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Write parametric and symmetric equations for the z-axis.

  1. Aug 23, 2008 #1
    Write parametric and symmetric equations for the z-axis.

    I'm not sure i am on the right track; here is my attempt to an answer.

    [0, 0, z] where z can equal any number.

    a = [0, 0, 1]
    b = [0, 0, z]

    Parametric equations

    x = 0
    y = 0
    z = 1 + tz

    Symmetric equations
    [tex]\frac{x - 0}{0}[/tex] = [tex]\frac{y - 0}{0}[/tex] = [tex]\frac{z - 1}{z}[/tex]
    0 = 0 = -1

    I'm not certain that my attempt to this question is correct; if anyone can guide me in the right direction if needed i would be grateful!
    Thanks :)
     
  2. jcsd
  3. Aug 23, 2008 #2

    Defennder

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    The parametric equations cannot contain the terms x,y,z on the RHS. So z = 1 + tz is not correct. Think in terms of vectors. We want to obtain the equation of a line for the z-axis in vector form: [tex]OP + t\vec{v}[/tex] where v is the direction of the line and OP any point on the line.

    For the 2nd one, take notes of the values of x,y along the z-axis.
     
  4. Aug 24, 2008 #3
    so would the parametric equation for z = a_z + tb_z

    and for the second one since the values of x and y are always equal to zero on the z-axis; would the symmetric equations be:

    x = y = z
    0 = 0 = [tex]\frac{z - a_z}{b_z}[tex]

    Thanks
     
  5. Aug 24, 2008 #4

    Defennder

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    What's a_z and b_z?
     
  6. Aug 24, 2008 #5

    HallsofIvy

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    Two points on the z axis are [0, 0, 1] and [0, 0, 2] but certainly not "[0, 0, z]" because z is not a specific number.

     
  7. Aug 25, 2008 #6
    ok so using the points [0,0,1] and [0,0,2] on the z-axis will i get the parametric and symmetric equations as follows:

    parametric equations
    x = 0
    y = 0
    z = 1 + 2t

    and the symmetric equations

    [tex]\frac{x - 0}{0}[/tex] = [tex]\frac{y - 0}{0}[/tex] = [tex]\frac{z - 1}{2}[/tex]
    0 = 0 = [tex]\frac{z - 1}{2}[/tex]

    is this anywhere near correct?
    Thanks
     
    Last edited by a moderator: Aug 25, 2008
  8. Aug 25, 2008 #7

    Defennder

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    Well your parametric equations looks ok, though more complicated than necessary. The Cartesian equations (what you name "symmetric") doesn't appear correct.
     
  9. Aug 25, 2008 #8

    HallsofIvy

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    x= 0, y= 0, z= t describes exactly the same line.

    Well, first, (x-0)/0 and (y- 0)/0 are NOT equal to 0! Multiply the entire set of equations by 0.

    the "symmetric equations" describing the z axis are x= y= 0.
     
  10. Aug 25, 2008 #9
    Great i think i got it!
    THANKS!! :D
     
  11. Jan 7, 2009 #10
    Im struggling with this problem as well. I got parametric no problem, but I dont quite understand how to put the symmetric equation on paper. Also, I thought that symmetric equation only exists when the denominator does not equal to zero. How exactly would I write out the equation. Thanks
     
  12. Jan 7, 2009 #11

    Defennder

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    The denominator of what, to be exact? There shouldn't be any denominator equating to 0.
     
  13. Jan 8, 2009 #12

    HallsofIvy

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    If in setting up the formulas "blindly", you get
    [tex]\frac{x- 0}{0}= \frac{y- 0}{0}= \frac{z-1}{2}[/tex]
    Then, because you can't have a "0" denominator, as Defennder said, you must have x-0= 0 and y- 0= 0. Of course, that gives you nothing for the "(z-1)/2" to be equal to so the equations are simply x= y= 0.
     
  14. Jan 8, 2009 #13
    Okay I see. Just to make it clear for me, say you had an example where you symmetric equation was (x-1)/2=(y-2)/3=(z-4)/0 would that mean the final equation is (x-1)/2=(y-2)/3=z?

    Thanks again!
     
  15. Jan 9, 2009 #14

    HallsofIvy

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    No, why would you think so? That contradicts what Defennder and I just said! Because you cannot divide by 0, you must have z-4= 0 or z= 4. The equations for the line are (x-1)/2= (y-2)/3 and z= 4. Taking the joint value of (x-1)/2 and (y-2)/3 as t, that would correspond to parametric equations x= 1+ 2t, y= 2+ 3t, z= 4.
     
  16. Jan 9, 2009 #15
    Indeed it does! haha. I wrote that comment around 1am..just about time my brains stopped working haha. I understand what you mean. Thanks for you help guys! I can see clearly now!
     
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