Home › UP Board Class 12 › Mathematics › Surface Areas and Volumes › The distance of the point (x₀, y₀, z₀) from the …
The 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: NCERT Class 12 Maths Part 2 Ch 11 "Three Dimensional Geometry", §11.9_
Related questions
Painting the outside of a toy needs to be worked out from its:Breaking a hard problem into solved smaller ones is the method:A cylindrical tube with a rounded end is a cylinder with a:Pouring a liquid into a container of another shape keeps its:Joining two solids at their flat faces hides those faces:Which solid is not among the basic ones met earlier?The word frustum comes from Latin, meaning a piece:The two circular ends of a frustum have radii that are: