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Find properties of a plane given three points that determine it

Let M be the plane determined by the three points (1,1,-1), (3,3,2), and (3,-1,-2). Find the following:

  1. A normal vector to the plane.
  2. A Cartesian equation for the plane.
  3. The distance between the plane and the origin.

  1. Denote the points by P = (3,-1,-2), Q = (1,1,-1) and R = (3,3,2), then we can compute a normal vector by

        \begin{align*}  N &= (P-Q) \times (P-R) \\   &= ((3,-1,-2) - (1,1,-1)) \times ((3,-1,-2) - (3,3,2)) \\  &= (2,-2,-1) \times (0,-4,-4) \\  &= (4,8,-8). \end{align*}

    Therefore, N = (1,2,-2) is a normal vector to the plane.

  2. Since (1,2,-2) is normal to the plane we have a Cartesian equation of the form

        \[ x + 2y - 2z = d. \]

    Then, since (1,1,-1) is on the plane we have d = 5. Hence, the Cartesian equation is

        \[ x + 2y - 2z = 5. \]

  3. The distance from the origin is

        \[ d = \frac{5}{ \lVert N \rVert} = \frac{5}{3}. \]

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