Answer :
Answer:
[tex]$ e_1 = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}[/tex] [tex]$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}[/tex] [tex]$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{2}}}{\textbf{2}}\\\\0\end{pmatrix}[/tex]
Step-by-step explanation:
we have to orthonormalize the vectors:
[tex]v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}[/tex] [tex]v_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}[/tex] [tex]$ v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}[/tex]
According to Gram - Schmidt process, we have:
[tex]u_k = v_k - \sum_{j = 1} ^ {k - 1} proj_{uj} (v_k)[/tex] where, [tex]$ proj_u (v) = \frac{u . v}{u . u}u[/tex]
The normalized vector is: [tex]$ e_k = \frac{u_k}{\sqrt{u_k.u_k}} $[/tex]
Now, the first step.
[tex]v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}[/tex] = u₁
Therefore, e₁ = [tex]$ \frac{u_1}{\sqrt{u_1.u_1}} $[/tex]
[tex]$ = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\\\\\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}[/tex]
Now, we find e₂.
[tex]$ u_2 = v_2 - \frac{u_1.v_2}{u_1.u_1}u_1 $[/tex]
[tex]$ = \begin{pmatrix} \frac{1}{3}\\ \\ 0 \\\\ \frac{1}{3}\\\\ \frac{-2}{3} \end{pmatrix}[/tex]
Therefore, [tex]$ e_2 = \frac{u_2}{\sqrt{u_2.u_2}} $[/tex]
[tex]$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{6}}}{\textbf{6}}\\\\\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}[/tex]
To find e₃:
[tex]$ u_3 = v_3 - \frac{u_1. v_3}{u_1.u_1}u_1 - \frac{u_2. v_3}{u_2.u_2} u_2 $[/tex]
[tex]$ = \begin{pmatrix} \frac{-1}{2} \\\\ 0\\ \\ \frac{1}{2} \\\\ 0 \\\end{pmatrix}[/tex]
[tex]$ e_3 = \frac{u_3}{\sqrt{u_3.u_3}} $[/tex]
[tex]$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\\\\\textbf{0} \\\\\frac{\sqrt{\textbf{2}}}{\textbf{2}}\\\\0\end{pmatrix}[/tex]
So, we have the orthonormalized vectors [tex]$ e_1, e_2, e_3 $[/tex].
Hence, the answer.