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3 Orthogonal Vectors Problem

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data

    A.
    Given unit vectors a, b, c in the x, y-plane such that a · b = b · c = 0,
    let v = a + b + c; what are the possible values of |v|?
    B.
    Repeat, except a, b, and c are unit vectors in 3-space

    2. Relevant equations



    3. The attempt at a solution

    I have solutions for both that I'm reasonably sure of, I'd just like a 2nd opinion to make sure I solved the problem correctly.

    For part A I got |v| = [itex]\sqrt{2}[/itex], [itex]\sqrt{5}[/itex]
    I took c = <0,1>, a = <1,0> b = <0,1> in this combination |v| = [itex]\sqrt{2}[/itex].
    The only other special case is when a = b in this case, |v| = [itex]\sqrt{5}[/itex]


    For Part B I took c = <0,0,1>, a = <1,0,0> and b = [itex]<cos\theta,sin\theta,0>[/itex]

    Therefore, |v| = [itex]\sqrt{2cos\theta + 3}[/itex], [itex]0\leq\theta\leq 2\pi[/itex]
     
  2. jcsd
  3. Sep 15, 2011 #2

    LCKurtz

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    The problem with your argument is that you have picked specific, very handy, vectors a,b, and c that satisfy your properties. So you have answered the question for only those three vectors. You have to argue it for any three such vectors.
     
  4. Sep 15, 2011 #3
    Yes, but since the vectors have to be unit vectors and both vectors a and c are orthogonal to b, |v| is going to be the same regardless. You can rotate a vector by a specific angle and the others have to stay orthogonal so they will move as well by the angle maintaining |v|. That's why I chose the most convenient vectors to work with.
     
    Last edited: Sep 15, 2011
  5. Sep 15, 2011 #4

    LCKurtz

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    Everything you say is true. If this is homework, I would at the minimum include this additional argument. Still, it is just as easy and more elegant to do it with arbitrary vectors in the first place.
     
  6. Sep 16, 2011 #5
    i don't understand,

    if b.c = 0, then how come b = c? i.e how can b=<0,1> = c ?
     
  7. Sep 17, 2011 #6
    why issn't part 1

    v = a + b + c
    |v| = [itex]\sqrt{a^2 + b^2 + c^2}[/itex] = [itex]\sqrt{3}[/itex] ?
     
  8. Sep 17, 2011 #7

    ehild

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    So you took c=b. Being unit vectors, their scalar product is 1 instead of zero. Which two vectors can be identical instead?

    ehild
     
  9. Sep 17, 2011 #8

    ehild

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    You do not have 3 independent, mutually orthogonal vectors in 2D. One of the vectors a, b, c is linear combination of the other two.

    ehild
     
  10. Sep 17, 2011 #9

    vela

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    Another reason for doing the problem in general is that your specific vectors might not cover every possibility, which has indeed happened here.
     
  11. Sep 17, 2011 #10
    its something like this right?

    *****B
    **** ^
    **** |
    A<---|--->C

    since they are 3 independ orthogonal vectors, how can one of them be a linear combination of the other two?

    how does he get sqrt 2 or sqrt 5?
     
  12. Sep 17, 2011 #11

    ehild

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    They are not three independent vectors.

    a=c or a=-c, so either v=2a+b or v=b

    The magnitude of v can not be sqrt(2).

    ehild
     
  13. Sep 17, 2011 #12

    HallsofIvy

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    You have already been told that there aren't three independent vectors in two dimensional space! The problem does NOT say that they are independent. In particular the problem, does NOT say that c is orthogonal to a. You are given that b is orthogonal to a and c and they are all of unit length which means that either a= c or a= -c.

    If a= c, then v= 2a+ b and, since a and b are unit length and orthogonal, the length of v is [itex]\sqrt{2^2+ 1}= \sqrt{5}[/itex]. If a= -c, then v= b which has length 1.
     
  14. Sep 17, 2011 #13
    ohh.h... isee.... thanks!
     
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