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Coplanar vs Collinear

  1. Dec 19, 2015 #1
    Like the title says i'm not sure when the result is coplanar or collinear
    1. The problem statement, all variables and given/known data
    Are the following vectors in triple product coplanar or collinear?
    u = i + 5j -2k
    v= 3i - J
    w= 5i + 9j - 4k

    2. Relevant equations
    Cross product - place into a 3x3 matrix grid

    3. The attempt at a solution
    lateximg.png

    After placing them into a 3x3 grid the result is 0. Now i'm not sure if this means they are coplanar or collinear.
     
  2. jcsd
  3. Dec 19, 2015 #2

    Student100

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    Does the set of vectors span a line or a plane?
     
  4. Dec 19, 2015 #3
    Not sure honestly, I changed the question around for my own use, but in the original question it says to verify why it is coplanar, so id assume it to span across a plane
     
  5. Dec 19, 2015 #4

    Student100

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    Load the vectors into a matrix and row reduce, then look at how you can write the result as a linear combination. That should be a good give away to whether you're dealing with a line or a plane.
     
  6. Dec 19, 2015 #5
    Seems I may be missing something, i still can't tell when its a plane or line
     
  7. Dec 19, 2015 #6

    Student100

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    Go ahead and row reduce the above, what's a basis for the subspace? How can you graphically represent the subspace?
     
  8. Dec 19, 2015 #7

    Student100

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    If what I'm saying sounds like gibberish, then by taking the scalar triple product and obtaining a result of zero lets you know the signed area of the parallelepiped is zero, thus the vectors are coplanar. You didn't do that correctly in the above, however.

    It also can be misleading, as collinear subspaces are coplanar, but coplanar subspaces aren't always collinear. The geometric approach of finding a basis for the subspace spanned by the vectors is a better approach to the question as posed.

    I'm also tired, so maybe I'm not doing a good job here articulating my thoughts.
     
  9. Dec 19, 2015 #8

    HallsofIvy

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    This is the "determinant", not just a "3x3 grid"!
     
  10. Dec 19, 2015 #9

    ehild

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    Go back to the definitions. Two vectors are collinear when they have the same direction, only the magnitudes differ. You get one vector with multiplying the other vector with a scalar. Is any of the given vectors u, v, w, a multiple of an other one?
    Three vectors are co-planar if they are in the same plane. A plane has two dimensions, so two independent vectors. The third one is linear combination of the other two:
    w=a u+b v. Write out this equation in components. Can you solve it?
    The three vectors are independent (span 3 dimensions) if you can not combine one of them from the other two, that is, a linear combination can be zero, au +bv +cw =0 only when all coefficients a, b, c, are zero. You get a non-zero solution only when the vectors are dependent, span a plane. Arranging the vector components so as they form the rows of a determinant, the value of the determinant is zero if and only if one row can be combined from the other two. You calculated the determinant and got zero. So are u,v,w dependent or independent vectors?
     
  11. Dec 21, 2015 #10
    Here's a helpful hint that may help you. The determinant
    $$\begin{vmatrix}
    1&5&-2 \\
    3&-1&0 \\
    5&9&-4
    \end{vmatrix}
    = \begin{vmatrix}
    u_x&u_y&u_z \\
    v_x&v_y&v_z \\
    w_x&w_y&w_z
    \end{vmatrix}$$
    may be rewritten as
    $$\begin{vmatrix}
    u_x&u_y&u_z \\
    v_x&v_y&v_z \\
    w_x&w_y&w_z
    \end{vmatrix} = \vec u \cdot ( ~ \vec v \times \vec w~ )$$
    You have already figured out the determinant is zero. What does this mean?
     
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