Question:
A plane passes through the point \( (1,-2,1) \) and is perpendicular to planes \(2x-2y+z=0\) and \(x-y+2z=4\). Then the equation of the plane is
A plane passes through the point \( (1,-2,1) \) and is perpendicular to planes \(2x-2y+z=0\) and \(x-y+2z=4\). Then the equation of the plane is
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Plane perpendicular to two planes ⇒ normal is cross product.
Updated On: Apr 30, 2026
- \(x+y+1=0\)
- \(x-y+1=0\)
- \(x+2y+1=0\)
- \(x-2y+1=0\)
- \(x-y-1=0\)
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The Correct Option is D
Solution and Explanation
Concept:
Required plane normal is perpendicular to both given normals ⇒ cross product.
Step 1: Normals of planes. \[ \vec{n_1}=(2,-2,1), \vec{n_2}=(1,-1,2) \]
Step 2: Find cross product. \[ \vec{n} = \vec{n_1} \times \vec{n_2} \] \[ = \begin{vmatrix} i & j & k 2 & -2 & 1 1 & -1 & 2 \end{vmatrix} \] \[ = i(-4+1) - j(4-1) + k(-2+2) \] \[ = -3i -3j + 0k \] \[ = (-1,-1,0) \]
Step 3: Equation of plane. \[ -1(x-1) -1(y+2) = 0 \] \[ -x+1 -y-2 =0 \] \[ x+y+1=0 \]
Step 1: Normals of planes. \[ \vec{n_1}=(2,-2,1), \vec{n_2}=(1,-1,2) \]
Step 2: Find cross product. \[ \vec{n} = \vec{n_1} \times \vec{n_2} \] \[ = \begin{vmatrix} i & j & k 2 & -2 & 1 1 & -1 & 2 \end{vmatrix} \] \[ = i(-4+1) - j(4-1) + k(-2+2) \] \[ = -3i -3j + 0k \] \[ = (-1,-1,0) \]
Step 3: Equation of plane. \[ -1(x-1) -1(y+2) = 0 \] \[ -x+1 -y-2 =0 \] \[ x+y+1=0 \]
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