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Prove a formula for the perpendicular distance between a point and a line

  1. Assume that B \neq C. Prove that the perpendicular distance from A to the line passing through the points B and C is given by the formula

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

  2. Compute the distance in the case

        \[ A = (1,-2,-5), \qquad B = (-1,1,1), \qquad C = (4,5,1). \]


  1. Proof. We know the area of the parallelogram determined by (A-B) and (C-B) is given by

        \[ \lVert (A-B) \times (C-B) \rVert. \]

    But this is exactly twice the area of the triangle with base C-B and height the perpendicular line from A to the line through B and C. Hence, h is the perpendicular distance from A to the line through B and C, and

        \[ \lVert (A-B) \times (C-B) \rVert = 2 \cdot \left( \frac{1}{2} h \lVert B-A \rVert \right) \quad \implies \quad h = \frac{\lVert (A-B) \times (C-B) \rVert}{\lVert B-C \rVert}. \qquad \blacksquare \]

  2. When A = (1,-2,-5), \ B = (-1,1,1), \ C = (4,5,1) we have

        \begin{align*}  \frac{\lVert (A-B) \times (C - B) \rVert}{\lVert B-C \rVert} &= \frac{\lVert (2,-3,-6) \times (5,4,0) \rVert}{\lVert (-5,-4,0) \rVert} \\[9pt]  &= \frac{\lVert (24,-30,23) \rVert}{\lVert (-5,-4,0) \rVert} \\[9pt]  &= \frac{\sqrt{2005}}{\sqrt{41}}.  \end{align*}

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