Question:
If A and B are the feet of the perpendiculars drawn from \((1,2,3)\) to planes \(YZ\) and \(ZX\), then the equation of the plane passing through the points A, B and the origin is
If A and B are the feet of the perpendiculars drawn from \((1,2,3)\) to planes \(YZ\) and \(ZX\), then the equation of the plane passing through the points A, B and the origin is
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The normal of the required plane is the cross product of the two plane normals.
Updated On: Oct 1, 2026
- \(6x+3y+2z = 0\)
- \(6x-3y-2z = 0\)
- \(6x+3y-2z = 0\)
- \(3x+6y+2z = 0\)
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The Correct Option is C
Solution and Explanation
Step 1: Normal vector:
Normals of the given planes: \(\vec n_1 = (1, 2, 2)\) and \(\vec n_2 = (3, 3, 2)\). The required plane is perpendicular to both, so its normal is \(\vec n = \vec n_1\times\vec n_2\).
Step 2: Cross product:
\[ \vec n = (2\cdot2 - 2\cdot3,\ 2\cdot3 - 1\cdot2,\ 1\cdot3 - 2\cdot3) = (-2, 4, -3) \]
So the plane is \(2x - 4y + 3z + d = 0\).
Step 3: Use the point:
\((1, 2, 1)\): \(2 - 8 + 3 + d = 0\), so \(d = 3\). The plane is \(2x - 4y + 3z + 3 = 0\).
Final Answer:
The required plane is \(2x - 4y + 3z + 3 = 0\), option (B). \[ \boxed{2x - 4y + 3z + 3 = 0} \]
Normals of the given planes: \(\vec n_1 = (1, 2, 2)\) and \(\vec n_2 = (3, 3, 2)\). The required plane is perpendicular to both, so its normal is \(\vec n = \vec n_1\times\vec n_2\).
Step 2: Cross product:
\[ \vec n = (2\cdot2 - 2\cdot3,\ 2\cdot3 - 1\cdot2,\ 1\cdot3 - 2\cdot3) = (-2, 4, -3) \]
So the plane is \(2x - 4y + 3z + d = 0\).
Step 3: Use the point:
\((1, 2, 1)\): \(2 - 8 + 3 + d = 0\), so \(d = 3\). The plane is \(2x - 4y + 3z + 3 = 0\).
Final Answer:
The required plane is \(2x - 4y + 3z + 3 = 0\), option (B). \[ \boxed{2x - 4y + 3z + 3 = 0} \]
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