Question

The equation of any two parallel planes are of the form ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 and the distance between them is equal to

|d₁ - d₂| /√(a² + b² + c²) 

Answer 1

Let 

E₁ = a₁x + b₁y + c₁z + d₁ = 0 

and E₂ = a₂x + b₂y + c₂z + d₂ = 0

be two parallel planes so that vector n1 = (a₁, b₁, c₁) and vector n2 (a₂, b₂, c₂) are the normal vectors to E₁ and E₂, respectively. Since E₁ and E₂ are parallel, their normals vector (n₁ and n₂) are parallel. Hence, let vector n₂ = k vector n₁ so that a₂ = ka₁, b₂ = kb₁ and c₂ = kc₁. Thus, the equation of E₂ is k(a₁x + b₁y + c₁z) + d = 0. Dividing by k, we have E₂ = a₁x + b₁y + c₁z + d₂ = 0 (here, in the place of d₂ /k, we take d₂). Let E₁ = ax + by + cz + d₁ = 0 and E₂ = ax + by + cz + d₂ = 0 be the two parallel planes. Let A(x₁, y₁, z₁) be a point in E₂ so that 

ax₁ + by₁ + cz₁ + d₂ = 0 ...(1)

Now, Distance between E₁ and E₂ = Distance of E₁ from point A 

 

From theorem (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

)

Notation: If E denotes ax + by + cz + d, then we denote ax₁ + by₁ + cz₁ + d by E₁₁ and ax₂ + by₂ + cz₂ + d₂ by E₂₂.