Question:
A plane which bisects the angle between the two given planes \(2x - y + 2z - 4 = 0\) and \(x + 2y + 2z - 2 = 0\), passes through the point :
A plane which bisects the angle between the two given planes \(2x - y + 2z - 4 = 0\) and \(x + 2y + 2z - 2 = 0\), passes through the point :
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Both planes have normals of the same length (3), so the bisecting planes are simply the sum and the difference of the two plane equations.
Updated On: Nov 11, 2026
- \((1, -4, 1)\)
- \((1, 4, -1)\)
- \((2, 4, 1)\)
- \((2, -4, 1)\)
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The Correct Option is D
Solution and Explanation
Normals: \((2,-1,2)\) and \((1,2,2)\), both of magnitude \(3\).
Bisector planes: \(\frac{2x-y+2z-4}{3} = \pm\frac{x+2y+2z-2}{3}\).
With \(+\): \(x - 3y - 2 = 0\). With \(-\): \(3x + y + 4z - 6 = 0\).
Test the options:
(1,-4,1): \(x-3y-2 = 11\neq 0\); \(3x+y+4z-6 = -3\neq 0\).
(1,4,-1): \(x-3y-2=-13\); \(3x+y+4z-6=-3\). No.
(2,4,1): \(x-3y-2=-12\); \(3x+y+4z-6=8\). No.
(2,-4,1): \(x-3y-2 = 12\neq 0\) but \(3x+y+4z-6 = 6-4+4-6 = 0\). Yes.
Answer: (4) \((2, -4, 1)\)
Bisector planes: \(\frac{2x-y+2z-4}{3} = \pm\frac{x+2y+2z-2}{3}\).
With \(+\): \(x - 3y - 2 = 0\). With \(-\): \(3x + y + 4z - 6 = 0\).
Test the options:
(1,-4,1): \(x-3y-2 = 11\neq 0\); \(3x+y+4z-6 = -3\neq 0\).
(1,4,-1): \(x-3y-2=-13\); \(3x+y+4z-6=-3\). No.
(2,4,1): \(x-3y-2=-12\); \(3x+y+4z-6=8\). No.
(2,-4,1): \(x-3y-2 = 12\neq 0\) but \(3x+y+4z-6 = 6-4+4-6 = 0\). Yes.
Answer: (4) \((2, -4, 1)\)
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