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3d - Vector question

  1. Feb 4, 2009 #1
    where is the flaw in the following:
    it is possible to define a 3d vector using the form:
    (x,y,z) = (x_1,y_1,z_1)+t(a,b,c)
    this can then be parameterized to become

    x = x_1 + a*t
    y = y_1 + b*t
    z = z_1 +c*t

    so, for example, given the vector (x,y,z) = (1,2,3) + t(5,6,7)
    one can parametrize it to become:

    x = 1 + 5t
    y = 2 + 6t
    z = 3 + 7t

    solving the first equation for t gives
    t = (x-1)/5

    plugging that into the other two equations gives:

    y = 2 + 6/5 * (x-1)
    z = 3 + 7/5 * (x-1)

    solving the second one for x gives:

    x = (5*z-8)/7

    changing y = 2 + 6/5 * (x-1) to y = 2 + 6/5 * (2x-x-1)
    and in the -x spot putting -(5*z-8)/7 yeilds:

    y = 2 + 6/5 * (2x-(5*z-8)/7-1)

    simplifying gets:
    76 = 35 y + 30 z - 84 x

    which is the equation of a plane.
    this cannot possibly be right as i started out with a 2-d object (the vector)

    where did i mess up?
    (p.s. i tried to do all of this algebraically but it got too ugly for me)

  2. jcsd
  3. Feb 4, 2009 #2


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    First of all, that is not the equation of a vector, it is a vector equation of a line.

    In that equation there exists 2 vectors, the position vector r and direction vector v

    so in your example

    r = <1,2,3>
    v = <5,6,7>

    so your line starts at (1,2,3) and points in the direction <5,6,7>
  4. Feb 4, 2009 #3
    sorry if i used the wrong words. the fact remains that i began with a line/line segment/vector and ended up with an equivalent plane. this is not possible. where is the mistake?
  5. Feb 5, 2009 #4
    You think you have made a mistake where you haven't (except possibly a conceptual one).

    You get the equation of a plane containing the line.

    Your calculations are correct. All you have done is introduced extra solutions. What you have essentially proven is that if (x, y, z) = (1, 2, 3) + t(5, 6, 7) for some t, then 76 = 35y + 30z - 84x. It, of course, is not true that if 76 = 35y + 30z - 84x, then (x, y, z) = (1, 2, 3) + t(5, 6, 7) for some t, but you haven't proven this (which is good, since it's false).

    I imagine with some slight variations, you will find an equation for a different plane. However, it will still be a plane containing the original line; the two planes you found would intersect at this line.
  6. Feb 5, 2009 #5


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    There are only 3 forms of a line in 3d. The vector, parametric and symmetric forms.

    Taking your example gives:

    vector form: [tex]\vec{r}=<1,2,3> + t<5,6,7>[/tex]

    symmetric form: [tex]\frac{x-1}{5}=\frac{y-2}{6}=\frac{z-3}{7}[/tex]

    parametric form: [tex]x = 1+5t[/tex] , [tex]y = 2+6t[/tex] , [tex]z = 3+7t[/tex]

    Of course you can figure out how the line behaves in each plane (which I presume you are searching for) by solving the symmetric form gives you the trivial equation [tex]y=mx+c[/tex] on a particular plane. For example [tex]z=my+c[/tex] (zy plane) or [tex]z=mx+c[/tex] (zx plane) etc...

    Edit I left out the 2 point form which looks like: [tex]\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}[/tex] so there are 4 (viewed at http://www.cs.fit.edu/~wds/classes/cse5255/thesis/lineEqn/lineEqn#2ptEqn)
    Last edited: Feb 5, 2009
  7. Feb 5, 2009 #6
    let me try to explain:
    i got three parametric equations.
    i simplified them into one equation
    i got the equation of a plane.
    therefore, i know something went wrong.
    read the PDF to see exactly what i did.

    Attached Files:

  8. Feb 5, 2009 #7
    Read my post. Then read it again.
  9. Feb 5, 2009 #8


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    adriank is correct. All you have shown is a plane that the line lies upon. There are infinitely many.
  10. Feb 5, 2009 #9
    that is not the case, as i have locked it down to one plane. what happened to all of the others?
  11. Feb 5, 2009 #10


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    I will answer you question with another, how many equations did you solve for t? I see 3 different solutions for t

    Also, how many lines of the same slope and direction lie upon a single plane ?
  12. Feb 5, 2009 #11
  13. Feb 5, 2009 #12
    In response to djeitnstine, i only solved one.
    doe that mean anything?
  14. Feb 8, 2009 #13
    show that the line containing the points (0,0,5) and (1,-1,4) is perpendicular to the line with equations:


    i need help with this question..
  15. Feb 8, 2009 #14
    Find vectors parallel to each line, and show that their dot product is zero.
  16. Feb 8, 2009 #15
    thank you adriank:)


    solving the first equation for t gives;

    x/7=t so x=7t

    and other two:

    y-3/4=t so y=4t+3.
    z+9/3=t so z=3t-9.

    u=(7,4,3) and P= (0,3,-9) is on l.
    u is paralell to l.

    then we find another point on l.
    x=2 so y=29/7 and z= -57/7.


    as we have Q (0,0,5) and Q1 (1,-1,4) ; Q1-Q= (1,-1,-1).

    P(0,0,-9) and P1( 2,29/7,-57/7) ; P1-P= (2,8/7,6/7)

    P1P*Q1Q=2-8/7-6/7=0 so we can say Q1Q is perpendiclar to l.

    is it true? could you pls explain? i'm not sure..
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