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posted by  MoOzY on 9/14/2009 10:16:41 PM  |  status: Closed  |  Earned Karma: 135

4-(16) help i will rate as a life saver if expained well, thanks

Course Textbook Chapter Problem Needs by
Calculus N/A N/A N/A 9/15/2009 at 10:00:00 PM
Question Details:
L_{1}: x = -2t, y = 1 + 2t, z = 3t and
L_{2}: x = -5 + 1\!s,y = $ e + 4\!s, z = 3 + 3\!s

Find the point of intersection of the two lines.
P = (, , )

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posted by Jasminek on 9/14/2009 11:52:23 PM  |  status: Live
Clarification Details:
post the question correctly.

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Sage
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posted by steb on 9/15/2009 12:01:41 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
x = -2t = -5 + s
z = 3t = 3 + 3s

Multiply the top equation by 3 and the bottom equation by 2 to get:

-6t = -15 + 3s
 6t = 6 + 6s

Add to get 0 = -9 + 9s, so s = 1. Plug this into the top equation to get:

-2t = -5 + 1 = -4, so t = 2.

Thus x = -2(2) = -4, y = 1+ 2(2) = 5 and z = 3(2) = 6.

So the point is: (-4, 5, 6).
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