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Facing problem in a questions on Vectors

  1. Feb 13, 2014 #1
    If |vector a+vector b|= |vector a- vector b| show that vector a is perpendicular to vector b.





    A friend of mine suggested me to square both the sides of the equation. At the end, the result was vector a. vector b=0, from which it could be proved that the vectors are perpendicular. But why do we use the dot product and not the cross product?
     
    Last edited: Feb 13, 2014
  2. jcsd
  3. Feb 13, 2014 #2
    How did you use the dot product here? And how would use the cross product?
     
  4. Feb 13, 2014 #3

    PhysicoRaj

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    He squared them and expanded in (a[itex]\pm[/itex]b)2 and applied dot product at 2(a.b)
     
  5. Feb 13, 2014 #4

    PhysicoRaj

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    How come? After squaring, the squared magnitudes get cancelled and you get 4abcosθ=0.
     
  6. Feb 13, 2014 #5
    Please understand that the questions are intended for the original poster (OP). The purpose of those questions is to let the OP show what the OP knows and whether the OP knows that correctly. Your answering those questions for the OP does not help me or the OP in any way, and is not appreciated. No offence, please.

    ritik.dutta3, please answer my original questions.
     
  7. Feb 13, 2014 #6
    but why do we have to use dot product and not cross product?
     
  8. Feb 13, 2014 #7

    jedishrfu

    Staff: Mentor

    The definition of A.B = |A||B|cos(theta) where theta is the angle between the two vectors so if A is perpendicular to B then theta = 90 degrees and the cos(theta) = 0.

    In contrast, the definition of AxB = |A||B|sin(theta) so for A is perpendicular to B then sin(theta) = 1 and
    AxB = |A||B|

    So given A and B, we can create a parallelogram with A and B as sides so that one diagonal is A+B and the other is A-B. Geometrically the diagonals of a parallelogram are always perpendicular and for them to be the same length then the parallelogram must be a square and hence A is perpendicular to B.

    I can't see why you couldn't have used the cross product except for the fact that it is further defined to create a vector perpendicular to both and this would confuse the issue a bit.
     
  9. Feb 13, 2014 #8
    voko, here is what i did:

    (a+b)2= a2+b2+2(a).(b).

    (a-b)2=a2+b2-2(a).(b).

    on equating them, the square terms get cancelled and what remains is this:
    4(a).(b)=0
    (a).(b)=0

    since in a dot product if the product is equal to 0, the angle between them should be 90 degrees.
    but why do we have to use the dot product over here?
     
  10. Feb 13, 2014 #9

    PhysicoRaj

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    Oops! sorry..:redface:
     
  11. Feb 13, 2014 #10
    You do not have to use it. But you did, and it worked. Can you think of any other way to solve the problem?
     
  12. Feb 19, 2014 #11
    No voko, i cannot.
     
  13. Feb 19, 2014 #12
    So you have a simple solution involving the dot product and no other solution without the dot product. I think it is reason enough why the dot product was used. I would stop worrying, because there is really nothing to worry about.
     
  14. Feb 19, 2014 #13
    it is a rectangle...
     
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