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How to prove if 3 arbitrary vectors are coplanar

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Let s and t be scalars. show that u, v and su + tv are coplanar.
    u and v are arbitrary vectors,

    2. Relevant equations


    3. The attempt at a solution
    Well, i suppose if they are coplanar, triple scalar product of them should be 0, right? and su+tv is evidently a linear combination of u and v. but i got a little confused when vectors are arbitrary and not really sure how to show it properly. any help would be appreciated, thanks.
     
  2. jcsd
  3. Dec 9, 2015 #2

    jedishrfu

    Staff: Mentor

    Try using the triple scalar product and some vector algebra to reduce the expression.
     
  4. Dec 9, 2015 #3
    ok i already went down that road and got lost, if i had numbers, it would be easier but how can i prove that they are coplanar if all i have is arbitrary vectors?
     
  5. Dec 9, 2015 #4

    Mark44

    Staff: Mentor

    Here's how I would do it:
    1. Calculate n = u X v. (I'm assuming that u and v are vectors in R3)
    2. Show that ##n \cdot (su + tv) = 0##.

    Geometrically, for all real values of s and t, su + tv defines the plane that u and v lie in.
     
  6. Dec 9, 2015 #5
    That is the general idea, yes but still, how to calculate n=u X v when i do not even have numbers, i think i'm missing your point here, can you be just a little more specific? i tried defining the vectors as u=(u1,u2,u3) and v=(v1,v2,v3) and then calculating the triple scalar product. it got me nothing.
     
  7. Dec 9, 2015 #6

    Mark44

    Staff: Mentor

    Using the coordinates of u and v, as you show above, what do you get for u x v? You know how to calculate a cross product, right?

    Once you get that, show that ##n \cdot (su + tv) = 0##. I don't think you need to use a scalar triple product.
     
  8. Dec 9, 2015 #7

    jedishrfu

    Staff: Mentor

    U x V . ( sU + tV) = U x V . sU + U x V . tV

    Using some more vector identities you can demonstrate that the expression evaluates to zero hence sU + tV is what?
     
  9. Dec 9, 2015 #8
    even if i dont use triple scalar product and do what you suggest, i'd still have to calculate cross product (which i know how to do it), correct me if i'm wrong but;

    u X v= det |(i,j,k),(u1,u2,u3),(v1,v2,v3)| right? that gives you a gigantic equation of nothing when calculated, which cannot be simplified or arranged, let alone taking the product of it with another arbitrary (su+tv).
     
  10. Dec 9, 2015 #9
    again, i cannot calculate the cross product of two arbitrary vectors in r3. it becomes a giant vector equation which is non-arrangeable. cross product seems like a dead end.
     
  11. Dec 9, 2015 #10

    Mark44

    Staff: Mentor

    It gives you a vector expression with six terms, so it's a gross exaggeration to call it a "gigantic equation." Instead of dreaming up reasons why you can't do this, why don't you put some effort into actually trying to work the problem?
    When you calculate ##n \cdot (su + tv)## all of the terms drop out, which is what you want to happen.
     
    Last edited: Dec 10, 2015
  12. Dec 11, 2015 #11
    Maybe you know that uxv is ortogonal to u hence you could deduce u. (uxv) ?
     
  13. Dec 11, 2015 #12

    Mark44

    Staff: Mentor

    I don't see how that gets the OP closer to showing that u, v and su + tv are coplanar.
     
  14. Dec 12, 2015 #13

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @cptcook: You don't need to mess with components. The above posts have shown the way. You yourself noted you just need to show the triple scalar product ##u\times v \cdot(su+tv)=0##. Post #11 reminds you that ##u \times v## is perpendicular to both ##u## and ##v##. Post # 7 shows you how to expand ##u\times v \cdot(su+tv)##. You should be able to see why that is zero without expanding by components.
     
  15. Dec 13, 2015 #14
    This is a bit too easy : you could use the definition of the plane spanned by u and v
     
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