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Stupidly Simple Vector Problem (I hope)

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Vectors [tex]\vec{A}[/tex], [tex]\vec{B}[/tex], and [tex]\vec{C}[/tex] satisy the vector equation [tex]\vec{A} + \vec{B} = \vec{C}[/tex]. Their magnitudes are related by [tex]\vec{A} - \vec{B} = \vec{C}[/tex]. Explain how this is possible.


    3. The attempt at a solution
    Vector [tex]\vec{B} = 0\hat{_i} + 0\hat{_j} + 0\hat{_k}[/tex]


    This was a question on my quiz. I received no credit for my solution. After class, I asked the professor why my solution was wrong and he then told me that since the vector I chose had a magnitude of zero, it did not exist. I argued my point a bit, but soon gave up. I was wondering, do zero vectors exist? I found another solution to the problem, but had not written it on my quiz:

    3. The attempt at a solution
    Vector [tex]\vec{B}[/tex] is antiparallel to [tex]\vec{A}[/tex]

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Sep 4, 2007 #2

    Dick

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    You mean |A|-|B|=|C|. Antiparallel certainly works, if |B|<=|A|. And yes, Virginia, there is are zero vectors. But there's only one in euclidean norm. Yes, your first solution is also correct. Maybe just not what they were expecting.
     
    Last edited: Sep 4, 2007
  4. Sep 4, 2007 #3
    Thanks for the help. I wrote the problem word for word, so no, I don't mean [tex]|\vec{A}| - |\vec{B}| = |\vec{C}|[/tex].

    Also, can anyone give me proof of the existence of zero vectors?
     
    Last edited: Sep 4, 2007
  5. Sep 4, 2007 #4

    Dick

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    [tex]\vec{A} - \vec{B} = \vec{C}[/tex]? Those aren't magnitudes. Possibly the exam wasn't written well either.
     
  6. Sep 4, 2007 #5
    It wasn't, at least in my opinion. The last question was ([tex]\vec{A} \cdot \vec{B}[/tex]) + [tex]\vec{C}[/tex] with given components. But that is off-topic.


    Is there any proof zero vectors can exist?
     
    Last edited: Sep 5, 2007
  7. Sep 4, 2007 #6

    Dick

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    You already listed one, 0i+0j+0k. A vector space always has a unique zero vector because it's an additive group. Depending on the definition of the norm, other non-zero vectors can also have zero norm.
     
    Last edited: Sep 5, 2007
  8. Sep 5, 2007 #7
    But do those exist? I mean I wrote it, but I just made it up on the spot to solve the problem. I've never actually come across a zero vector, and don't see when or where I would.
     
  9. Sep 5, 2007 #8

    Dick

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    The displacement vector in moving from A to A is A-A=0. That's the zero vector. (1,1,1)-(1,1,1)=(0,0,0). I just ran across it.
     
  10. Sep 5, 2007 #9

    learningphysics

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    A badly written question... I would definitely have been confused by the way it is written

    I'd probably have done it the same way as you.

    [tex]\vec{A} + \vec{B} = \vec{C}[/tex]
    [tex]\vec{A} - \vec{B} = \vec{C}[/tex]

    subtract the first equation from the second implies:

    [tex]2\vec{B} = 0[/tex]
    so
    [tex]\vec{B} = 0[/tex]

    Yeah, he should have written the question as:
    [tex]\vec{A} + \vec{B} = \vec{C}[/tex]
    [tex]|\vec{A}| - |\vec{B}| = |\vec{C}|[/tex]
     
  11. Sep 5, 2007 #10

    learningphysics

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    I think your prof might have been looking for a solution like this...

    [tex]\vec{A} + \vec{B} = \vec{C}[/tex]

    take the dot product of each side with itself

    [tex](\vec{A} + \vec{B})\cdot(\vec{A} + \vec{B}) = \vec{C}\cdot\vec{C}[/tex]

    then evaluating the left side

    [tex]|\vec{A}|^2 + 2\vec{A}\cdot\vec{B} + |\vec{B}|^2 = |\vec{C}|^2[/tex] (equation 1)

    And then square the second equation you should have been given:

    [tex]|\vec{A}| - |\vec{B}| = |\vec{C}|[/tex]
    [tex]|\vec{A}|^2 - 2|\vec{A}||\vec{B}| + |\vec{B}|^2 = |\vec{C}|^2[/tex] (equation 2)

    then using equations 1 and 2

    [tex]\vec{A}\cdot\vec{B} = -|\vec{A}||\vec{B}|[/tex]

    from which you get that A and B are antiparallel...
     
  12. Sep 5, 2007 #11

    nrqed

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    Yes, the zero vector exists! If it did not, the operation [itex] \vec{A} - \vec{A} [/itex] would not be defined.

    A few comments:


    a) The way the question is written in yoru initial post, it simply does nto make any sense at all. As someone mentioned, he probably meant an equation for the difference of magnitude. I am amazed that a prof would write a question like this and not admit that it is worded incorrectly.



    b) the question you mentioned just above where a vector is added to a scalar product does not make any sense either. One can't add vectors to scalars. I hope this is a typo and not a second key mistake in the same test!

    c) I don't understand your answer. You said that B is zero but you also said that [itex] \vec{A} = \vec{C} [/itex] ? Or you said that ll three vectors are zero?
    Your answer is indeed correct, although slightly "trivial". It's tricky to argue but to me it seems that if the prof does not specify that some cases must be excluded, they should be accepted. But it sounds as if he (or she) does not accept zero vectors as possible vectors. Which is wrong.
     
  13. Sep 5, 2007 #12
    [​IMG] [​IMG]
    I had problems with questions 2, 5, and 7c. For 2 I used [D] instead of [L] in my dimensional analysis, 5 is the one listed above, and 7c I got right. After I turned in my quiz I asked the prof how to solve number 7 c and a few minutes later it was concluded that it is not a very easy task to add a vector to a scalar. As a side note, this is my first physics course =/
     
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