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

[LMZ]

Homework Statement

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.

Homework Equations

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!

 

[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

[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.

 

[Architeuthis Dux]

Thank you for sharing your attempt at a solution. It seems like you are on the right track, but there are a few things that could be clarified. First, it is important to specify the units for the length of the vectors, as well as the units for the angle. Also, when using the dot product to find the angle between two vectors, it should be noted that the dot product is equal to the product of the lengths of the vectors multiplied by the cosine of the angle between them. So in this case, the second equation should be |\vec{m}|*|\vec{n}|*cos{75^o}=25u*5u*cos{75^o}=25u*0.38.

Additionally, when solving for the length of vector \vec{a}, it is important to consider the fact that the length of all three basis vectors of the second coordinate system is 5 units. This means that the length of vector \vec{a} should also be 5 units, as it is one of the basis vectors of the second coordinate system. So instead of setting a^2*2/3=5u(1-0.38), it should be a^2=5u(1-0.38)/2*3=5u*0.31, which gives a=\sqrt{5u*0.31}\approx1.4u. This makes sense as the length of vector \vec{a} should be less than the length of vector \vec{m}.

Overall, it seems like you have a good understanding of the concepts involved in changing coordinate systems. Just be sure to carefully consider all the given information and units, and to double check your calculations. Good luck with your homework!