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Finding all vectors <x,z>=<y,z>=0

  1. Sep 13, 2015 #1
    5iZ64h0.png

    Hi Guys, that's what i got

    <x,z>=<y,z>
    <x,z>-<y,z>=0
    <x,z>+<-y,z>=0
    <x-y,z>=0

    x-y = [0,2,0]

    <2*[0,1,0],Z>=0
    2<[0,1,0],z> = 0
    <[0,1,0],z>=0

    So 'im stuck at that. Any ideas?
     
  2. jcsd
  3. Sep 13, 2015 #2

    mfb

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    If you write z with its components, what do you get as result?

    Unrelated:
    You didn't use the "=0" part yet.
     
  4. Sep 13, 2015 #3
    <[0,1,0],[z1,z2,z3]>=0
    Can i assume that the inner product is <x,y>=x1*y1+x2*y2+x3*y3 because it's a Euclidean Space?
     
  5. Sep 13, 2015 #4

    mfb

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    Sure.
     
  6. Sep 13, 2015 #5
    That's it then! Thank you very much.
     
  7. Sep 13, 2015 #6

    mfb

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    Well, it is the first part of the solution, yes.
     
  8. Sep 13, 2015 #7

    mathman

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    How about z=a(xXy), where X is cross product and a is any real number?
    This gives you z=(2a,0,-2a).
     
  9. Sep 13, 2015 #8
    0*z1+1*z2+0*z3=0
    z2=0

    <x,z>=0
    z1+z3=0
    z2=-z3

    z is all vectors in the form (x,0,-x)
     
  10. Sep 13, 2015 #9

    mfb

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    Correct (I wouldn't use x here as x is used as vector in the problem statement).
     
  11. Sep 13, 2015 #10

    Mark44

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    One thing I don't see mentioned in this thread is that the notation <x, z> represents the inner product of x and z, I believe. If z is an arbitrary vector with z = <z1, z2, z3>, then <x, z> = 0 means that x and z are perpendicular. Also, <x, z> = ##x_1z_1 + x_2z_2 + x_3z_3 = z_1 + z_2 + z_3##, and similarly for <y, z>.
     
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