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Prove some equations about distances between points and planes

  1. Prove that the distance from the point (x_0, y_0, z_0) to the plane

        \[ ax + by + ca + d = 0 \]

    is given by the formula

        \[ \frac{| ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}. \]

  2. Find the point P on the plane 5x - 14y + 2z + 9 = 0 which is nearest to the point Q = (-2,15,-7).

  1. Proof. By Theorem 13.6 (page 476 of Apostol) we know that the distance from a point to a plane is given by

        \begin{align*}  d &= \frac{| N \cdot P |}{\lVert N \rVert} \\[9pt]  &= \frac{ | (a,b,c) \cdot (x_0, y_0, z_0) + d|}{\sqrt{a^2+b^2+c^2}} \\[9pt]  &= \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2+b^2+c^2}}. \qquad \blacksquare \end{align*}

  2. A normal to the plane is given by N = (5,-14,2). So, P \cdot N = 9 for any point P \in M. Further, the distance from M to a point Q not on M is minimal when X = tN where

        \[ t = \frac{P \cdot N}{N \cdot N} = \frac{9}{225} = \frac{1}{25}. \]


        \[ X = \frac{1}{25} (5,-14,2). \]

    Naming X to be the point P we have

        \[ P = \frac{1}{25} (5,-14,2). \]


    • Van Gogh says:

      For b) I got a different answer… P = (3,1,-5). The given answer 1/25(5,-14,2) does not even seem to be on the plane of 5x-14y+2z=-9.

      • Artem says:

        Got the same answer: (3, 1, -5). Answer for (b) is incorrect in both Apostol, and above – the found point is not on the plane.

  1. tom says:

    a) correct statement is distance equals projection of P-Q onto N, not projection P onto N. In Cartesian equation then, d is -N⋅P while N⋅Q is the value of given point.

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