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Determining a path to represent a plane?

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a path [itex]\gamma (t)[/itex] representing the triangle (1, 0, 0) to (0, 2, 0) to (0, 0, 3) to (1, 0, 0)

    2. Relevant equations
    None


    3. The attempt at a solution
    Well I followed the steps on how to get an equation of a plane that contains the 3 vertices and I got [itex]6x + 3y + 2z - 6 = 0[/itex].

    Solving for z, I have [itex]z = 3 - (3/2)y - 3x[/itex] so the path along the triangle can be represented as:

    [itex]f(x,y) = (x, y, 3 - 3x - (3/2)y)[/itex]

    But this is not a function of t. How do I make it only a function of t? Do I set both x and y = t so that [itex]\gamma (t) = (t, t, 3 - 3t - (3/2)t)[/itex]?
     
  2. jcsd
  3. Feb 4, 2013 #2

    LCKurtz

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    What you have described is a parametric equation of the plane containing the three points. Surfaces need two parameters which, in your example, are x and y. You don't need the plane to parameterize the boundary. Remember if P and Q are two points, then the vector equation of the line from P to Q is$$
    \vec \gamma(t) = (1-t)P + tQ$$
    The point will move from P to Q as t goes from 0 to 1. You will need three pieces to represent the three sides.
     
  4. Feb 5, 2013 #3
    Hmm, I don't think we make line equations like that in my class. We do it in the form l(t) = p + tv, where p is one of the two points and v is the direction vector.

    Finding the direction vectors for each, I get:

    Let $$\vec v_1 = (0,2,0) - (1,0,0) = (-1,2,0)$$
    $$\vec\gamma_1(t) = (1,0,0) + t(-1,2,0)$$

    Let $$\vec v_2 = (0,0,3) - (0,2,0) = (0,-2,3)$$
    $$\vec\gamma_2(t) = (0,2,0) + t(0,-2,3)$$

    Let $$\vec v_3 = (1,0,0) - (0,0,3) = (1,0,-3)$$
    $$\vec\gamma_3(t) = (0,0,3) + t(1,0,-3)$$

    Now that I have line equations for each part of the triangle, what would I do to combine them to make ##\vec\gamma(t)##?
     
  5. Feb 5, 2013 #4

    HallsofIvy

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    Okay, this goes from (0, 2, 0) to (1, 0, 0) as t goes from 0 to 1. If you were to replace t with 3s, getting [itex]\vec{\gamma(s)}= (1, 0, 0)+ 3s(-1, 2, 0= (1, 0, 0)+ s(-3, 6, 0)[/itex], you have a vector function that goes from (0, 2, 0) to (1, 0, 0) as s goes from 0 to 1/3.

    Okay, this goes from (0, 2, 0) to (0, 0, 3) as t goes from 0 to 1. Since we already have "from (0, 2, 0) to (1, 0, 0)", in order to be "moving" around the triangle, we would prefer to go from (0, 0, 3) to (0, 2, 0). We could do that by replacing t with r= 1- t, with r from 0 to 2. But, further, because that will attach to the previous segment, we want it to end the chain- that is we want to have the parameter from 2/3 to 1. s will go from 2/3 to 1 as r goes from 0 to 2 if s= 2/3+ (1/3)r which is the same as r= 3(s- 2/3)= 3s- 2 . That means that we must have t= 1- (3s- 2)= 2- 3x. [itex]\vec{\gamma(s)}= (0, 2, 0)+ (2- 3s)(0, -2, 3)= (0, 0, 6)+ (0, 6s, 9s)[/itex] for s from 2/3 to 1.

    Okay, this goes from (1, 0, 0) to (0, 0, 3) as t goes from 0 to 1 and this side is the middle of the chain. We want it to go from (1, 0, 0) to (0, 0, 3) as s goes from 1/3 to 2/3 so we want t= 3(s- 1/3)= 3s- 1. [itex]\vec{\gamma(s)}= (0, 0, 3)+ (3s-1)(1, 0, -3)= (-1, 0, 6)+ (3s, 0, -9s)[/itex] for s from 1/3 to 2/3.

    I you don't like piecewise defined functions, you can use the "Heaviside function", H(x) which is 0 for x< 0, 1 for [itex]x\ge 0[/itex]. If F(x)= f(x) for x< a, g(x) for [itex]a\le x< b[/itex], and h(x) for [itex]b\le x[/itex], then you can write
    F(x)= f(x)+ H(x- a)(g(x)- f(x))+ H(x- b)(h(x)- g(x)).
     
    Last edited: Feb 5, 2013
  6. Feb 5, 2013 #5
    EDIT: I think I've figured out why you split the interval into three.

    So the piecewise function would look like:

    $$\vec\gamma(s) =
    \begin{cases}
    (1,0,0)+s(−3,6,0), & 0 \le t < \frac{1}{3}\\
    (−1,0,6)+s(3,0,−9), & \frac{1}{3} \le t < \frac{2}{3}\\
    (0,0,6)+s(0,6,9), & \frac{2}{3} \le t \le 1
    \end{cases}$$
     
    Last edited: Feb 5, 2013
  7. Feb 5, 2013 #6

    LCKurtz

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    In your second equation replace ##t## by ##t-1## and let ##t## go from 1 to 2. Similarly in the third equation so the result is ##t## varying from 0 to 3.
     
  8. Feb 5, 2013 #7
    Thank you both. I see now that there are infinitely many ways to parametrize the path. I think the most easiest would be to go from 0 to 1, 1 to 2, then 2 to 3 as LCKurtz described it.
     
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