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Find the coordinates of a point where a line intersects the y-axis.

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    The following equation describes a straight line:
    ⟨x, y, z⟩ = ⟨−1, 0, −2⟩ + t⟨1, 2, 2⟩
    Find the coordinates of the point where this line intersects the y-axis.

    2. Relevant equations

    Equation of a Line: r = ro + tv

    3. The attempt at a solution

    I'm not really sure how y-intercepts work in 3d and I'm having trouble even attempting a solution. My guess was the ro point <-1,0,-2> because the y coordinate is zero but I don't think that it would be that easy. Also, this is my first time posting a question on the Physics Forum site so please go easy on me if I messed anything up :)
     
  2. jcsd
  3. Jan 21, 2013 #2

    Dick

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    It's almost that easy. <-1,0,-2> is in the x-z plane, it isn't on the y-axis. A point on the y-axis looks like <0,y,0>.
     
  4. Jan 21, 2013 #3
    So then it has to be the point when the x and z coordinates are zero?
     
  5. Jan 21, 2013 #4

    Dick

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    Yes. What is it?
     
  6. Jan 21, 2013 #5
    Well the equation multiplied out would be

    <t-1,2t,2t-2>

    so to make the x and z zero, x has to be 1 and y has to be 1.

    So would <1,1,1> be a point that this line intersects the y axis?
     
  7. Jan 21, 2013 #6

    Dick

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    Not quite sure what you are thinking there. I would say to put x=0 you should pick t=1.
     
  8. Jan 21, 2013 #7
    Okay, so where t=1 the coordinates are <0,0,0> and thus intersect the y-axis at that point?
     
  9. Jan 21, 2013 #8
    Wait, no it would be <0,2,0>
     
  10. Jan 21, 2013 #9

    Dick

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    ???Can you explain how putting t=1 into <t-1,2t,2t-2> gives you <0,0,0>???
     
  11. Jan 21, 2013 #10
    Sorry, bad algebra. <0,2,0>
     
  12. Jan 21, 2013 #11

    Dick

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    Ok. Yes, it's <0,2,0>.
     
  13. Jan 21, 2013 #12
    Thanks for the help, I really appreciate it!
     
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