Question

The distance of the plane ax + by + cz + d = 0 from a point A(x₁, y₁, z₁) is |ax₁ + by₁ + cz₁ + d| /√(a² + b² + c²)  

Answer 1

Let E be the given plane and vector (r . n) = p be its equation. Let B (vector b) be a point which is not equal to N (see Fig.) where N is the foot of the perpendicular from A onto E so that vector (b . n) = p. Therefore

Now, vector a = (x₁, y₁, z₁), vector n = (a, b, c) and p = - d imply