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Vector Algebra: Finding a parallel vector

  1. Apr 10, 2016 #1

    squelch

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    1. The problem statement, all variables and given/known data

    A line is given by the equation ##x + 2y - 3z = 7##.
    Find any vector in the direction parallel to this line in the Cartesian coordinate system.

    2. Relevant equations

    I imagine that there are some fundamental relationships I am missing here that would make this more comprehensible to me.

    3. The attempt at a solution

    Looking at the problem at the surface it would make much more sense if he had described the equation as a "plane" rather than as a "line," and if he were asking for a vector either orthogonal or parallel to the plane. As it is, it's quite confusing. I've been sifting through materials but I can't get past that this looks like the equation for a plane, not a line, and it's worded (and written) very specifically -- the notation above is the notation the professor used (no vectors, "hats," etc, but that notation appears in other questions immediately around this one, so I feel he omitted it on purpose).

    What I think I want to do is convert this to a vector equation of a line in form
    [itex]\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over r} = \left\langle {{x_0},{y_0},{z_0}} \right\rangle + t\left\langle {a,b,c} \right\rangle [/itex], but I'm not sure how to do that from this form.
     
  2. jcsd
  3. Apr 10, 2016 #2

    Math_QED

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    What are the coordinates of the vector that describe the direction of this line? Once you have this, the problem is trivial. You might want to use parametric equations.
     
  4. Apr 10, 2016 #3

    squelch

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    Finding that is, I think, precisely my problem. I have a candidate relation:
    [tex]\hat x{A_x} + \hat y{A_y} + \hat z{A_z} = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over A} [/tex]
    But I'm not completely sure if I am using it correctly if I say:
    [tex]\hat x + \hat y2 - \hat z3 = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over A} [/tex]
     
  5. Apr 10, 2016 #4

    Math_QED

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    The coordinate of this vector is any (x,y,z) for which: x + 2y - 3z = 0

    This is the line going through (0,0,0), any coordinates on this line are coordinates of the direction of the vector you seek. Do you understand why?
     
  6. Apr 10, 2016 #5

    squelch

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    I do not, precisely. I'm still not getting past the idea that the equation seems to describe a plane to me. It seems that if I pick any z, there would be infinitely many x and y that satisfy the equation, and that I wouldn't be able to pick out a particular direction.
     
  7. Apr 10, 2016 #6

    Math_QED

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    As far as I can tell, this is a plane and not a line. Forget what I posted before (it's been a long time since I used this). The intersection of two planes would be the line, the equation represents an infinite amount of lines, aka a plane.
     
  8. Apr 10, 2016 #7

    Mark44

    Staff: Mentor

    I agree. The equation above is definitely that of a plane.
     
  9. Apr 10, 2016 #8

    squelch

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    So I suppose my only recourse here is to go ask for clarification from the actual professor.

    Thanks for the sanity check.
     
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