### 1.30. Applying the Gram-Schmidt orthogonalization process, construct the orthonormal basis $\left\{e_{1}, e_{2}, e_{3}\right\}$ from the given basis $\left\{g_{1}, g_{2}, g_{3}\right\}:$
$$
g_{1}=(1 ; 1 ; 0), \quad g_{2}=(2 ; 0 ; 1), \quad g_{3}=(0 ; 1 ;-2)
$$
Solution.

\[
\begin{aligned}
& f_{1}=g_{1}=(1 ; 1 ; 0), \\
& C_{1}^{2}=\left(g_{2}, f_{1}\right) /\left(f_{1}, f_{1}\right)=(2 \cdot 1+0 \cdot 1+1 \cdot 0) /\left(1^{2}+1^{2}+0^{2}\right)=1, \\
& f_{2}=g_{2}-C_{1}^{2} f_{1}=(2 ; 0 ; 1)-1 \cdot(1 ; 1 ; 0)=(1 ;-1 ; 1), \\
& C_{1}^{3}=\left(g_{3}, f_{1}\right) /\left(f_{1}, f_{1}\right)=(0 \cdot 1+1 \cdot 1+(-2) \cdot 0) /\left(1^{2}+1^{2}+0^{2}\right)=0.5, \\
& C_{2}^{3}=\left(g_{3}, f_{2}\right) /\left(f_{2}, f_{2}\right)=(0 \cdot 1+1 \cdot(-1)+ \\
& +(-2) \cdot 1) /\left(1^{2}+(-1)^{2}+1^{2}\right)=-1, \\
& f_{3}=g_{3}-C f_{1}^{3} f_{1}-C C_{2}^{3} f_{2}=(0 ; 1 ;-2)- \\
& -0.5 \cdot(1 ; 1 ; 0)-(-1) \cdot(1 ;-1 ; 1)=(0.5 ;-0.5 ;-1) . \\
& e_{1}=f_{1} / \sqrt{\left(f_{1}, f_{1}\right)}=\left(\frac{\sqrt{2}}{2} ; \frac{\sqrt{2}}{2} ; 0\right) ;
\end{aligned}
\]

\[
\begin{aligned}
& e_{2}=f_{2} / \sqrt{\left(f_{2}, f_{2}\right)}=\left(\frac{\sqrt{3}}{3} ;-\frac{\sqrt{3}}{3} ; \frac{\sqrt{3}}{3}\right) \\
& e_{3}=f_{3} / \sqrt{\left(f_{3}, f_{3}\right)}=\left(\frac{\sqrt{6}}{6} ;-\frac{\sqrt{6}}{6} ;-\frac{\sqrt{6}}{6}\right)
\end{aligned}
\]