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Hint for dot product proof

  1. Jul 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Let A be a vector perpendicular to every vector X. Show that A = O Edit: it is O not 0. (OH not zero) ha


    2. Relevant equations
    So, we know if A and X are perpendicular then A(dot)X = 0
    I see no reason why A would have to be equal to 0.
    Could X (not equal) 0? Could it be both?


    3. The attempt at a solution

    Above is my attempt. I get a hint please?
     
  2. jcsd
  3. Jul 18, 2013 #2
    Think about the trigonometric definition of the dot product.
     
  4. Jul 18, 2013 #3

    LCKurtz

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    You might start by telling us what vector space your vectors are from. Is it ##{\mathbb R}^n##? What happens if you dot A with the standard basis vectors?
     
  5. Jul 18, 2013 #4

    Dick

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    If a vector A is perpendicular to EVERY vector, it must be perpendicular to itself. So?
     
  6. Jul 19, 2013 #5

    Simon Bridge

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    ... assuming A is in the same space as X.
    X could be a set of vectors lying in a plane.

    But I agree - they key to this is to realize that that A must be perpendicular to EVERY vector.
     
  7. Jul 19, 2013 #6
    I wrote the questions exactly as is. Doesn't say anything about the vector space. If I dot A with the standard basis vectors then....I'm not sure because all that I'm told is that A is perpendicular to X I do not know anything else.
    How can a vector be perpendicular to itself?
     
  8. Jul 19, 2013 #7
    You are told that A is perpendicular to every vector. Not just some vector.

    There is only one way, and it proves your result.

    |A||X|cos(theta) = A (dot) B

    Is the trigonometric definition.

    If A is perpendicular to all vectors, it is perpendicular to itself. All vectors means all vectors, A is a vector, therefore A is perpendicular to A. Therefore

    |A||X|cos(theta) = 0

    Even whenever X = A.

    What is the angle between X and A if X=A?

    How can this equation always be 0 (perpendicular to any vector) if I allow one magnitude and theta to vary freely?
     
  9. Jul 19, 2013 #8
    Only if A = 0. But wouldn't theta not vary freely if you are told it is perpindicular then the angle must always be 90 degree.
    So why do you say theta varies freely. It would seem to me that X would vary and theta would always be 90.
     
  10. Jul 19, 2013 #9
    How about this...can you just say ..The only way that A can be perpindicular to it's self is if it is the 0 vector?
     
  11. Jul 19, 2013 #10
    No.

    Two vectors are perpendicular if and only if their dot product vanishes. That is the definition. If I tell you that two vectors are perpendicular, then you can tell me that their dot product vanishes, and that is all. The dot product of O (dot) X = 0 for any X, it doesn't matter what I call the angle between O and X, nor can I say what it is (because it is arbitrary).

    That two perpendicular nonzero vectors have an angle of 90* is a direct corollary to that definition. It has no effect on this proof, because we aren't explicitly talking about nonzero vectors.

    If you do not let theta vary freely then you are not considering every vector. The theorem you are proving states that X is perpendicular to every vector, not just ones that make an angle of 90* with it.


    You have to show why that assertion is true.
     
    Last edited: Jul 19, 2013
  12. Jul 19, 2013 #11
    0?

    So
    So just from the trig defination like you say the only way that it can be 0 is if A is 0 because X and theta vary.
    I don't know how to show that the only way to be perpindiuclar to it's self is if it is 0.
    Your proof is mostly words like stating conditons not really like showing it. (you are though if you get what I mean)
    So I suppose that saying A can only be perpindicular to its self if it is the zero vector would be sort of the same. I can't think of it yet.
    Thanks
     
  13. Jul 19, 2013 #12

    Dick

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    What's your definition of 'dot product'?
     
  14. Jul 19, 2013 #13
    If A is perpendicular to B, then

    |A||B|cos(theta) = 0

    Consider the case where A = B.


    If I say anything else, I've done it for you completely.
     
  15. Jul 19, 2013 #14
    X dot A = |X||A|cos(theta)
     
  16. Jul 19, 2013 #15
    Yes, so now write that for X (dot) X and show, algebraically, that |X| is 0 if the dot product is 0.
     
  17. Jul 19, 2013 #16
    Is just
    A dot A = |A||A|cos(theta)
    A dot A = |A||A|cos(0)
    cos(0) = 1 so A must be 0
    I don't like this
    ?
     
  18. Jul 19, 2013 #17
    You need to set the equation equal to 0.

    |A||A|cos(0)=0
    |A|^2 =0
    |A|=0

    What do you mean you don't like it? It concisely proves the theorem by literally, simplifying and dividing once. Itdoesn't get much cleaner than that.
     
  19. Jul 19, 2013 #18
    "Show that if A is perpendicular to all vectors, A is the 0 vector"

    Proof:

    If A is perpendicular to all vectors, then it must be perpendicular to itself, thus

    A(dot)A= 0
    |A||A|cos(0) = 0
    |A|^2 = 0
    |A|=0

    Since the magnitude of A is 0, A is the zero vector.

    That's all there is to it.
     
  20. Jul 19, 2013 #19
    What is wrong iwth how I said it?
     
  21. Jul 19, 2013 #20
    You didn't set the equation equal to 0, all you said was that A(dot)A = |A||A| and then asserted that A was 0. Reading that, no one would know what you were talking about. You can't solve for A in the equation A(dot)A = |A||A| because it is tautological.
     
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