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Equation of a line

  1. Oct 14, 2003 #1
    I tried to do this but I dont get the answer as in the book...


    A line passes through point (0,4). Its first direction angle is 60°, meaning alpha is 60°.

    I found that the second direction angles are 30° and 150° thats is beta is 30° and 150°. But what are the parametric equations of the line for each set of direction angles?

    Thx any help appreciated
     
  2. jcsd
  3. Oct 15, 2003 #2

    HallsofIvy

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    My understanding of "direction angles" for a line is that they are the angles the line makes with each of the coordinate axes. In a two dimensional problem, there are only two direction angles. In this case, since one angle is given as 60 degrees (the angle the line makes with the x-axis) the other angle has to be 30 degrees (in two dimensions, the angles have to add to 90 degrees- direction angles (and direction cosines) are more often used in three or more dimensions). I don't understand why "second direction angles" is plural. I also do not understand what you mean by " the parametric equations of the line for each set of direction angles".
     
  4. Oct 15, 2003 #3
    I agree with HallsofIvy that a third direction angle makes no sense in this problem. So ignoring that ...

    A vector along that line would be
    v = i + √(3)j
    or <x,y> = <1,&radic;(3)>

    So a vector equation for the line in the form
    r = r0 + tv (where r0 is the position vector of your point (0,4) and t is the parameter might be
    <x,y> = <0,4> + t<1,&radic;(3)> or
    <x,y> = <0+t, 4+t&radic;(3)>

    and then the parametric equations of the line would be
    x = t
    y = 4 + t&radic;(3)

    Is that the answer given in your book?
     
    Last edited: Oct 15, 2003
  5. Oct 15, 2003 #4
    That was probably too involved. Maybe this is a better answer:

    Since you know that &alpha; is 60o, you know that the slope of the line is tan60o = &radic;(3)

    So, an equation for the line through point (0,4) with slope &radic;(3) is
    y - 4 = &radic;(3) * (x - 0)
    y = x&radic;(3) + 4

    To parametrise this, let the parameter be t.
    Now, since there are no restrictions on x, we can simply let
    x = t
    and then, since we require that y = x&radic;(3) + 4, substitute the parameter t for x and you get
    y = t&radic;(3) + 4

    So those are the parametric equations.
     
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