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Linear Algebra (Linearly Dependent)

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    a = [2, 2, 2]
    b = [3, 0, 1]

    Find all vectors c so that the vectors a, b, c are linearly dependent.


    2. Relevant equations

    (a x b) . c
    ( . = dot product)....is this how i'm supposed to get started?

    linearly dependent would mean make it equal to zero right?



    3. The attempt at a solution

    when i use the triple product i get 2c1 + 4c2 - 6c3 = 0
    i dont know where to go from there
     
  2. jcsd
  3. Jan 25, 2010 #2

    Dick

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    That's one way to express it. The set of all vectors c=[c1,c2,c3] such that 2c1 + 4c2 - 6c3 = 0. If you want to go one step further you solve your equation for, say c1. Then you could express c in terms of just the two parameters c2 and c3. If you think about it a little bit, you could probably find a way to express c in terms of two parameters without even going through the triple product.
     
  4. Jan 25, 2010 #3
    Thanks =)

    So if I was going to solve for c1 would it be something like

    c1 = -2c3 + 3c3
     
  5. Jan 25, 2010 #4

    Dick

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    What do you mean 'something like'?? Sure! So c=[-2c3+3c3,c2,c3] for any choice of c2 and c3. Now can you find a way to bypass the triple product?
     
  6. Jan 25, 2010 #5
    okay sweet, thank you! =)

    Not really, i'm not sure how it would work the other way without the triple product.
     
  7. Jan 25, 2010 #6

    Dick

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    How about c=s*[2, 2, 2]+t*[3, 0, 1]?
     
  8. Jan 25, 2010 #7

    Mark44

    Staff: Mentor

    What Dick is suggesting here is the simplest, most straightforward approach. If you want to find all vectors c so that {a, b, c} is a linearly dependent set, and you can tell by inspection that a and b are linearly independent, then c must be a linear combination of a and b. This is exactly what Dick's equation represents.
     
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