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Prove a formula for the distance between a plane determined by three points and a point

  1. If a plane is determined by the points A,B,C prove that the distance from a point Q to this plane is given by

        \[ \frac{|(Q-A) \cdot (B-A) \times (C-A)|}{\lVert (B-A) \times (C-A) \rVert}. \]

  2. Using part (a) compute the distance in the case

        \[ Q = (1,0,0), \quad A = (0,1,1), \quad B = (1,-1,1), \quad C = (2,3,4). \]


  1. Proof. We know the distance from a plane containing a point P to a point Q not on the plane is given by the formula

        \[ \frac{(Q-P) \cdot N}{\lVert N \rVert}. \]

    Since the plane through A,B,C is the set of points

        \[ M = \{ A + s(B-A) + t(C-A) \} \]

    we have N = (B-A) \times (C-A). Thus, the distance from Q to M is

        \[ d = \frac{|(Q-A) \cdot (B-A) \times (C-A)|}{\lVert (B-A) \times (C-A) \rVert}. \]

  2. For the given points Q,A,B,C we have

        \[ d = \frac{|((1,0,0) - (0,1,1)) \cdot ((1,-1,1) - (0,1,1)) \times ((2,3,4) - (0,1,1))|}{\lVert ((1,-1,1) - (0,1,1)) \times ((2,3,4) - (0,1,1)) \rVert} = \frac{|-6+3-6|}{9} = 1. \]

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