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Quick Vector intersection question

  1. Oct 29, 2006 #1
    So, I am to find a parametric equation for the line through [tex]P_1[/tex] and [tex]P_2[/tex] and then determine (if possible) the points at which the line intersects each of the coordinate planes.

    [tex]P_1[/tex] (5,-2,4)
    [tex]P_2[/tex] (2,6,1)

    [tex]\vec{P_{1}P_{2}}[/tex] <-3,8,-3>

    x = 5-3t
    y = -2+8t
    z = 4-3t

    So I got this far, but now I am not sure on what to do to find out if the line intersects the coordinate planes. I want to say that when the x, y, or z values are 0 in the parametric equations, is when the line intersects that plane, but I am not sure how to put it. If anyone can help me out there that would be great. Thanks!
     
  2. jcsd
  3. Oct 29, 2006 #2
    So I looked through my notes again to try and understand this, and I see that I found [tex]\vec{P_{2}P_{1}}[/tex] = <3,-8,3> and then found the parametric equation for the line from [tex]P_{2}[/tex] which gave me:

    x = 2+3s
    y = 6-8s
    z = 1+3s

    Seeing that I then set each one equal to eachother and solved for t and s and found that they all came out to the same equation of 1 = t+s. Can anyone tell me what this means and if I am on the right track here?
     
  4. Oct 30, 2006 #3

    HallsofIvy

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    Just as, on the x-axis, y is 0 and on the y-axis, x= 0, on the coordinate planes, the remaining variable is 0.

    On the xy-plane, z= 0. On the xz-plane, y= 0. On the yz-plane, x= 0.
    If x= 2+ 3s= 0, then s= -2/3. What are y and z?

    I have no idea what you mean by "set each one equal to each other"!
     
  5. Oct 30, 2006 #4
    So, yeah... I don't know what I was doing either in my second post. That step was totally not needed. I mean, the two sets of parametric equations that I have here in Post #1 and Post #2 are the same, but in different directions, right? So I could use either one to find the points at which this line crosses the three planes.

    If a line passes through a coordinate plane, then at that point in time when it crosses that plane, either the x, y, or z coordinate must be zero. So I can use this information to set either x, y, or z to zero, solve for t, or s, what have you, and find the remaining coordinates to construct the point at which the line passes through that coordinate plane. I think I got it.

    So for this question, where you have solved for s = -2/3 with the x coordinate, the point where the line:

    x = 2+3s
    y = 6-8s
    z = 1+3s

    Passes through the yz plane is (0, 34/3, -1) How does that look?
     
  6. Oct 30, 2006 #5

    radou

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    Actually, there's no time involved. But, if this makes it easier for you to think about parameters that way, then ok...
     
  7. Oct 30, 2006 #6

    HallsofIvy

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    Yes, that's correct.
     
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