Home » Blog » Establish properties of four planes with given Cartesian equations

Establish properties of four planes with given Cartesian equations

by RoRi

Consider four planes with the Cartesian equations:

    \begin{align*} x + 2y - 2z & = 5 \\ 3x - 6y + 3z & = 2 \\ 2x + y + 2z &= -1 \\ x- 2y + z &= 7. \end{align*}

  1. Establish that two of them are parallel and the other two are perpendicular.
  2. For the two parallel planes, find the distance between them.
  1. The second and fourth planes are parallel since they have the same normal vector, (1,-2,1).
    To see that the first and third are perpendicular, we denote the normal vectors by N_1 and N_3, respectively, and compute

        \[ N_1 \cdot N_3 = (1,2,-2) \cdot (2,1,2) = 2 + 2 - 4 = 0. \]

    Hence, they are perpendicular.

  2. Denoting the second and fourth planes by M_2 and M_4, respectively we have Cartesian equations

        \[ M_2: \ x - 2y + z = \frac{2}{3}, \qquad M_4: \ x - 2y + z = 7. \]

    Therefore, the distance d between them is

        \[ d = \frac{7- \frac{2}{3}}{\lVert N \rVert} = \frac{19}{3 \sqrt{6}} = \frac{19 \sqrt{6}}{18}. \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):