1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Chnage coordinate system

  1. Aug 23, 2009 #1


    User Avatar

    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


    User Avatar
    Science Advisor
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook