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Chnage coordinate system

  1. Aug 23, 2009 #1

    LMZ

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    1. The problem statement, all variables and given/known data
    2 coordinate systems are given:
    1st: [tex]\vec{a}, \vec{b}, \vec{c}[/tex]
    2nd: [tex]\vec{m}, \vec{n}, \vec{p}[/tex]
    in system [tex]\vec{a}, \vec{b}, \vec{c}[/tex] basis vectors of 2nd system have values:
    [tex]\vec{m}=\{2/3, 1/3, 1/3\}, \vec{n}=\{-1/3, 1/3, 1/3\}, \vec{p}=\{-1/3, -2/3, 1/3\}[/tex]
    also known that all 3 basis vectors of 2nd coordinate system have length 5 units and angle between each 2 vectors of 2nd system is 75 Grades.

    2. Relevant equations



    3. The attempt at a solution
    1st equation: distance of vector [tex]\vec{m}[/tex] is [tex]|\vec{m}|=\sqrt{a^2*4/9 + b^2*1/9 + c^2*1/9}=5u[/tex]
    2nd equation: dot product [tex](\vec{m}, \vec{n})[/tex] is [tex]|\vec{m}|*|\vec{n}|\cos{75^o}=25u*0.38=a^2*(-2/9) + b^2*1/9 + c^2*1/9}[/tex]
    next multiply 2nd with (-1) and substract from 1st 2nd:
    [tex]a^2*2/3=5u(1-0.38); a \approx 2.15u[/tex], BUT length of the vector a should be greater then leght of the vector m, i guess...

    Hope you'll understand what I mean ;)
    thanks for your help!
     
    Last edited: Aug 23, 2009
  2. jcsd
  3. Aug 25, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    I don't think that the vectors in the first coordinate system are orthogonal. If they are, then the dot product of m and n would be zero which makes the angle between them 90o not 75o. Therefore, you cannot say that

    [tex]|\vec{m}|=\sqrt{a^2*4/9 + b^2*1/9 + c^2*1/9}[/tex]

    You are missing the cross terms in the dot product

    [tex]m ^{2} = \vec{m}\cdot \vec{m}[/tex]

    I assume you are looking for the magnitudes of a, b and c.
     
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