Question

Find the distance between the planes \[ x - 2y + 2z = 6 \quad \text{and} \quad 3x - 6y + 6z = 2. \]

Hint:

Distance between parallel planes is the absolute difference of their constants divided by the magnitude of the normal vector.

Answer 1

First, check if the planes are parallel. The normal vectors are: \[ \vec{n}_1 = (1, -2, 2), \quad \vec{n}_2 = (3, -6, 6). \] Since \(\vec{n}_2 = 3 \vec{n}_1\), the planes are parallel. Rewrite the second plane to match the first plane’s normal vector by dividing by 3: \[ x - 2y + 2z = \frac{2}{3}. \] Distance \(d\) between two parallel planes \[ Ax + By + Cz + D_1 = 0, \quad Ax + By + Cz + D_2 = 0 \] is given by: \[ d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}. \] Rewrite planes as: \[ x - 2y + 2z - 6 = 0, \] \[ x - 2y + 2z - \frac{2}{3} = 0. \] Calculate: \[ d = \frac{\left| -6 + \frac{2}{3} \right|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{\left| -\frac{16}{3} \right|}{\sqrt{1 + 4 + 4}} = \frac{\frac{16}{3}}{3} = \frac{16}{9}. \]
Final answer: \[ \boxed{ \frac{16}{9}. } \]