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Simple comlex vector problem

  1. Jul 28, 2009 #1
    1. The problem statement, all variables and given/known data
    let x=(i,1,1) and y=(1,i,2)

    find the magnitude of ix+y


    3. The attempt at a solution

    i(i,1,1)+(1,i,2)=(0,2i,2+i)

    therefore mag sqrd = 0^2+(2i+2(-i))+(2+i)(2-i)=9 => magnitude = 3???

    is this rite? i have completely forgotten when to take the conjugates... idk if im not meant to take the conjugate of 2i, but i have to for 2+i.... plz can anyone reassure me.
     
  2. jcsd
  3. Jul 28, 2009 #2

    Office_Shredder

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    It doesn't look like you did it right... why don't you start by writing down the definition of the magnitude of a vector? Then remember that [itex]|a|^2 = a \bar{a}[/itex]
     
  4. Jul 28, 2009 #3
    magnitude of a vector is its length, normally would do pythagouras, but for complex the magnitude is sqr root of the vector*conjugate. i think
     
  5. Jul 28, 2009 #4

    Office_Shredder

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    So why don't you write that out explicitly? That's not what you did in your original post. Work it out slowly
     
  6. Jul 28, 2009 #5
    ok well i(i,1,1)dotproduct(1,i,2)=(-1,i,i)dotproduct(1,i,2)=(-1+1)i+(i+i)j+(2+i)k=(0,2i,2+i)

    sqr root of that taking conjugates..(which not not sure about) = root(0^2+(2i*-2i)+((2+i)(2-i)))=root(0+4+5)=root(9)=3????? i cant take it any slower then that, plz can u tell me where i am going wrong
     
  7. Jul 28, 2009 #6
    or r u sayingi have to take the conjugate when i multiply x by i???

    then it would be root((1,i,i)dot(1,-i,-i)+(1,i,2)dot(1,-i,2))=root((1+1+1)+(1+1+4))=root(9)=3??????? omg i dont know
     
  8. Jul 28, 2009 #7
    anyone have any ideaS?
     
  9. Jul 29, 2009 #8
    plz anyone?
     
  10. Jul 31, 2009 #9
    no one?
     
  11. Aug 1, 2009 #10

    Mark44

    Staff: Mentor

    I don't believe so. Your ix + y is the vector you want the magnitude of, so it's the one you want to multiply by its complex conjugate.
    I also get 3 for the magnitude of ix + y.
     
  12. Aug 1, 2009 #11
    ok thankyou very much = ) yes i thought so, just havnt done this stuff in so long, lost confidence.
     
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