Let a line in three dimensions be specified by two points and lying on it, so a vector along the line is given by
(1)
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The squared distance between a point on the line with parameter and a point is therefore
(2)
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To minimize the distance, set and solve for to obtain
(3)
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where denotes the dot product. The minimum distance can then be found by plugging back into (2) to obtain
(4)
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(5)
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(6)
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Using the vector quadruple product
(7)
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where denotes the cross product then gives
(8)
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and taking the square root results in the beautiful formula
(9)
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(10)
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(11)
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Here, the numerator is simply twice the area of the triangle formed by points , , and , and the denominator is the length of one of the bases of the triangle, which follows since, from the usual triangle area formula, .