Answer
$\{(\frac{1}{\sqrt 14},\frac{2}{\sqrt 14},\frac{3}{\sqrt 14}),(\frac{6}{\sqrt 45},\frac{-3}{\sqrt 45},0)\}$
Work Step by Step
Let $x_1=(1,2,3) \\
x_2=(6,-3,0)$
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,2,3)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(6,-3,0)-\frac{((6,-3,0),(1,2,3))}{||(1,2,3)||^2}(1,2,3) \\
=(6,-3,0)-\frac{6.1+(-3).2+0.3}{1^2+2^2+3^2}(1,2,3)\\
=(6,-3,0)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,2,3)}{\sqrt 1^2+2^2+3^2}=\frac{(1,2,3)}{\sqrt 14}=(\frac{1}{\sqrt 14},\frac{2}{\sqrt 14},\frac{3}{\sqrt 14})$
$\frac{v_2}{||v_2||}=\frac{(6,-3,0)}{\sqrt 6^2+(-3)^2+0^2}=\frac{(6,-3,0)}{\sqrt 45}=(\frac{6}{\sqrt 45},\frac{-3}{\sqrt 45},0)$
Hence, a corresponding orthonormal set of vector is :
$\{(\frac{1}{\sqrt 14},\frac{2}{\sqrt 14},\frac{3}{\sqrt 14}),(\frac{6}{\sqrt 45},\frac{-3}{\sqrt 45},0)\}$
