Question:
If the foot of perpendicular drawn from the origin to the plane is \( (3, 2, 1) \), then the equation of the plane is
If the foot of perpendicular drawn from the origin to the plane is \( (3, 2, 1) \), then the equation of the plane is
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When finding the equation of a plane, use the point and the normal vector to form the equation in the form \( ax + by + cz = d \).
Updated On: Jan 27, 2026
- \( 3x + 2y - z = 12 \)
- \( 3x + 2y - z = 14 \)
- \( 3x + 2y + z = 14 \)
- \( 3x - 2y - z = 12 \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the problem.
The foot of the perpendicular from the origin to the plane is given as \( (3, 2, 1) \). This gives us a point on the plane. The equation of the plane can be derived using the normal vector and the point on the plane.
Step 2: Deriving the equation of the plane.
The general form of the equation of a plane is \( ax + by + cz = d \). The coefficients \( a, b, c \) are the components of the normal vector to the plane. Using the given point \( (3, 2, 1) \) and the condition for the normal vector, we find that the equation of the plane is \( 3x + 2y - z = 14 \).
Step 3: Conclusion.
Thus, the equation of the plane is \( 3x + 2y - z = 14 \), which makes option (B) the correct answer.
The foot of the perpendicular from the origin to the plane is given as \( (3, 2, 1) \). This gives us a point on the plane. The equation of the plane can be derived using the normal vector and the point on the plane.
Step 2: Deriving the equation of the plane.
The general form of the equation of a plane is \( ax + by + cz = d \). The coefficients \( a, b, c \) are the components of the normal vector to the plane. Using the given point \( (3, 2, 1) \) and the condition for the normal vector, we find that the equation of the plane is \( 3x + 2y - z = 14 \).
Step 3: Conclusion.
Thus, the equation of the plane is \( 3x + 2y - z = 14 \), which makes option (B) the correct answer.
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