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Homework Help: Proving 3 vectors are coplanar

  1. Feb 11, 2007 #1
    Prove that the vectors a=3i+j-4k
    b= 5i-3j-2k
    c= 4i-j-3k are COPLANAR

    2. Relevant equations

    (axb)c=0





    3. The attempt at a solution

    If (axb)c=0 then c is orthogonal to axb and therefore c is in the plane perpendicular to axb since axb is perpendicular to both a and b, both a,b,c are in the same plane perpendicular to axb.
    My problem is that my answer when using the formula doesnt equal 0 meaning that it isn't coplanar which means its wrong because i have to prove it is. Can someone show me a walkthrough and how to use this formula because im doing something wrong :frown:
     
  2. jcsd
  3. Feb 11, 2007 #2

    radou

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    Another approach would be to test if they are linearly independent. If they are not, then they are coplanar.
     
  4. Feb 11, 2007 #3
    Thank you ill have a go at it... And i just realized something... this is in the wrong forum right?
     
  5. Feb 11, 2007 #4

    radou

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    Actually, it isn't. :smile:
     
  6. Feb 11, 2007 #5
    hmm then it means there is a mistake in my textbook...
    Grrr i hate when tht happens so i wasn't misusing the formula god :P just wasted 3 hours of my time but atleast i know the problem :)
    thanks for your help
     
  7. Feb 11, 2007 #6
    oh wait... i completely misread... i understood you said it isnt as in the question isnt coplanar :P lol sorry
     
  8. Feb 11, 2007 #7

    radou

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    The question states that you have to check if the vectors ARE coplanar, right? But it doesn't really matter, since all you need to know is that, as I already wrote, if they are linearly independent then they AREN'T coplanar. If they, of course, happen to be linearly dependent, then they ARE coplanar.
     
  9. Feb 11, 2007 #8
    I have been looking on the internet to find methods on how to work out linear independency but i haven't found any, any that i understand.
    do you happen to know any tutorials for this that are simple to understand
     
  10. Feb 11, 2007 #9

    Hootenanny

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    One method (the one I find easiest) is to place the three vectors into a 3x3 matrix and find the determinant. If the determinant is non-zero then the vectors are linearly independent. This is equivalent to using the formula you originally posted.
     
  11. Feb 11, 2007 #10

    radou

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    All you need is the definition of linear independency of a set of vectors. After you find it, you can create an equation from your vectors and the solution of this equation will tell you everything about their dependency/independency, i.e. if they're coplanar or not.

    Edit: what Hootenanny suggested is something that you'll need to solve after developing the vector equation (i.e. the system of equations) I was talking about.
     
  12. Feb 11, 2007 #11
    I dont know how to use Matrixs because i havent really learnt about them... But considering My problem how would i input the data into this formula (axb)c=0 my answer always gives something else :S
     
  13. Feb 11, 2007 #12

    radou

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    You don't need to use matrices, it's just a formality.

    OK, you have three vectors, a, b, and c. They are linearly independent if the equation

    a x + b y + c z = 0 implies x = y = z = 0, where x, y and z are scalars, i.e. real numbers in this case. Can you set up that equation and try to solve it for x, y, and z (or call them whatever you like)?
     
  14. Feb 11, 2007 #13
    Ill have a think about it ill get back to you later and thank you :)
     
  15. Feb 11, 2007 #14
    Isnt it true that A set of points is said to be COPLANAR if and only if they lie on the same geometric plane THREE points are ALWAYS COPLANAR.
    Could i use this statement to answer the question?
     
  16. Feb 11, 2007 #15
    If the three vectors are coplanar then the volume of the parallelepiped spanned by the vectors will be zero. This volume is given by the vector triple product, so the vector triple product - as given by your formula - will be zero.

    It is.

    http://en.wikipedia.org/wiki/Parallelepiped
     
  17. Feb 11, 2007 #16

    Hootenanny

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    You don't have three points, you have three lines.
     
  18. Feb 11, 2007 #17

    HallsofIvy

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    I'm coming in late to this but I just did a quick calculation of (axb).c and it DOES in fact, equal 0. If you are still having difficulty, show us exactly what you did.

    As Hootenanny said, the simplest way to do the triple product is to use the three vectors as the rows of a single determinant. That should be 0.
     
  19. Feb 11, 2007 #18
    Re:

    Could you tell me which values you put in the equation to help me understand it?
     
  20. Feb 11, 2007 #19

    Hootenanny

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    Okay, I am assuming you know how to calculate 2x2 determinants?
     
  21. Feb 11, 2007 #20

    HallsofIvy

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    YOU said you knew how to do this but were just getting the wrong result! Are you now saying you have no idea how to set it up?


    You said the vectors were a=3i+j-4k, b= 5i-3j-2k, c= 4i-j-3k .
    The cross product of a and b is, of course,
    [tex]\left|\begin{array}{ccc} i && j && k \\ 3 && 1 && -4 \\ 5 && -3 && -2\end{array}\right|[/tex]
    and you want to take the dot product of that with 4i- j- 3k.

    But since the dot product would just multiply corresponding components, that is, 4 times the i component, -1 times the j component and -3 times the k component, that is exactly the same as expanding
    [tex]\left|\begin{array}{ccc} 4 && -3 && k \\ 3 && 1 && -4 \\ 5 && -3 && -2\end{array}\right|[/tex]
    by the first row.

    Is that what you did? What did you get?
     
  22. Feb 12, 2007 #21
    Re:

    Hello, i asked my Math teacher about this question and showed him the various methods which you guys had shown me and he said that we haven't gotten so far yet in vectors and gave me a hint which was to put the three vectors in a equation. What i did was the following :

    a= | 3 | b= |5 | C= |4 |
    | 1 | |-3 | |-1 |
    | -4 | |-2 | |-3 |


    C= λa+μb

    ( 4 ) (3 ) (5 )
    (-1 )= λ(1 )+μ (-3 )
    (-3 ) ( -4) (-2 )

    giving the following equations:

    3λ+5μ-4=0 (1)
    λ -3μ+1=0 (2)
    -4λ -2μ+3=0 (3)

    using simultaneous equations: [ elimination]

    i got λ= 0.5
    μ=0.5

    The values are compatible for the 3 equations therefore the 3 lines are coplanar
    ---------------------------------------------------------------------
    Is this a correct way of doing it???
    By the way thank you people for taking your time with me i really appreciate it and also i think this forum is great has a lot from where you can learn :)
     
  23. Apr 7, 2010 #22
    Hi
    bxc = 7i +7j +7k that is the cross product between b and c
    [bxc].a = (7i+7j+7k).(3i+j-k)= 21+7-28=0
    the triple product =0 , therefore b, c and a are coplaner. as simple as this
    Adam Hanna
    Parade College
    Melbourne[/B]









     
  24. Apr 7, 2010 #23
    Hi
    bxc = 7i +7j +7k that is the cross product between b and c
    [bxc].a = (7i+7j+7k).(3i+j-k)= 21+7-28=0
    the triple product =0 , therefore b, c and a are coplaner. as simple as this
    Adam Hanna
    Parade College
    Melbourne[/b]
     
  25. Apr 9, 2010 #24
    Just out of curiosity: how do you compute (axb)c without using matrices? That's in fact the determinant method that Hootenanny suggested. And if you get that to be non-zero, then the three vectors are *not* coplanar. So you were right all the time, your teacher isn't (if your calculation is right).
     
  26. Apr 9, 2010 #25

    Gib Z

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    To calculate the triple product without matrices one would just need to know how to find cross products and dot products in general, and apply them here. However, seeing as the component form of the cross product is not all that easy to remember it may be prudent to learn how to evaluate determinants of 2x2 and 3x3 matrices as soon as possible, even if you don't know what they really mean yet. Cross and Triple products are easily evaluated this way.
     
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