Complete Example 4 to verify the rest of the components of the inertia tensor and the principal moments of inertia and principal axes. Verify that the three principal axes form an orthogonal triad.

Physics definitions are given.

In a right-handed coordinate system, the unit vectors $\mathit{i}\mathbf{,}\mathit{j}\mathbf{,}\mathit{k}$are vectors of unit magnitude that point in the direction of the $\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}$axes, respectively. The orthogonal triad of unit vectors, often known as basic vectors, is a set of three unit vectors that are orthogonal to each other.

Use the formula $I_{ij}=\sum _k{m}_k\left(r_k^2{\delta }_{ij}-r_{k,i}{r}_{k,j}\right)$to find the inertia.

$$\begin{gathered} I_{xx}=1\left(1^2+1^2\right)+2\left({\left(-1\right)}^2+0^2\right) \\ =4 \\ {l}_{yy}=1\left(0^2+1^2\right)+\left(1^2+0^2\right) \\ =3 \end{gathered}$$

Continue the process.

$$\begin{gathered} I_{zz}=1\left(0^2+1^2\right)+2\left(1^2+{\left(-1\right)}^2\right) \\ =5 \end{gathered}$$

Continue the process for other axes.

$$\begin{gathered} L_{xy}=I_{yx} \\ =-1·0·1-2·1\left(-1\right) \\ =2 \end{gathered}$$

Repeat the process.

Repeat the process a last time.

$$\begin{gathered} l_{yz}=l_{zy} \\ =-1·1·1-2·\left(-1\right)·0 \\ =-1 \end{gathered}$$

Write the solution in matrix form.

$l=\left(\begin{array}{ccc}4& 2& 0\\ 2& 3& -1\\ 0& -1& 5\end{array}\right)$

Find the principal moments of inertia.

$$\begin{gathered} l_1=6 \\ {l}_2=3+\sqrt{3} \\ {l}_3=3-\sqrt{3} \end{gathered}$$

Find the eigenvectors.

$$\begin{gathered} v_1=\left(1,1,-1\right) \\ {v}_2=\left(-1-\sqrt{3},2+\sqrt{},\right) \\ {v}_3=\left(-1+\sqrt{3,}2-\sqrt{3},1\right) \end{gathered}$$

Evaluate the dot product of eigenvectors.

$$\begin{gathered} v_1·v_2=-1-\sqrt{3}+2+\sqrt{3}-1 \\ =0 \\ {v}_1·v_3=-1+\sqrt{3}+2-\sqrt{3}-1 \\ =0 \end{gathered}$$

Hence, evaluate the dot product of the final eigenvectors.

$$\begin{gathered} v_2·v_3=1-3+4-3+1 \\ =0 \end{gathered}$$