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Need help with this equation of line

  1. Jan 28, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    I have been given a line in 3D space whose equation is x+y+2z-3=2x+3y+4z-4=0.
    But what kind of equation is this? This is not a standard equation and I have never seen such an equation of line. It seems to me a plane instead of a line. Can anyone help me convert this into standard form?
     
  2. jcsd
  3. Jan 28, 2014 #2

    vanhees71

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    You have two equations between the three coordinates, i.e., you have one independent parameter. You can take, e.g., [itex]x[/itex] as this independent parameter and solve for [itex]y[/itex] and [itex]z[/itex]. Then you get a parameter representation of a curve
    [tex]\vec{r}(x)=\begin{pmatrix} x \\ y(x) \\ z(x) \end{pmatrix}. [/tex]
    Because the system of equations is linear this curve is a straight line.
     
  4. Jan 28, 2014 #3

    ehild

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    x+y+2z-3=0 and 2x+3y+4z-4=0 are equations of two planes. All common points of both planes make the intersection of those planes, which is a straight line.

    ehild
     
  5. Jan 28, 2014 #4

    HallsofIvy

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    The question is what do you consider the "standard form" for a line? There are two forms usually used for lines in three dimensions:
    The "symmetric form" Ax+ By+ Cz+ D= Ex+ Fy+ Gz+ H= Ix+ Jy+ Kz+ L which is the form you already have and, as ehild says, can be interpreted as defining the line as the intersection of two planes.

    The "parametric form" where x= f(t), y= g(t), z= h(t) for some parameter t. This can be interpreted as an object moving along a line with t giving the time.
     
  6. Jan 28, 2014 #5

    Ray Vickson

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    It looks like you are given a badly-written pair of linear equations, namely, x+y+2z-3 = 0 and 2x+3y+4z-4=0. As others have said, this describes two (non-parallel) planes in 3D, so taken together they describe a line. You can get a parametric form by solving for two of the variables in terms of the third; for example, solve for x and y in terms of z. Again, as others have said, I don't know what you regard as a "standard form".
     
  7. Jan 28, 2014 #6

    utkarshakash

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    Thanks to all of you!
     
  8. Jan 29, 2014 #7

    utkarshakash

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    When I assume x=t z can be expressed in terms of t but what about y? By solving I get y=-2. How should I express it in terms of t?
     
  9. Jan 29, 2014 #8

    HallsofIvy

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    x+y+2z-3=2x+3y+4z-4=0.

    Taking x= t, we then have t+ y+ 2z- 3= 0 and 2t+ 3y+ 4z- 4= 0 or
    y+ 2z= 3- t and 3y+ 4z= 4- 2t.
    Multiplying the first equation by 2, 2y+ 4z= 6- 2t. Subtracting that from the second equation,
    y= (4- 2t)- (6- 2t)= -2 for all t. That is what you have. Excellent! That says that y is a contstant not depending on t. If you like, you can write it as y= -2+ 0t but just "y= -2" is correct.
     
  10. Jan 29, 2014 #9

    Ray Vickson

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    What is wrong with the simple parametric equation y = -2 + 0*t?
     
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