Questão 3 [2 pontos]
Considere os vetores no espaço \mathbb{R}^{3} :
1
\vec{v}_{1}=\left(\begin{array}{l} 1 \\\\ 1 \\\\ 0 \end{array}\right), \quad \vec{v}_{2}=\left(\begin{array}{l} 1 \\\\ 0 \\\\ 1 \end{array}\right), \quad \vec{v}_{3}=\left(\begin{array}{l} 0 \\\\ 1 \\\\ 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 pede para aplicar o processo de Gram–Schmidt ao conjunto \displaystyle \vec{v}_{1}=\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix},\quad \vec{v}_{2}=\begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix},\quad \vec{v}_{3}=\begin{pmatrix}0\\\ 1\\\ 1\end{pmatrix}
para obter um conjunto ortogonal \{\vec{u}_1,\vec{u}_2,\vec{u}_3\}. Vamos usar o produto interno usual em \mathbb{R}^3 e a fórmula da projeção
\mathrm{proj}_{\vec{u}}(\vec{v}) = \frac{\vec{v}\cdot\vec{u}}{\vec{u}\cdot\vec{u}}\,\vec{u}.

Passo 2

Definimos
\vec{u}_1 = \vec{v}_1 = \begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix}.
Calculamos
\vec{u}_1\cdot\vec{u}_1 = 1^2 + 1^2 + 0^2 = 2.

Passo 3

Para \vec{u}_2 subtraímos de \vec{v}_2 sua projeção sobre \vec{u}_1. Primeiro
\vec{v}_2\cdot\vec{u}_1 = 1\cdot1 + 0\cdot1 + 1\cdot0 = 1,
então
\mathrm{proj}_{\vec{u}_1}(\vec{v}_2) = \frac{1}{2}\,\vec{u}_1 = \frac12\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix} = \begin{pmatrix}1/2\\\ 1/2\\\ 0\end{pmatrix},
logo
\vec{u}_2 = \vec{v}_2 - \mathrm{proj}_{\vec{u}_1}(\vec{v}_2) = \begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix} - \begin{pmatrix}1/2\\\ 1/2\\\ 0\end{pmatrix} = \begin{pmatrix}1/2\\\ -1/2\\\ 1\end{pmatrix}.

Passo 4

Para \vec{u}_3 subtraímos de \vec{v}_3 as projeções sobre \vec{u}_1 e \vec{u}_2. Primeiro
\vec{v}_3\cdot\vec{u}_1 = 0\cdot1 + 1\cdot1 + 1\cdot0 = 1,
então
\vec{v}_3 - \frac{\vec{v}_3\cdot\vec{u}_1}{\vec{u}_1\cdot\vec{u}_1}\,\vec{u}_1 = \begin{pmatrix}0\\\ 1\\\ 1\end{pmatrix} - \frac12\begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix} = \begin{pmatrix}-1/2\\\ 1/2\\\ 1\end{pmatrix}.
Em seguida calculamos
\vec{u}_2\cdot\vec{u}_2 = \bigl(\tfrac12\bigr)^2 + \bigl(-\tfrac12\bigr)^2 + 1^2 = \tfrac34+\tfrac14 = \tfrac32,
\bigl(\vec{v}_3 - \tfrac12\vec{u}_1\bigr)\!\cdot\!\vec{u}_2 = (-\tfrac12)\tfrac12 + \tfrac12(-\tfrac12) + 1\cdot1 = -\tfrac14 -\tfrac14 +1 = \tfrac12,
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\mathrm{proj}_{\vec{u}_2}\!\bigl(\vec{v}_3 - \tfrac12\vec{u}_1\bigr) = \frac{\tfrac12}{\tfrac32}\,\vec{u}_2 = \frac13\,\vec{u}_2 = \frac13\begin{pmatrix}1/2\\\ -1/2\\\ 1\end{pmatrix} = \begin{pmatrix}1/6\\\ -1/6\\\ 1/3\end{pmatrix}.
Finalmente
\vec{u}_3 = \begin{pmatrix}-1/2\\\ 1/2\\\ 1\end{pmatrix} - \begin{pmatrix}1/6\\\ -1/6\\\ 1/3\end{pmatrix} = \begin{pmatrix}-2/3\\\ 2/3\\\ 2/3\end{pmatrix}.

Resposta

\vec{u}_1 = \begin{pmatrix}1\\\ 1\\\ 0\end{pmatrix},\quad \vec{u}_2 = \begin{pmatrix}1/2\\\ -1/2\\\ 1\end{pmatrix},\quad \vec{u}_3 = \begin{pmatrix}-2/3\\\ 2/3\\\ 2/3\end{pmatrix}.

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