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Perpendicular Vectors

  1. Dec 29, 2009 #1
    1. The problem statement, all variables and given/known data
    If a and b are perpendicular, simplify [tex](a - 2b) \cdot (3a + 5b)[/tex]


    3. The attempt at a solution

    Not really sure what they're asking, but to be perp. the angle between them is 90. So using the scalar

    so a.b = 0??

    The answer in the book it (3a² - 10b²) and I can see that they just multiplied the as and bs together, but why?

    Thanks
    Thomas
     
  2. jcsd
  3. Dec 29, 2009 #2

    rock.freak667

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    You are to expand it as follows [itex]3a \cdot (a - 2b)+5b \cdot(a-2b)[/itex] and then use the fact that a.b=0
     
  4. Dec 29, 2009 #3

    tiny-tim

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    Science Advisor
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    Hi Thomas! :smile:

    Yes, a.b = 0 …

    so just expand (a-2b).(3a+5b) in the usual way, and use a.b = 0 :wink:
     
  5. Dec 30, 2009 #4
    oh it's really easy ;) I was thinking that the each of the brackets were vectors when of course a and b are the vectors. I'm a silly billy.

    How about this one

    Use the definition of the scalar product to show that if two vectors are perpendicular, their scalar product is zero.

    well a.b = |a||b|Cosx

    and when x = 90° => cos x = 0 and that means

    a.b = |a||b|.0
    a.b = 0

    but that's a bit easy isn't it?

    Thanks
     
  6. Dec 30, 2009 #5

    rock.freak667

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    yes that would be correct
     
  7. Dec 30, 2009 #6

    Mark44

    Staff: Mentor

    Each of the quantities inside parentheses is a vector, made up of scalar multiples of a and b, which are themselves vectors.
     
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