Considere a base B de \mathbb{R}^{3} dada por B=\{(2,0,1),(-1,1,2),(0,0,1)\}.
(a) Determine uma base ortonormal para \mathbb{R}^{3} a partir da base B usando o processo de orto-
(b) Determine as coordenadas do vetor v=(-2,1,3) na base encontrada no item (a). normalização de Gram-Schmidt.

Passo 1

O enunciado dá a base B=\{b_1,b_2,b_3\} de \mathbb{R}^3, com
b_1=(2,0,1),\quad b_2=(-1,1,2),\quad b_3=(0,0,1),
e pede (a) uma base ortonormal obtida por Gram-Schmidt e (b) as coordenadas de v=(-2,1,3) nessa base. Usaremos o produto interno usual e projeções ortogonais.

Passo 2

Definimos
u_1 = b_1 = (2,0,1), \quad \|u_1\| = \sqrt{2^2+0^2+1^2}=\sqrt5,
logo
e_1 = \frac{u_1}{\|u_1\|} = \Bigl(\frac{2}{\sqrt5},\,0,\,\frac{1}{\sqrt5}\Bigr).

Passo 3

Projetamos b_2 em e_1:
\mathrm{proj}_{e_1}(b_2) = (b_2\cdot e_1)\,e_1, \quad b_2\cdot e_1 = \frac{-1\cdot2 + 1\cdot0 + 2\cdot1}{\sqrt5} = 0,
então
u_2 = b_2 - 0\cdot e_1 = b_2 = (-1,1,2), \quad \|u_2\|=\sqrt{1+1+4}=\sqrt6,
e
e_2 = \frac{u_2}{\|u_2\|} = \Bigl(-\frac{1}{\sqrt6},\,\frac{1}{\sqrt6},\,\frac{2}{\sqrt6}\Bigr).

Passo 4

Agora projetamos b_3 em e_1 e em e_2:
b_3\cdot e_1 = \frac{1}{\sqrt5},\quad \mathrm{proj}_{e_1}(b_3)=\frac{1}{\sqrt5}e_1=\Bigl(\frac{2}{5},0,\frac{1}{5}\Bigr),
b_3\cdot e_2 = \frac{2}{\sqrt6},\quad \mathrm{proj}_{e_2}(b_3)=\frac{2}{\sqrt6}e_2=\Bigl(-\tfrac{1}{3},\tfrac{1}{3},\tfrac{2}{3}\Bigr).
Então
u_3 = b_3 - \mathrm{proj}_{e_1}(b_3) - \mathrm{proj}_{e_2}(b_3) = \Bigl(-\tfrac{1}{15},\,-\tfrac{1}{3},\,\tfrac{2}{15}\Bigr).

Passo 5

Calculamos a norma de u_3:
\|u_3\| = \sqrt{\bigl(-\tfrac{1}{15}\bigr)^2+\bigl(-\tfrac{1}{3}\bigr)^2+\bigl(\tfrac{2}{15}\bigr)^2} = \frac{\sqrt{30}}{15},
logo
e_3 = \frac{u_3}{\|u_3\|} = \Bigl(-\tfrac{1}{\sqrt{30}},\,-\tfrac{5}{\sqrt{30}},\,\tfrac{2}{\sqrt{30}}\Bigr).

Passo 6

Para as coordenadas de v=(-2,1,3) nessa base, calculamos
a_1 = v\cdot e_1 = \frac{-4+0+3}{\sqrt5} = -\frac{1}{\sqrt5},
a_2 = v\cdot e_2 = \frac{2+1+6}{\sqrt6} = \frac{9}{\sqrt6},
a_3 = v\cdot e_3 = \frac{2 -5 +6}{\sqrt{30}} = \frac{3}{\sqrt{30}}.

Resposta

(a) Uma base ortonormal é
\bigl\{e_1,e_2,e_3\bigr\} = \Bigl\{\,(2/\sqrt5,0,1/\sqrt5),\;(-1/\sqrt6,1/\sqrt6,2/\sqrt6),\; (-1/\sqrt{30},-5/\sqrt{30},2/\sqrt{30})\Bigr\}.
(b) As coordenadas de v nessa base são
\bigl(a_1,a_2,a_3\bigr) = \Bigl(-\tfrac{1}{\sqrt5},\,\tfrac{9}{\sqrt6},\,\tfrac{3}{\sqrt{30}}\Bigr).

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