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November 29, 2014

November 29, 2014

Posted by **manny** on Tuesday, September 11, 2007 at 10:08pm.

I keep getting 0 and that's not the correct answer and I can't figure out what's going wrong.

- Calculus -
**drwls**, Wednesday, September 12, 2007 at 2:09amExpressed in terms of the paratmer t, the squared distance from P to any point on the line is

D^2 = (x1-x2)^2 + (y1-y2)^2 +(z1-z2)^2

=5^2 + (5-5t-5)^2 + (2+t-2)^2

= 25 + 25t^2 + t^2 = 26 t^2 + 25

Differentiate with respect to t and set the derivative equal to zero to find where the distance is a minimum

52t = 0

t = 0

At that value of t, the distance^2 is

D^2 = 25

so D = 5

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