Assinale a alternativa que contém a base de \mathbb{R}^{3} gerada a partir da base B=\{(-1,0,0),(0,2,1),(-1,-1,-2)\} utilizando 0 processo de Gram Schimidt.
a. \left\{(1,0,0),\left(0,-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}
b. \left\{(1,0,0),\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right),\left(0, \frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right)\right\}
c. \left\{(-1,0,0),\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right),\left(0, \frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right)\right\}
d. \left\{(-1,0,0),\left(0,-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right),\left(0, \frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right)\right\}
e. \quad\left\{(-1,0,0),\left(0, \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}
f. \left\{(1,0,0),\left(0,-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}}\right),\left(0,-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\right\}

Passo 1

O enunciado dá a base \displaystyle B=\{v_{1}=(-1,0,0),\;v_{2}=(0,2,1),\;v_{3}=(-1,-1,-2)\}
em \mathbb{R}^{3} e pede uma base ortonormal obtida pelo processo de Gram–Schmidt. Vamos calcular vetores ortogonais u_{1},u_{2},u_{3} e depois normalizá-los para obter e_{1},e_{2},e_{3}.

Passo 2

Definimos \displaystyle u_{1}=v_{1}=(-1,0,0).
O comprimento é \displaystyle \|u_{1}\|=\sqrt{(-1)^{2}+0^{2}+0^{2}}=1,
logo o primeiro vetor normalizado é \displaystyle e_{1}=\frac{u_{1}}{\|u_{1}\|}=(-1,0,0).

Passo 3

Agora
u_{2}=v_{2}-\mathrm{proj}_{e_{1}}(v_{2})
e \displaystyle \mathrm{proj}_{e_{1}}(v_{2})=(v_{2}\cdot e_{1})\,e_{1}=(0\cdot(-1)+2\cdot0+1\cdot0)\,e_{1}=0.
Então \displaystyle u_{2}=v_{2}=(0,2,1)
e \displaystyle \|u_{2}\|=\sqrt{0^{2}+2^{2}+1^{2}}=\sqrt{5}.
O segundo vetor ortonormal é \displaystyle e_{2}=\frac{u_{2}}{\|u_{2}\|}=\bigl(0,\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\bigr).

Passo 4

Por fim,
u_{3} =v_{3}-\mathrm{proj}_{e_{1}}(v_{3})-\mathrm{proj}_{e_{2}}(v_{3}).
Calculamos \displaystyle \mathrm{proj}_{e_{1}}(v_{3})=(v_{3}\cdot e_{1})\,e_{1}=(1)\,(-1,0,0)=(-1,0,0),
logo \displaystyle v_{3}-\mathrm{proj}_{e_{1}}(v_{3})=(-1,-1,-2)-(-1,0,0)=(0,-1,-2).
Em seguida
v_{3}'=(0,-1,-2),\quad v_{3}'\cdot e_{2}=-1\cdot\frac{2}{\sqrt{5}}-2\cdot\frac{1}{\sqrt{5}}=-\frac{4}{\sqrt{5}},
\mathrm{proj}_{e_{2}}(v_{3})=-\frac{4}{\sqrt{5}}\;e_{2} =\Bigl(0,-\frac{8}{5},-\frac{4}{5}\Bigr),
u_{3}=(0,-1,-2)-\Bigl(0,-\tfrac{8}{5},-\tfrac{4}{5}\Bigr) =\Bigl(0,\tfrac{3}{5},-\tfrac{6}{5}\Bigr), \displaystyle \|u_{3}\|=\frac{3\sqrt{5}}{5},
portanto \displaystyle e_{3}=\frac{u_{3}}{\|u_{3}\|}=\bigl(0,\tfrac{1}{\sqrt{5}},-\tfrac{2}{\sqrt{5}}\bigr).

Passo 5

A base ortonormal resultante é
\Bigl\{(-1,0,0),\;\bigl(0,\tfrac{2}{\sqrt{5}},\tfrac{1}{\sqrt{5}}\bigr), \;\bigl(0,\tfrac{1}{\sqrt{5}},-\tfrac{2}{\sqrt{5}}\bigr)\Bigr\},
que corresponde à alternativa (c).

Resposta

(c)

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