Questão 3 [2 pontos]

Considere os vetores no espaço \mathbb{R}^{3} :
\vec{v}_{1}=\left(\begin{array}{l} 1 \\\\ 1 \\\\ 1 \end{array}\right), \quad \vec{v}_{2}=\left(\begin{array}{l} 0 \\\\ 1 \\\\ 1 \end{array}\right), \quad \vec{v}_{3}=\left(\begin{array}{l} 0 \\\\ 0 \\\\ 1 \end{array}\right)
Utilize o processo de Gram-Schmidt para ortogonalizar esse conjunto de vetores, obtendo um conjunto ortogonal \left\{\vec{u}_{1}, \vec{u}_{2}, \vec{u}_{3}\right\}.

Passo 1

O enunciado nos dá três vetores em \mathbb{R}^3,
\vec{v}_{1}=\begin{pmatrix}1\\\ 1\\\ 1\end{pmatrix},\quad \vec{v}_{2}=\begin{pmatrix}0\\\ 1\\\ 1\end{pmatrix},\quad \vec{v}_{3}=\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix},
e pede para aplicar o processo de Gram–Schmidt, que usa produto interno e projeções para gerar um conjunto ortogonal \{\vec{u}_1,\vec{u}_2,\vec{u}_3\}.

Passo 2

Definimos
\vec{u}_1 = \vec{v}_1 = \begin{pmatrix}1\\\ 1\\\ 1\end{pmatrix}.
Esse é o primeiro vetor ortogonal.

Passo 3

Agora calculamos
\vec{u}_2 = \vec{v}_2 - \proj_{\vec{u}_1}(\vec{v}_2),
onde
\proj_{\vec{u}_1}(\vec{v}_2) = \frac{\vec{v}_2 \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}\,\vec{u}_1.
Primeiro,
\vec{v}_2 \cdot \vec{u}_1 = 0\cdot1 + 1\cdot1 + 1\cdot1 = 2, \qquad \vec{u}_1 \cdot \vec{u}_1 = 1^2+1^2+1^2 = 3.
Logo,
\proj_{\vec{u}_1}(\vec{v}_2) = \frac{2}{3}\,\begin{pmatrix}1\\\ 1\\\ 1\end{pmatrix} = \begin{pmatrix}2/3\\\ 2/3\\\ 2/3\end{pmatrix},
e portanto
\vec{u}_2 = \begin{pmatrix}0\\\ 1\\\ 1\end{pmatrix} - \begin{pmatrix}2/3\\\ 2/3\\\ 2/3\end{pmatrix} = \begin{pmatrix}-2/3\\\ 1/3\\\ 1/3\end{pmatrix}.

Passo 4

Finalmente,
\vec{u}_3 = \vec{v}_3 - \proj_{\vec{u}_1}(\vec{v}_3) - \proj_{\vec{u}_2}(\vec{v}_3),
com
\proj_{\vec{u}_1}(\vec{v}_3) = \frac{\vec{v}_3\cdot\vec{u}_1}{\vec{u}_1\cdot\vec{u}_1}\,\vec{u}_1 = \frac{1}{3}\,\begin{pmatrix}1\\\ 1\\\ 1\end{pmatrix} = \begin{pmatrix}1/3\\\ 1/3\\\ 1/3\end{pmatrix},
e
\vec{u}_2\cdot\vec{u}_2 = \bigl(-\tfrac23\bigr)^2+\bigl(\tfrac13\bigr)^2+\bigl(\tfrac13\bigr)^2 = \tfrac{6}{9}=\tfrac23, \qquad \vec{v}_3\cdot\vec{u}_2 = 0\cdot(-\tfrac23)+0\cdot\tfrac13+1\cdot\tfrac13 = \tfrac13.
Logo,
\proj_{\vec{u}_2}(\vec{v}_3) = \frac{\tfrac13}{\tfrac23}\,\vec{u}_2 = \tfrac12\,\begin{pmatrix}-\tfrac23\\tfrac13\\tfrac13\end{pmatrix} = \begin{pmatrix}-1/3\\\ 1/6\\\ 1/6\end{pmatrix}.
Portanto
\vec{u}_3 = \begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix} - \begin{pmatrix}1/3\\\ 1/3\\\ 1/3\end{pmatrix} - \begin{pmatrix}-1/3\\\ 1/6\\\ 1/6\end{pmatrix} = \begin{pmatrix}0\\\ -1/2\\\ 1/2\end{pmatrix}.

Resposta

O conjunto ortogonal obtido é
\vec{u}_1 = \begin{pmatrix}1\\\ 1\\\ 1\end{pmatrix},\quad \vec{u}_2 = \begin{pmatrix}-\tfrac{2}{3}\\[4pt]\tfrac{1}{3}\\[2pt]\tfrac{1}{3}\end{pmatrix},\quad \vec{u}_3 = \begin{pmatrix}0\\[2pt]-\tfrac{1}{2}\\[2pt]\tfrac{1}{2}\end{pmatrix}.]

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