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.