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Orthogonal vectors

  1. Apr 30, 2010 #1
    If x is ⊥ u and v, then x is ⊥ u - v.

    I know this is true because u - v is in the same place as u and v; therefore, x is orthogonal. How can this be written better?
     
  2. jcsd
  3. Apr 30, 2010 #2
    Use the property of the dot product that a dot (b+c) = a dot b + a dot c
     
  4. Apr 30, 2010 #3

    Mark44

    Staff: Mentor

    "is in the same plane as u and v..."

    Dot product?
     
  5. Apr 30, 2010 #4
    I understand the dot product needs to be used to do this (VeeEight and Mark) but I don't know how to implement it correctly.
     
  6. Apr 30, 2010 #5
    If two vectors are orthogonal, what is their dot product?
     
  7. Apr 30, 2010 #6
    I got it <x,v>=0, <x,u>=0

    <x,u-v>=<x,u>-<x,v>=0-0=0
     
  8. Apr 30, 2010 #7

    Mark44

    Staff: Mentor

    If x ⊥ u, how can you say the same thing using the dot product. You don't have to "implement" anything - just use the dot product.
     
  9. Apr 30, 2010 #8

    Mark44

    Staff: Mentor

    OK, then what does this say about x and u - v?
     
  10. Apr 30, 2010 #9
    :smile:
     
  11. Apr 30, 2010 #10
    I was thinking more of proof then doing.
     
  12. Apr 30, 2010 #11
    :confused:
     
  13. Apr 30, 2010 #12

    Mark44

    Staff: Mentor

    I second that.
     
  14. Apr 30, 2010 #13
    I was thinking of a formal proof then just doing <x..> ..... =0
     
  15. Apr 30, 2010 #14

    Mark44

    Staff: Mentor

    That is a formal proof. Don't overthink this stuff at the expense of understanding what's going on.
     
  16. Apr 30, 2010 #15
    Use the linearity of the dot product ( , ) so,

    (x,u)=(u,x)=0 and (x,v)=(v,x)=0,
    So consider,
    (u-v,x)=(u,x)-(v,x)=0 => (x,u-v)=0.
     
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