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Is this graph correct?

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Write the parametric and vector equations of:

    x + 2 / 3 = y − 1 / 2 = z + 5/ -3

    x=-1+5t
    and plot a graph of this line when t = −2, −1, 0, 1 and 2.

    2. Relevant equations

    vector:
    r= [-2,1,-5]+t[3,2,-3]
    parametric
    x=3t-2
    y=2t+1
    z=-3t-5

    using parametric equation:: i drew the graph, please check if the graph is correct at -2,-1,0,1,,2

    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2012 #2

    LCKurtz

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    Not sure where that x = -1+5t came from. But your parametric equations look correct. I didn't check all your points, but if you plotted the points to scale as apparently you did, and they came out on a straight line, aren't you pretty confident in the answers?
     
  4. Dec 2, 2012 #3
    actually i was worried about the axis ... in specific the z axis.. i just had this gut feeling like i took -z the wrong direction ...
     
  5. Dec 2, 2012 #4
    x=-1+5t is part of the question .
     
  6. Dec 2, 2012 #5

    Mark44

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    Based on your subsequent work, your symmetric equations need parentheses.

    Here are the equations, using LaTeX.

    $$ \frac{x + 2}{3} = \frac{y - 1}{2} = \frac{z + 5}{-3}$$

    When you write them in plain text, they should look like this:

    (x + 2)/3 = (y - 1)/2 = (z + 5)/(-3)

    What you wrote means the following:
    x + ##\frac{2}{3}## = y - ##\frac{1}{2}## = z + ##\frac{5}{-3}##
     
  7. Dec 2, 2012 #6

    Dick

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    Then doesn't it seem odd that you are saying x=3t-2 in the solution? I assume when you are writing "x + 2 / 3" you mean (x+2)/3, not x+(2/3). But even assuming that, your parametric form looks screwed up. Put x=(-1)+5t into (x+2)/3=(y-1)/2 and solve for y in terms of t. The 't' in your equations is the wrong parameter. Your are going to get the same line, but the points corresponding to a given value of t will be different.
     
    Last edited: Dec 2, 2012
  8. Dec 17, 2012 #7
    hey , sorry for late response but if i put x=-1+5t in [x+2]/3= [y-1]/2=[z+5]/-3 what will i get after... how do i solve for y
     
  9. Dec 17, 2012 #8

    SammyS

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    Are you asking how to solve
    [itex]\displaystyle \frac{(-1+5t)+2}{3}=\frac{y-1}{2}[/itex]​
    for y ?

    Give it a try.

    What do you get?
     
  10. Dec 18, 2012 #9
    i didnt mean how i will solve it..rather i was concerned about what too do after i get the Y value


    y =[5+10t]/3
     
  11. Dec 18, 2012 #10

    SammyS

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    That depends.

    What is it you're trying to do next?
     
  12. Dec 18, 2012 #11

    HallsofIvy

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    Then you have two different expessions for x a function of t. What are you supposed to use?
     
  13. Dec 18, 2012 #12
    next i need to find vector and parametric equation and then plot for t ...|
     
  14. Dec 18, 2012 #13

    LCKurtz

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    This seems like a very strange way to phrase the problem, making me wonder what the exact statement of the problem is.

    Taking the problem as you have quoted, don't use ##t## here. You apparently should write:
    x=3s-2
    y=2s+1
    z=-3s-5
    using s for your parameter. Now if x = -1+5t you can put that in and get s in terms of t. Then use that in these last three equations. Of course, this all assumes I understand what you are trying to do.
     
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