E\) is perpendicular to the two planes \(2 x -2 y + z =0\) and \(x - y +2 z =4\), and passes through the point \(P (1,-1,1)\). If the distance of the plane \(E\) from the point \(Q(a, a, 2)\) is \(3 \sqrt{2}\), then \(( PQ )^{2}\) is equal to
- A \(9\)
- B \(12\)
- C \(21\)
- D \(33\)
Answer & Solution
Correct Answer
(C) \(21\)
Step-by-step Solution
Detailed explanation
First plane, \(P_{1}=2 x-2 y+z=0\), normal vector \(\equiv \bar{n}_{1}=(2,-2,1)\) Second plane, \(P_{2} \equiv x-y+2 z=4\), normal vector \(\equiv \bar{n}_{2}=(1,-1,2)\) Plane perpendicular to \(P_{1}\) and \(P_{2}\) will have normal vector \(\bar{n}_{3}\) Where…
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