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Find geometric properties of a plane with given Cartesian equation

Consider a plane with the Cartesian equation

    \[ x + 2y - 2z + 7 = 0. \]

Find each of the following:

  1. A normal vector to the plane with length 1.
  2. The x,y,z intercepts of the plane.
  3. The distance from the plane to the origin.
  4. The point on the plane, Q, nearest to the origin.

  1. From the Cartesian equation we know that a normal vector to the plane is N' = (1,2,-2). Making this of unit length, we have a unit normal N given by

        \[ N = \frac{N'}{\lVert N \rVert} = \frac{1}{3} (1,2,-2) = \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right). \]

  2. Since the Cartesian equation gives us x + 2y - 2z = -7 we have the x-intercept is -7, the y-intercept is -\frac{7}{2} and the z-intercept is \frac{7}{2}.
  3. The distance from the plane to the origin is given by

        \[ d = \frac{7}{\lVert (1,2,-2) \rVert} = \frac{7}{3}. \]

  4. The point Q on the plane which is nearest to the origin is in the direction of the normal vector, and at a distance of \frac{7}{3}, therefore,

        \[ Q = \frac{7}{3} \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right) = \left( \frac{7}{9}, \frac{14}{9}, -\frac{14}{9} \right). \]

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