順に計算していくと

$\overrightarrow{e_1} = \dfrac{\overrightarrow{a_1}}{|\overrightarrow{a_1}|} = \dfrac{1}{\sqrt{3}} $$\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$ $

$ $$\begin{eqnarray*} \overrightarrow{v_2} & = & \overrightarrow{a_2} - \left( \overrightarrow{a_2} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1}\\[1em] & = & \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - 0\cdot \dfrac{1}{\sqrt{3}} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\\[1em] & = & \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \end{eqnarray*}$$ $

$\overrightarrow{e_2} = \dfrac{\overrightarrow{v_2}}{|\overrightarrow{v_1}|} = \dfrac{1}{\sqrt{2}} $$\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$ $

$ $$\begin{eqnarray*} \overrightarrow{v_3} & = & \overrightarrow{a_3} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_1} \right) \overrightarrow{e_1} - \left( \overrightarrow{a_3} \cdot \overrightarrow{e_2} \right) \overrightarrow{e_2}\\[1em] & = & \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} - 0\cdot \dfrac{1}{\sqrt{3}} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} - \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\\[1em] & = & \dfrac{1}{2}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \end{eqnarray*}$$ $

$\overrightarrow{e_3} = \dfrac{\overrightarrow{v_3}}{|\overrightarrow{v_3}|} = \dfrac{1}{\sqrt{6}} $$\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$ $

よって $\overrightarrow{e_3} = \dfrac{1}{\sqrt{6}} $$\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$ $ である。