1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)

    thanks.
     
  2. jcsd
  3. Feb 4, 2009 #2

    djeitnstine

    User Avatar
    Gold Member

    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

    djeitnstine

    User Avatar
    Gold Member

    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

    djeitnstine

    User Avatar
    Gold Member

    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

    djeitnstine

    User Avatar
    Gold Member

    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:

    x/7=y-3/4=z+9/3

    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:)
    so;

    x/7=y-3/4=z+9/3=t

    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.

    P1(2,29/7,-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..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: 3d - Vector question
  1. 3D Plane question (Replies: 3)

  2. 3D geometry (Replies: 18)

  3. Trig in 3D (Replies: 1)

Loading...