Answer
See below
Work Step by Step
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,1,-1,0)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(-1,0,1,1)-\frac{)}{||(1,1,-1,0)||^2}(1,1,-1,0)\\
=(-1,0,1,1)-\frac{(-1).1+1.0+1.(-1)+0.1}{1^2+1^2+(-1)^2+0^2}(1,1,-1,0)\\
=(-1,0,1,1)-\frac{-1+0-1+0}{2}(1,1,-1,0)\\
=(-1,0,1,1)-\frac{2}{3}(1,1,-1,0)\\
=\frac{1}{3}(-1,2,1,3)$
$v_3=x_3-\frac{}{||v_1||^2}v_1-\frac{}{||v_2||^2}v_2-(-6,0,-2)\\
=(2,-1,2,1)-\frac{)}{||(1,1,-1,0)||^2}(1,0,-1,0)-\frac{)}{||\frac{1}{3}(-1,2,1,3)||^2}\frac{1}{3}(-1,2,1,3)\\
=(2,-1,2,1)-\frac{1+0+0+1}{1+4+0+1}(1,2,0,1)-\frac{4+0+6-2}{16+1+9+4}(4,-1,3,-2)\\
=(2,-1,2,1)+\frac{1}{3}(1,1,-1,0)-\frac{1}{15}(-1,2,1,3)\\
=\frac{4}{5}(3,-1,2,1)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,1,-1,0)}{\sqrt 1^2+1^2+(-1)^2+0^2}=\frac{(1,1,-1,0)}{\sqrt 3}=\frac{1}{\sqrt 3}(1,1,-1,0)$
$\frac{v_2}{||v_2||}=\frac{(-1,2,1,3)}{\sqrt 15}=\frac{1}{\sqrt 15}(-1,2,1,3)$
$\frac{v_3}{||v_3||}=\frac{(3,-1,2,1)}{\sqrt 20}=\frac{1}{\sqrt 15}(3,-1,2,1)$
