The distance between (5,1,3) and the line x=3, y=7+t, z=1+t is

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Solution

The correct option is C 6
$\begin{matrix} x = 3, y = 7 + t, z = 1 + t A (3, 7 + t, 1 + t) 0. (3 - 5) + 1 (7 + t - 1) + 1 (1 + t - 3) = 0 6 + t + t - 2 = 0 2 t = - 4 t = - 2 A : (3, 5, - 1) d i s tan c e = \sqrt{4 + 16 + 16} = 6 O p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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