How Do You Convert Between Two Coordinate Systems with Different Basis Vectors?

[LMZ]

Homework Statement

2 coordinate systems are given:
1st: \vec{a}, \vec{b}, \vec{c}
2nd: \vec{m}, \vec{n}, \vec{p}
in system \vec{a}, \vec{b}, \vec{c} basis vectors of 2nd system have values:
\vec{m}=\{2/3, 1/3, 1/3\}, \vec{n}=\{-1/3, 1/3, 1/3\}, \vec{p}=\{-1/3, -2/3, 1/3\}
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.

Homework Equations

The Attempt at a Solution

1st equation: distance of vector \vec{m} is |\vec{m}|=\sqrt{a^2*4/9 + b^2*1/9 + c^2*1/9}=5u
2nd equation: dot product (\vec{m}, \vec{n}) is |\vec{m}|*|\vec{n}|\cos{75^o}=25u*0.38=a^2*(-2/9) + b^2*1/9 + c^2*1/9}
next multiply 2nd with (-1) and substract from 1st 2nd:
a^2*2/3=5u(1-0.38); a \approx 2.15u, 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!

 

[kuruman]

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 90^o not 75^o. Therefore, you cannot say that

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

You are missing the cross terms in the dot product

m ^{2} = \vec{m}\cdot \vec{m}

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