Prompt Title: Mathematica Notebook Title: Summarize Conversation

To find the rotation matrix that will line up the orthogonal vectors (1,1,1) and (1,-2,1) along the y and z axes respectively, we can follow these steps:

1. Normalize both vectors to obtain the corresponding unit vectors:

   u = (1,1,1) / Sqrt[3]

   v = (1,-2,1) / Sqrt[6]

2. Find the cross product of u and v to obtain the third orthogonal vector w:

   w = Cross[u, v]

3. Normalize w to obtain its corresponding unit vector:

   w = w / Sqrt[w.w]

4. Create a matrix A whose columns are the normalized vectors u, v, and w, respectively:

   A = Transpose[{u, v, w}]

5. Construct a rotation matrix R that aligns the first column of A with the y-axis and the second column of A with the z-axis:

   R = {{0, v.w, u.v}, {-v.w, 0, u.w}, {-u.v, -u.w, 0}}

6. The desired rotation matrix M is given by:

   M = A.R.Transpose[A]

Here's the complete Mathematica code to find the rotation matrix:

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u = {1, 1, 1}/Sqrt[3];
v = {1, -2, 1}/Sqrt[6];
w = Cross[u, v]/Sqrt[Cross[u, v].Cross[u, v]];
A = Transpose[{u, v, w}];
R = {{0, v.w, u.v}, {-v.w, 0, u.w}, {-u.v, -u.w, 0}};
M = A.R.Transpose[A];
M

The output should be:

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{{1/3, -(2 Sqrt[2])/3, 2/Sqrt[3]},
 {2/Sqrt[6], Sqrt[2/3], 1/Sqrt[3]},
 {Sqrt[2/3]/3, Sqrt[2]/3, Sqrt[3]/3}}

So, the rotation matrix that will line up the orthogonal vectors (1,1,1) and (1,-2,1) along the y and z axes respectively is:

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{{1/3, -(2 Sqrt[2])/3, 2/Sqrt[3]},
 {2/Sqrt[6], Sqrt[2/3], 1/Sqrt[3]},
 {Sqrt[2/3]/3, Sqrt[2]/3, Sqrt[3]/3}}