Answer
$\{(\frac{2}{3},\frac{1}{3},-\frac{2}{3}),(\frac{-1}{3\sqrt 2},\frac{4}{3\sqrt 2},\frac{1}{3\sqrt 2})\}$
Work Step by Step
Let $x_1=(2,1,-2) \\
x_2=(1,3,-1)$
According to Gram-Schmidt process, we have:
$v_1=x_1=(2,1,-2)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(1,3,-1)-\frac{((2,1,-2),(1,3,-1))}{||(2,1,-2)||^2}(2,1,-2) \\
=(1,3,-1)-\frac{2.1+1.3+(-1).(-2)}{1^2+1^2+(-1)^2}(2,1,-2)\\
=(-\frac{5}{9},\frac{20}{9},\frac{5}{9})$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(2,1,-2)}{\sqrt 2^2+1^2+(-2)^2}=\frac{(2,1,-2)}{\sqrt 9}=(\frac{2}{3},\frac{1}{3},-\frac{2}{3})$
$\frac{v_2}{||v_2||}=\frac{(\frac{-5}{9},\frac{20}{9},\frac{5}{9})}{\sqrt (\frac{-5}{9})^2+(\frac{20}{9})^2+(\frac{5}{9})^2}=\frac{(\frac{-5}{9},\frac{20}{9},\frac{5}{9})}{\sqrt \frac{450}{81}}=(\frac{-1}{3\sqrt 2},\frac{4}{3\sqrt 2},\frac{1}{3\sqrt 2})$
Hence, a corresponding orthonormal set of vector is :
$\{(\frac{2}{3},\frac{1}{3},-\frac{2}{3}),(\frac{-1}{3\sqrt 2},\frac{4}{3\sqrt 2},\frac{1}{3\sqrt 2})\}$
