Homework Help: Calculus

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

Find the distance from point P<5,5,2> to the line x=0 , y=5-5t, z=2+t.

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 2:09am
Expressed 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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