alleryThe distance of the point (x₀, y₀, z₀) from the plane ax + by + cz + d = 0 is
A|ax₀ + by₀ + cz₀ + d| / (a + b + c)
B|ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)
C(ax₀ + by₀ + cz₀ + d) / √(a + b + c)
D|ax₀ + by₀ + cz₀ + d| · √(a² + b² + c²)
Answer & Solution
Correct answer: B. |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)
1. The normal to the plane is (a, b, c); its magnitude is √(a² + b² + c²).
2. The signed projection of (x₀, y₀, z₀) onto the unit normal gives the distance.
3. Use d = (ax₀ + by₀ + cz₀ + d) / |n|.
4. Take absolute value for unsigned distance.
_Source: syllabus Maths Part 2 Ch 11 "Three Dimensional Geometry", §11.9_
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