Answer
See below
Work Step by Step
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,0,-1,0)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(1,1,-1,0)-\frac{)}{||(1,0,-1,0)||^2}(1,0,-1,0)\\
=(1,1,-1,0)-\frac{1.1+1.0+(-1).(-1)}{1^2+0^2+(-1)^2+0^2}(1,1,-1,0)\\
=(1,1,-1,0)-\frac{1+0+1+0}{2}(1,0,-1,0)\\
=(1,1,-1,0)+(1,0,-1,0)\\
=(0,1,0,0)$
$v_3=x_3-\frac{}{||v_1||^2}v_1-\frac{}{||v_2||^2}v_2-(-6,0,-2)\\
=(-1,1,0,1)-\frac{)}{||(1,0,-1,0)||^2}(2,0,1)-\frac{)}{||(0,1,0,0)||^2}(0,1,0,0)\\
=(-1,1,0,1)-\frac{-1.1+1.0+0.(-1)+1.0}{(-1)^2+0^2+1^2+0^2}(1,0,-1,0)-\frac{0.(-1)+1.1+0.0+1.0}{||0^2+1^2+0^2+0^2}(0,1,0,0)\\
=(-1,1,0,1)+(\frac{1}{2},0,-\frac{1}{2},0)-2.(0,1,0,0)\\
=(-\frac{1}{2},0,-\frac{1}{2},1)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,0,-1,0)}{\sqrt 1^2+0^2+(-1)^2+0^2}=\frac{(1,0,-1,0)}{\sqrt 2}=(\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2},0)$
$\frac{v_2}{||v_2||}=\frac{(0,1,0,0)}{\sqrt 1}=(0,1,0,0)$
$\frac{v_3}{||v_3||}=\frac{(-\frac{1}{2},0,-\frac{1}{2},0)}{\sqrt \frac{3}{2}}=\frac{2}{\sqrt 6}(-\frac{1}{2},0,-\frac{1}{2},1)=(-\frac{1}{\sqrt 6},0,-\frac{1}{\sqrt 6},\frac{2}{\sqrt 6})$
Hence, a corresponding orthonormal set of vector is :
$\{(\frac{1}{\sqrt 2},0,-\frac{1}{\sqrt 2},0);(0,1,0,0);(-\frac{1}{\sqrt 6},0,-\frac{1}{\sqrt 6},\frac{2}{\sqrt 6})\}$
