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Proof using dot product?

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data

    If |A+B|^2=|A|^2+|B|^2, prove that A is perpendicular to B.


    2. Relevant equations

    A^2=|A||A|

    3. The attempt at a solution

    All I can think of to do is expand the equation to get (A.B)(A.B)=A.A+B.B. I know that iff a.b=0, the two are perpendicular but I can't figure out how to show that because I don't really even understand what the question is asking. I'm in desperate need of some help!
     
  2. jcsd
  3. Aug 26, 2008 #2

    Defennder

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    As said before, start with the LHS, not the RHS. Your approach uses the RHS. What is another way of writing |A+B|^2 using the dot product?
     
  4. Aug 26, 2008 #3
    Does it involve taking the square root of both sides?
     
  5. Aug 26, 2008 #4

    Defennder

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    Nope, not at all.
     
  6. Aug 26, 2008 #5
    Yes, I just tried it and it was definitely the wrong idea. When you say I need to begin with the LHS, though, do you mean I need to write it out like (A.B)(A.B) to obtain 2AB? If that's not it, could you give me a hint to get me going in the right direction?
     
  7. Aug 26, 2008 #6
    Wait, 2AB is the wrong thing, I looked at the something else I'd written down that looked similar. The dot product there would be 2A+2B, right?
     
  8. Aug 26, 2008 #7

    Defennder

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    Well I gave you a hint in the other thread. This is the third thread with the same question. You ought to stick to one thread and certainly only one forum. [tex]|A|^2 = \mathbf{A} \cdot \mathbf{A}[/tex]
     
  9. Aug 26, 2008 #8

    Ben Niehoff

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    |A+B|^2 is NOT equal to (A.B)^2.

    Try letting C = A+B. Then what is |C|^2?
     
  10. Aug 26, 2008 #9
    oh, whoops, I messed that last thing up entirely. I'm way too tired, this prof gave way too much homework tonight. I didn't think to use a third variable, that seems like a good idea. defennder, you said to begin with the LHS so I thought that |A|^2=A.A wouldn't yet be relevant.


    also, defennder: someone moved my first post to a different forum, so I couldn't find it and reposted it in another one. then I had so many questions in the thread that I thought it'd be less intimidating and I'd get more answers if I asked just the one question. sorry! patience is a virtue god failed to give me. =)
     
  11. Aug 26, 2008 #10
    I keep coming back to the fact that 0=2AB. I'm not sure if this has any significance, but if I multiply out |A+B|^2 I get A^2+2AB+B^2, and if I set that equal to the RHS and simplify, I get 2AB=0.
     
  12. Aug 26, 2008 #11
    no, that seems like it would prove the two are NOT perpendicular. why is this so hard for me????
     
  13. Aug 26, 2008 #12
    actually, it was the right train of thought, am I right that 2(A.B)=0 is significant?
     
  14. Aug 26, 2008 #13

    Ben Niehoff

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    Yes. What does it mean if the dot product between two vectors is zero?
     
  15. Aug 26, 2008 #14
    that they are perpendicular!! I think I've finally got a viable proof--thank you both so much for your help! now it is off to bed for a couple hours before class starts... good night!
     
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