# Transformation of coordinates

1. Oct 14, 2015

### tub08918

So Im going through the exam guide for my exam tomorrow and there is a second problem that stump me. We transform the cartesian axis to <1/√3,1/√3,1√3> and <1/√2,0,-1/√2> which are orthogonal and we find the third axis by taking the cross product which gives <-881/2158,881/1079,-881/2158>

Then we have to find a vector that is the same in both coordinate systems

I plotted in MATLAB and the only vector I can find that matches the criteria is <0,0,0> is there anything Im missing?

2. Oct 14, 2015

### DEvens

If you have two orthonormal coordinate systems, then they are related by a rotation. (I did not check if your basis vectors are indeed orthonormal. You should check that third one pretty carefully. I don't think it is correct.)

A rotation has an axis. Under this rotation, that axis does not change. You need to find the axis of rotation between the two coordinate systems. However, unless you have done quite a bit of geometry, that might not be particularly helpful for you.

Instead, what you want is a vector that projects onto both sets of coordinates with the same first, second, and third values. If that vector is V=(x,y,z), then the projection on the original coordinates is easy. It's just x, y, and z, in the first, second, and third directions, respectively.

Now you need to find an x, y, and z such that the following are satisfied.
V dot <1/√3,1/√3,1√3> = x
V dot <1/√2,0,-1/√2> = y
V dot <-881/2158,881/1079,-881/2158> = z

Then, the components of the vector V in the new coordinate system will be the same as in the old.

3. Oct 14, 2015

### tub08918

Hi DEvens

vector <1/√3,1/√3,1√3> and <1/√2,0,-1/√2> have a dot product equal to zero, and the vector <-881/2158,881/1079,-881/2158> was the result of crossing a and b sso te should be all orthogonal to one another but I could be wrong. Your three formulas are very interesting but I am confused as to what to plu g in where.

So I have three rotated vectors <1/√3,1/√3,1√3> <1/√2,0,-1/√2> <-881/2158,881/1079,-881/2158>
and the original vectors
<1,0,0> <0,1,0> <0,0,1>
so do I do
<1,0,0>dot<<1/√3,1/√3,1√3>> = x that is the same in both coordinate system?

4. Oct 14, 2015

### davidmoore63@y

appears to be a rational approximation to the correct answer < -1/sqrt 6, 2/sqrt6, -1/sqrt6 >

5. Oct 15, 2015

### tub08918

So I spent yesterday emulating a video on khan academy to my data:
C=[a' b' c'] where
a=[1\3.5,1/3.5,1/3.5]
b=[1/2.5,0,2/6.5]
c=[-1/6.5, 2/6.5,-1/6.5]

D=[a' b' c'] where a'=[1 0 0] b =[0 1 0] c=[0 0 1]

So I find the transformation matrix as

T=CDC-1

which just gives me an identity vector. Is v and v' the same then?