Hello, chrisduluk!

I don't know if you've had Calculus yet.

. . I'll give you a non-Calculus solution.

Let point Q be on the line: .Q (2t, t-3, 2t+2)Find the distance between the point P(1,-2,4) and the line given by:

. . x .= .2t,. . y .= .t - 3,. . z .= .2t + 2

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The distance from P to Q is: . d . = . √[(2t) - 1]² + [(t - 3) + 2]² + [(2t + 2) - 4]²

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. . which simplifies to: . d .= .√9t² - 14t + 6 . [1]

And we want tominimize

If we minimize thesquareof , we will minimize

Let .= . .= .9t² - 14t + 6

This is an up-opening parabola; its minimum point is at itsvertex.

The formula for the vertex is: .v .= .-b/2a

. . We have: .a = 9, .b = -14

. . The vertex is: .t .= .-(-14)/(2·9) .= .7/9

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Substitute into [1]: . d . = . √9(7/9)² - 14(7/9) + 6 . = . √5/9

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Therefore: . d .= .√5/3