Question:
Equation of a plane passing through \((-1,1,1)\) and \((1,-1,1)\) and perpendicular to \(x + 2y + 2z - 5 = 0\) is
Equation of a plane passing through \((-1,1,1)\) and \((1,-1,1)\) and perpendicular to \(x + 2y + 2z - 5 = 0\) is
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Plane containing a line and perpendicular to another → use cross product of direction and normal.
Updated On: Apr 16, 2026
- \(2x + 3y - 3z + 3 = 0\)
- \(x + y + 3z - 5 = 0\)
- \(2x + 2y - 3z + 3 = 0\)
- \(x + y + z - 3 = 0\)
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The Correct Option is C
Solution and Explanation
Concept:
A plane perpendicular to another plane has its normal vector perpendicular to the given plane’s normal.
Step 1: Normal of given plane.
\[ x + 2y + 2z - 5 = 0 \Rightarrow \vec{n_1} = (1,2,2) \]
Step 2: Direction vector of line joining given points.
\[ \vec{AB} = (1 - (-1), -1 - 1, 1 - 1) = (2, -2, 0) \]
Step 3: Normal of required plane.
Required plane contains line AB and is perpendicular to given plane. So its normal is perpendicular to both \(\vec{AB}\) and \(\vec{n_1}\): \[ \vec{n} = \vec{AB} \times \vec{n_1} \] \[ = \begin{vmatrix} i & j & k 2 & -2 & 0 1 & 2 & 2 \end{vmatrix} = (-4)i - (4)j + (6)k \] \[ \Rightarrow \vec{n} = (-4,-4,6) \propto (2,2,-3) \]
Step 4: Equation of plane.
Using point \((-1,1,1)\): \[ 2(x+1) + 2(y-1) -3(z-1) = 0 \] \[ 2x + 2y - 3z + 3 = 0 \]
Step 1: Normal of given plane.
\[ x + 2y + 2z - 5 = 0 \Rightarrow \vec{n_1} = (1,2,2) \]
Step 2: Direction vector of line joining given points.
\[ \vec{AB} = (1 - (-1), -1 - 1, 1 - 1) = (2, -2, 0) \]
Step 3: Normal of required plane.
Required plane contains line AB and is perpendicular to given plane. So its normal is perpendicular to both \(\vec{AB}\) and \(\vec{n_1}\): \[ \vec{n} = \vec{AB} \times \vec{n_1} \] \[ = \begin{vmatrix} i & j & k 2 & -2 & 0 1 & 2 & 2 \end{vmatrix} = (-4)i - (4)j + (6)k \] \[ \Rightarrow \vec{n} = (-4,-4,6) \propto (2,2,-3) \]
Step 4: Equation of plane.
Using point \((-1,1,1)\): \[ 2(x+1) + 2(y-1) -3(z-1) = 0 \] \[ 2x + 2y - 3z + 3 = 0 \]
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