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Homework Help: Vector dependence, please help

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that vectors [tex]\vec{u}[/tex], [tex]\vec{v}[/tex]and [tex]\vec{w}[/tex] are coplanar if and only if vectors [tex]\vec{u}[/tex], [tex]\vec{v}[/tex]and [tex]\vec{w}[/tex]are linearly dependent

    2. Relevant equations



    3. The attempt at a solution

    This may sound like an awkward question, but I am having much difficulty proving this. I tried to multiply each vector by a scalar, but got stuck, below is my solution, up until I got stuck.Any help is much appreciated!

    We need to write [tex]\vec{u}[/tex]= c1 [tex]\vec{v}[/tex]+ c2 [tex]\vec{w}[/tex]as:



    [x1, y1, z1] = c1 [x2, y2, z2] + c2 [x3,y3, z3]


    [x1, y1, z1] = [c1 x2, c1 y2, c1 z2] + [c2 x3, c2 y3, c2 z3]


    x1 = c1 x2 + c2 x3
    y1 = c1 y2 + c2 y3
    z1 = c1 z2 + c2 z3


    This is where I got stuck, and Im not sure what to do next.

    Thanks,
     
  2. jcsd
  3. Apr 10, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Hmm.. I think the easiest way to prove this might be to realize that if 3 vectors are coplanar, then there will be some normal vector [itex]\textbf{n}[/itex] (the normal to the plane) that is perpendicular to all three and then look at some dot (and perhaps also cross) products.
     
  4. Apr 10, 2010 #3
    Thats basically the problem, I'm supposed to prove this without using either dot or cross products Also, do you think that my working is fine, or will not work?

    Thank you very much
     
  5. Apr 10, 2010 #4

    Mark44

    Staff: Mentor

    This is an if and only if proof, which means you have to prove two things:
    1. u, v, and w are coplanar ==> u, v, and w are linearly dependent.
    2. u, v, and w are linearly dependent ==>u, v, and w are coplanar.

    Maybe you already knew that.

    To get you started, if u, v, and w are coplanar, then u = c1v + c2 w for some nonzero constants c1 and c2. So what can you say about the equation u - c1v - c2 w = 0?
     
  6. Apr 10, 2010 #5
    That means that they are linearly dependent...I completely understand the concept, but am stuck on the proof part; I'm a little confused on how to prove this using c1 and c2..

    Thanks.
     
  7. Apr 10, 2010 #6

    gabbagabbahey

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    Gold Member

    Not necessarily, if v and w are collinear, this need not be true. You need to examine two cases:

    (1) At least two of the vectors are collinear
    (2) None of the vectors are collinear.
     
  8. Apr 10, 2010 #7

    Mark44

    Staff: Mentor

    I was afraid that might not be enough. It's a real restriction not having access to the dot product and cross product.
     
  9. Apr 10, 2010 #8
    I think you're on the right track, could you please elaborate on your method..

    Thanks!
     
  10. Apr 11, 2010 #9

    Mark44

    Staff: Mentor

    Per gabbagabbahey's suggestion, look at two cases for u, v, and w.
    1. At least two of the vectors are collinear.
    2. None of the vectors is collinear.

    In the first case, you can assume without loss of generality that u and v are collinear, i.e., that u = kv, for some constant k. Use that to show that the equation c1u + c2v + c3w = 0 has a solution where not all the constants are zero.

    In the second case, use what I said toward the end of post 4 to show that the vectors are linearly dependent (i.e., that the equation c1u + c2v + c3w = 0 has a solution where not all the constants are zero).
     
  11. Apr 12, 2010 #10
    ok, so I get it up till c1u + c2v + c3w = 0...How exactly do I show that c1u + c2v + c3w is equal to 0, for non zero scalars c1, and c1..

    I would really appreciate it if you could please help me with this,

    thanks for your help
     
  12. Apr 12, 2010 #11

    Mark44

    Staff: Mentor

    Depends on which case you're working on. If it's the first case, u = kv, so replace u in the c1u + c2v + c3w = 0 equation.

    If it's the second case, see what I said in post 4.
     
  13. Apr 12, 2010 #12
    it is the second case, where u = c1v + c2w, for non zero scalars, I don't understand how to actually prove what you said in post 4..or writing that would be sufficient..

    Thanks,
     
  14. Apr 12, 2010 #13
    ok, I get it...I was wondering how you got the u - c2v - c3w?

    sorry, I just started vectors, and am a little confused with them..
     
  15. Apr 12, 2010 #14

    Mark44

    Staff: Mentor

    If u = c1v + c2w, then u - c1v - c2w = 0.

    So here I have a solution to the equation c1u + c2v + c3w = 0 where not all the constants are zero.
     
  16. Apr 12, 2010 #15

    ok, so I get 0 when I substitute u = c1v + c2w into the equation: is this correct?, since these combinations need to be linearly dependent
    inorder to be coplanar, which we proved, as u = c1v + c2w...
     
  17. Apr 12, 2010 #16

    Mark44

    Staff: Mentor

    No, that's not it at all. Please tell me the definition of linear dependence, in the context of three vectors u, v, and w.
     
  18. Apr 12, 2010 #17
    Linear dependence:

    c1u + c2v + c3w= 0
     
  19. Apr 12, 2010 #18

    Mark44

    Staff: Mentor

    Haven't you noticed that exactly the same equation is used to define linear independence? IOW, there is more to dependence or independence than just that equation. What does the complete definition say?
     
  20. Apr 12, 2010 #19
    ok, so from what I've been taught, if you can express one of the vectors as a linear combination of the other vectors, then they are linearly dependent vectors...please remember that this is high school level math. I'm really confused on how to prove this, like I asked a tutor online , and he said that since these vectors are coplanar, so all we need to prove is that u - c1v + c2w = 0, so that c1, and c2 are non zero scalars...How could you prove this?
    help is much appreciated..

    Thanks!
     
  21. Apr 12, 2010 #20

    Mark44

    Staff: Mentor

    This is not how linear dependence is defined.

    For example, consider u = <1, 0, 0>, v = <0, 1, 0> and w = <1, 1, 0>.

    I can set c1u + c2v + c3w = 0, and that alone doesn't tell me anything, because this equation is always true if c1 = c2 = c3 = 0. For these vectors, however, it turns out that 1u + 1v - 1w = 0 is also true, so I have found a solution for the equation c1u + c2v + c3w = 0, where not all of the constants are zero.

    Here's another example, this time with u and v the same, but w = <1, 0, 1>.

    I can again set c1u + c2v + c3w = 0, which I can always do, no matter what vectors I'm talking about. If I solve for the constants this time, I have
    c1<1, 0, 0> + c2<0, 1, 0> + c3<1, 0, 1> = <0, 0, 0>
    ==> c1 + c3 = 0
    c2 = 0
    c2 + c3 = 0

    Substituting 0 for c2 in the 3rd equation gives me c3 = 0. Substituting 0 for c3 in the first equation gives me c1 = 0.

    The only solution to the equation c1u + c2v + c3w = 0 is c1 = c2 = c3 = 0, and there are no other solutions.

    The upshot of this is that for an equation c1u + c2v + c3w = 0, u, v, and w are linearly independent if the only solution is c1 = c2 = c3 = 0, AKA the trivial solution. For the same equation, u, v, and w are linearly dependent if there is some solutions with at least one of the constants not equal to zero. The solution with all three constants equal to 0 applies in both cases.

    Now, if you have u = av + bw, is there a solution to the equation u1 - av - bw = 0 other than the trivial solution? There obviously is.
     
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