Considere o produto interno usual em R^{4} e os seguintes dois vectores ortogonais e normalizados:
\mathrm{u}_{1}=\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}, 0\right) \quad \text { e } \quad \mathrm{u}_{2}=\left(\frac{2}{3}, \frac{-2}{3}, 0, \frac{1}{3}\right)
Considere ainda \mathrm{v}=(1,0,0,0). Determine um vector \mathrm{u}_{3} de modo que u_{1}, u_{2} e u_{3} formem uma base ortonormada do subespaço \mathrm{L}\left(\mathrm{u}_{1}, \mathrm{u}_{2}, \mathrm{v}\right).

Passo 1

O enunciado dá dois vetores ortonormais
\mathrm{u}_1=(\tfrac{2}{3},\tfrac{2}{3},\tfrac{1}{3},0) e
\mathrm{u}_2=(\tfrac{2}{3},-\tfrac{2}{3},0,\tfrac{1}{3})
e o vetor \mathrm{v}=(1,0,0,0). Queremos um \mathrm{u}_3 tal que
\{\mathrm{u}_1,\mathrm{u}_2,\mathrm{u}_3\} seja uma base ortonormal de
\mathrm{L}(\mathrm{u}_1,\mathrm{u}_2,\mathrm{v}).
Para isso, usamos o processo de Gram–Schmidt: projetar \mathrm{v} em
\mathrm{span}\{\mathrm{u}_1,\mathrm{u}_2\}, subtrair e normalizar o resto.

Passo 2

Calculemos as projeções de \mathrm{v} em \mathrm{u}_1 e \mathrm{u}_2:
(\mathrm{v},\mathrm{u}_1) =1\cdot\tfrac{2}{3}+0+0+0 =\tfrac{2}{3}, \quad (\mathrm{v},\mathrm{u}_2) =1\cdot\tfrac{2}{3}+0+0+0 =\tfrac{2}{3}.

Passo 3

Formamos
\mathrm{w} =\mathrm{v} - (\mathrm{v},\mathrm{u}_1)\,\mathrm{u}_1 - (\mathrm{v},\mathrm{u}_2)\,\mathrm{u}_2 \;=\; (1,0,0,0) -\tfrac{2}{3}( \tfrac{2}{3},\tfrac{2}{3},\tfrac{1}{3},0) -\tfrac{2}{3}( \tfrac{2}{3},-\tfrac{2}{3},0,\tfrac{1}{3}) \;=\; \bigl(\tfrac{1}{9},0,-\tfrac{2}{9},-\tfrac{2}{9}\bigr).

Passo 4

Calculamos a norma de \mathrm{w}:
\|\mathrm{w}\|^2 =\Bigl(\tfrac{1}{9}\Bigr)^{2} +\Bigl(-\tfrac{2}{9}\Bigr)^{2} +\Bigl(-\tfrac{2}{9}\Bigr)^{2} =\tfrac{1+4+4}{81} =\tfrac{9}{81} =\tfrac{1}{9},
logo \|\mathrm{w}\|=\tfrac{1}{3}.
Portanto,
\mathrm{u}_3 =\frac{\mathrm{w}}{\|\mathrm{w}\|} =\frac{1}{\tfrac{1}{3}}\bigl(\tfrac{1}{9},0,-\tfrac{2}{9},-\tfrac{2}{9}\bigr) =\bigl(\tfrac{1}{3},0,-\tfrac{2}{3},-\tfrac{2}{3}\bigr).

Resposta

\mathrm{u}_3=\bigl(\tfrac{1}{3},0,-\tfrac{2}{3},-\tfrac{2}{3}\bigr).

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