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posted by  jarodbarnum on 9/11/2007 11:25:47 PM  |  status: Live  

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Course Textbook Chapter Problem
Calculus N/A N/A N/A
Question Details:
   Find the distance from the point <5,5,2> to the line x=0,  y=5+5t, z=1+t
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posted by Anonanon on 9/11/2007 11:35:20 PM  |  status: Live
Asker's Rating: Helpful   
jarodbarnum's comment:
"its wrong"
Response Details:
The distance between point 1 and point 2, where Point1=(5,5,2) Point2=(0,5,1)
 
=sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)
 
Therefore sqrt((0-5)^2 + (5-5)^2 + (1-2)^2)
=sqrt(-5^2 + 0^2 + -1^2)
 
=sqrt(26)
Tags: Math, Calculus
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(Edith Cowan University)
posted by KaVkaZ86 on 3/24/2008 1:01:13 PM  |  status: Live
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Response Details:
: . (x - a)/ l = (y - b) /m = (z - c) / n = t
-------------------------------------------------
 
(x-a)/l = t  
x = lt + a  < = > x = 0
: . a = 0  < = > t = 0    
 
(y-b)/m = t
y = mt + b   <=> y = 5+5t
: . m = 5
   b = 5
 
(z-c)/n = t
z = nt + c  <=> z = 1 + t
: . n = 1
    c = 1
-------------------------------------------------
: . P2 = (0,5,1) & P1=(5,5,2)
 
: . D = √[(0-5)2 + (5-5)2 + ( 1-2)2 ]
 
        = √ [(-5)2 + (-1)2 ]
 
        = √26
 
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Goodluck!
"Science is a differential equation. Religion is a boundary condition."
Tags: Math, Calculus
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