Question:
The equation of the plane passing through the point \((1,2,1)\) and perpendicular to the planes \(x+2y+2z-7 = 0\) and \(3x+3y+2z-5 = 0\) is
The equation of the plane passing through the point \((1,2,1)\) and perpendicular to the planes \(x+2y+2z-7 = 0\) and \(3x+3y+2z-5 = 0\) is
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The normal of the required plane is the cross product of the normals of the two given planes.
Updated On: Oct 1, 2026
- \(2x+y-2z-2 = 0\)
- \(2x-4y+3z+3 = 0\)
- \(2x+4y-5z-5 = 0\)
- \(x+4y-6z-3 = 0\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
If a plane is perpendicular to two planes, its normal is perpendicular to both of their normals. So it is along \(\vec n_1 \times \vec n_2\).
Step 2: Normal
\(\vec n_1 = (1, 2, 2)\), \(\vec n_2 = (3, 3, 2)\).
\[ \vec n_1 \times \vec n_2 = (4 - 6)\hat i - (2 - 6)\hat j + (3 - 6)\hat k = -2\hat i + 4\hat j - 3\hat k \]
Step 3: Equation
The plane through \((1, 2, 1)\):
\[ -2(x - 1) + 4(y - 2) - 3(z - 1) = 0 \Rightarrow -2x + 4y - 3z - 3 = 0 \]
\[ 2x - 4y + 3z + 3 = 0 \]
Option (B). A quick check: substituting \((1, 2, 1)\) gives \(2 - 8 + 3 + 3 = 0\).
Final Answer:
The plane is \(2x - 4y + 3z + 3 = 0\), option (B). \[ \boxed{2x - 4y + 3z + 3 = 0} \]
If a plane is perpendicular to two planes, its normal is perpendicular to both of their normals. So it is along \(\vec n_1 \times \vec n_2\).
Step 2: Normal
\(\vec n_1 = (1, 2, 2)\), \(\vec n_2 = (3, 3, 2)\).
\[ \vec n_1 \times \vec n_2 = (4 - 6)\hat i - (2 - 6)\hat j + (3 - 6)\hat k = -2\hat i + 4\hat j - 3\hat k \]
Step 3: Equation
The plane through \((1, 2, 1)\):
\[ -2(x - 1) + 4(y - 2) - 3(z - 1) = 0 \Rightarrow -2x + 4y - 3z - 3 = 0 \]
\[ 2x - 4y + 3z + 3 = 0 \]
Option (B). A quick check: substituting \((1, 2, 1)\) gives \(2 - 8 + 3 + 3 = 0\).
Final Answer:
The plane is \(2x - 4y + 3z + 3 = 0\), option (B). \[ \boxed{2x - 4y + 3z + 3 = 0} \]
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