Encontre uma base ortonormal \mathcal{B} à partir da base \mathcal{A}=\{(1,1,1),(1,1,0),(1,0,0)\}. Calcule os coeficientes de Fourier c_{1}, c_{2} e c_{3} de u=[1,-3,2] na base \mathcal{B}. A soma s=c_{1}+c_{2}+c_{3} vale:
a. s<0
b. 0 \leq s<0,5
c. 0,5 \leq s<1,0
d. 1,0 \leq s<2,0
e. s \geq 2,0

Passo 1

Resumindo o enunciado: temos a base \mathcal{A}=\{(1,1,1),\,(1,1,0),\,(1,0,0)\} em \mathbb{R}^3 e o vetor u=(1,-3,2).
O que precisamos fazer:
  1. Usar o processo de Gram–Schmidt para obter uma base ortonormal \mathcal{B}=\{e_1,e_2,e_3\}.
  2. Calcular os coeficientes de Fourier c_i=\langle u,e_i\rangle.
  3. Somar s=c_1+c_2+c_3 e decidir em qual intervalo ele se encaixa.

Passo 2

Aplicando Gram–Schmidt em
v_1=(1,1,1),\quad v_2=(1,1,0),\quad v_3=(1,0,0),
definimos
u_1=v_1,\qquad e_1=\frac{u_1}{\|u_1\|}=\frac{(1,1,1)}{\sqrt{3}}.
Em seguida
u_2 = v_2 - \frac{\langle v_2,u_1\rangle}{\|u_1\|^2}\,u_1 = (1,1,0)-\frac{2}{3}(1,1,1) =\bigl(\tfrac13,\tfrac13,-\tfrac23\bigr),
e
e_2 = \frac{u_2}{\|u_2\|} =\frac{\bigl(\tfrac13,\tfrac13,-\tfrac23\bigr)}{\sqrt{6}/3} =\frac{(1,1,-2)}{\sqrt{6}}.

Passo 3

Para o terceiro vetor, primeiro tiramos a projeção em u_1 e u_2:
w = v_3-\frac{\langle v_3,u_1\rangle}{\|u_1\|^2}u_1 = (1,0,0)-\tfrac13(1,1,1) =\bigl(\tfrac23,-\tfrac13,-\tfrac13\bigr),
u_3 = w - \frac{\langle w,u_2\rangle}{\|u_2\|^2}u_2 =\bigl(\tfrac23,-\tfrac13,-\tfrac13\bigr)-\tfrac12\bigl(\tfrac13,\tfrac13,-\tfrac23\bigr) =\bigl(\tfrac12,-\tfrac12,0\bigr),
e_3 = \frac{u_3}{\|u_3\|} =\frac{\bigl(\tfrac12,-\tfrac12,0\bigr)}{1/\sqrt2} =\frac{(1,-1,0)}{\sqrt{2}}.

Passo 4

Agora calculamos os coeficientes de Fourier
c_i = \langle u,e_i\rangle,\quad u=(1,-3,2).
Isso dá
c_1 = \frac{1-3+2}{\sqrt{3}} = 0,\qquad c_2 = \frac{1-3-4}{\sqrt{6}} = -\sqrt{6},\qquad c_3 = \frac{1+3+0}{\sqrt{2}} = 2\sqrt{2}.

Passo 5

A soma é
s = c_1+c_2+c_3 = 0 - \sqrt{6} + 2\sqrt{2}\approx 2{,}828 - 2{,}449 = 0{,}379.
Logo 0\le s<0,5.

Resposta

b. 0 \le s < 0,5

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