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Homework Help: Parametric representation of a straight line

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data

    There are two questions,
    1) straight line through (2, 0, 4) and (-3, 0, 9)
    2) straight line y = 2x + 3, z = 7x

    2. Relevant equations

    r(t) = a + tb = [a1 + tb1, a2 + tb2, a3 + tb3]

    The book also explains how to calculate the line if b is a unit vector, but I don't understand what it is trying to say (directional cosines?). For instance, the straight line in the xy-plane through A: (3,2) having slope 1 is:

    r(t) = [3, 2, 0] + t[1, 1, 0] = [3 + t, 2 + t, 0]

    I don't understand how exactly does having slope of 1 translate to [1, 1, 0].

    3. The attempt at a solution

    For (1), I get the vector (-5, 0, 5) by subtracting the vertices, but I have no idea how to get the second vector to plug into the equation above.

    For (2), (x, y, z) = (x, 2x + 3, 7x). Suppose x = 1, then (x, y, z) = (1, 5, 7). Suppose x = 0, then (x, y, z) = (0, 3, 0). Subtracting those to get a vector, I get (1, 2, 7). Like in (1), I don't know how to get a second vector.
     
  2. jcsd
  3. Sep 15, 2010 #2

    Mark44

    Staff: Mentor

    If the line has a slope of 1 and goes through (3, 2), then it also goes through (4, 3). A 3D vector in this direction is <1, 1, 0>.
    Think about it in terms of the sum of two vectors. The first goes from the origin to some point on the line. The second goes from that point to an arbitrary point (x, y, z) on the line.

    So r = <vector from origin to (2, 0, 4)> + <vector from (2, 0, 4) to (x, y, z)>

    You know that the vector <-5, 0, 5> has the same direction as the line, so the second vector above has to be a scalar multiple of <-5, 0, 5>.


    Let x = t. Then what would y be? What would z be?
     
  4. Sep 16, 2010 #3
    Suppose the second vector is (-1, 0, 1).

    Then r = a + tb = (-5, 0, 5) + t(-1, 0, 1) = (-5 - t, 0, 5 + t)
    Is this correct?

    What exactly is t? I don't know what it represents.
     
  5. Sep 16, 2010 #4

    Mark44

    Staff: Mentor

    No, because you are confusing points on the line with vectors parallel to the line. Your equation above says that (-5, 0, 5) is a point on the line, and I'm pretty sure it's not. Instead, use one of the points that were given - (2, 0, 4) or (-3, 0, 9).
    The title of this thread is Parametric representation of a straight line. t is the parameter.
     
  6. Sep 16, 2010 #5
    The formula uses bold letters for a and b, so I thought they were supposed to be vectors. Anyway, should it be this then?

    (2, 0, 4) + t(-5, 0, 5) = (2 - 5t, 0, 4 + 5t)

    The title of this thread is Parametric representation of a straight line. t is the parameter.[/QUOTE]

    Yes, but I don't understand exactly how it works. The book I'm using is very brief on that. Is the parameter something like the gradient for a two-dimensional line equation?
     
    Last edited: Sep 16, 2010
  7. Sep 16, 2010 #6

    Mark44

    Staff: Mentor

    Almost. The last part should be <2 - 5t, 0, 4 + 5t>. As a function, it's r(t) = <2, 0, 4> + t<-5, 0, 5>. The first vector, <2, 0, 4> gets you from the origin to the point (2, 0, 4) on the line. The second vector is parallel to the line. The parameter t scales the vector along the line so that you can get to any point on the line.

    As a check, notice that r(0) = <2, 0, 4>, the vector that takes you to the point (2, 0, 4). r(1) = <2, 0, 4> + 1<-5, 0, 5> = <-3, 0, 9>, the vector that takes you to the point (-3, 0, 9).

    The parametric equations for x, y, and z are x = 2 - 5t, y = 0, and z = 4 + 5t. These equations give you the x, y, and z values for any point on the line in terms of the parameter t.
    Yes, but I don't understand exactly how it works. The book I'm using is very brief on that. Is the parameter something like the gradient for a two-dimensional line equation?[/QUOTE]
    No. For lines in space (three dimensions), there really isn't the concept of slope (or gradient), at least not in the simple form of rise/run as it appears for lines in the plane.
     
  8. Sep 16, 2010 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The parameter is just some variable that labels where you are on the curve. For example, suppose the curve is a circle centered at the origin with radius 2 in the xy plane. You could express the circle parametrically as

    [tex]\vec{r} = (2\cos \theta, 2\sin\theta)[/tex]

    where the angle θ measured from the x-axis is the parameter. You could also write it as

    [tex]\vec{r} = (2\cos (s/2), 2\sin (s/2))[/tex]

    where the parameter s corresponds to the arc length along the circle from the point (2,0). Or you could come up with a parameter which has no physical/geometric interpretation. In all cases, though, as you vary the parameter, you'll move along the curve.
     
  9. Sep 16, 2010 #8
    Thank you both, I think I'm getting the hang of it. That last bit of my answer for question one was a silly calculation error.

    Now on to question two.

    (x, y, z) = (x, 2x + 3, 7x).

    Suppose x = 1, then (x, y, z) = (1, 5, 7).
    Suppose x = 0, then (x, y, z) = (0, 3, 0).
    Subtracting those, I get (1, 2, 7).

    r(0) = (1, 5, 7) + t(1, 2, 7)
    r(0) = (1 + t, 5 + 2t, 7 + 7t)

    According to question one, this is how I should do it, but since you mentioned

    Suppose x = 1, then (x, y, z) = (1, 5, 7).
    Suppose x = t, then (x, y, z) = (t, 2t + 3, 7t).
    (1, 5, 7) - (t, 2t + 3, 7t) = (1 - t, 2 - 2t, 7 - 7t) would be the vector.

    r(t) = (1, 5, 7) + t(1 - t, 2 - 2t, 7 - 7t)
    r(t) = (1, 5, 7) + (t - t^2, 2t -2t^2, 7t - 7t^2)
    r(t) = (1 + t - t^2, 5 + 2t - 2t^2, 7 + 7t - 7t^2)

    Is this correct? I'm not sure it is since the coordinates are quadratics.
     
  10. Sep 16, 2010 #9

    Mark44

    Staff: Mentor

    You're way off here. You didn't include the question, so I don't know what this problem is asking for.

    We can rewrite this as r(t) = <t, 2t + 3, 7t>. The parametric equations are
    x = t
    y = 2t + 3
    z = 7t

    r(0) = <0, 3, 0>.
    r(1) = <1, 5, 7>.

    These vectors extend from the origin to the points (0, 3, 0) and (1, 5, 7) on the line. A vector from one of these points to the other is <1, 2, 7>.

    Notice
    No, those are both expressions for r(t), not r(0). You have already figured out r(0) = <0, 3, 0>.
    No, it's not. See above.
     
  11. Sep 16, 2010 #10
    The question was to find a parametric representation of the straight line y = 2x + 3, z = 7x.

    So is it correct that the answer is r(t) = (1 + t, 5 + 2t, 7 + 7t)?

    By the way, is there any good online tutorial that teaches the basics of these?
     
  12. Sep 16, 2010 #11

    Mark44

    Staff: Mentor

    No, it's not. Please reread my previous post carefully. I answered this question in my last post.
     
  13. Sep 17, 2010 #12
    Ah, so it's r(t) = (t, 2t + 3, 7t)?
    Shouldn't the vector from (1, 5, 7) to (0, 3, 0) be (-1, -2, -7)?

    Also, for the first question, is it alright to reduce the vector (-5, 0, 5) to (-1, 0, 1) since we only need the direction (I think the magnitude is determined by t)?
     
  14. Sep 17, 2010 #13

    Mark44

    Staff: Mentor

    Yes. You have to admit that that problem was a "gimme."
    My vector goes in the opposite direction; that is, from (0, 3, 0) to (1, 5, 7). The two vectors are scalar multiples of one another.
    Sure, since all you need is a vector (of whatever length) in the right direction.
     
  15. Sep 17, 2010 #14
    Thanks for all the help.
     
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