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Scalar Equations!

  1. Mar 25, 2004 #1
    Scalar equations such as y=2x+3 actually generate POINTS which are collinear. A vector equation, as the name implies, generates VECTORS, and these vectors are definitely NOT COLLINEAR.

    How then can we say that an equation such as
    r = (2,1,3) + t(1,2,4) is the "equation of a line"?

    Also, why is it not possible to produce a scalar equation for a line in 3-D?
     
  2. jcsd
  3. Mar 25, 2004 #2
    Consider the physical meaning of the vector r. That is called the position vector. It represents a spatial displacement from a point called the origin. The tip of the vector defines a point and it is that point we are refering to as the position.

    Since r is the position vector which traces out a line, i.e. the tip of the vector traces out a line, the its called the equation of a line. Likewise the tip of the vector

    [tex] \mathbf{r} = cos \theta \mathbf{i} + sin \theta \mathbf{j}[/tex]

    traces out a circle. Therefore one can say that this is the equation of a circle.
     
  4. Mar 26, 2004 #3

    matt grime

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    It is possible to produce a set of scalar equations that generate a line in R^3

    eg the line (1,2,3) + t(2,3,1) is also described as

    (x- 1)/2 = (y-2)/3 = z-3


    just as the original scalar equation you gave is expressible as a vector equation:

    L = (0,3) +t(1,2)
     
    Last edited: Mar 26, 2004
  5. Mar 26, 2004 #4

    selfAdjoint

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    This is called a parametric rep, with t as the parameter. For three dimension look up "direction cosines".
     
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