(2,5 pontos) Sejam os vetores \boldsymbol{u}=(1,0,-2), \boldsymbol{v}=(1,3,1) e \boldsymbol{w}=(-1,1,1) e considere as seguintes matrizes:
U=\left[\begin{array}{lll} x & 1 & 2 \\\\ 0 & 3 & 1 \\\\ 2 & 1 & 0 \end{array}\right], \quad V=\left[\begin{array}{lll} 1 & 1 & 2 \\\\ 0 & y & 1 \\\\ 1 & 3 & 2 \end{array}\right], \text { e } \quad W=\left[\begin{array}{lll} 2 & 1 & 2 \\\\ 1 & 3 & 1 \\\\ 2 & 0 & z \end{array}\right] .
Utilizando a regra de Cramer, determine x, y e z de forma que os vetores \tilde{\boldsymbol{u}}=U \boldsymbol{u}, \tilde{\boldsymbol{v}}=V \boldsymbol{v} e \tilde{\boldsymbol{w}}=W \boldsymbol{w} sejam ortogonais. Encontre os vetores unitários correspondentes para \tilde{\boldsymbol{u}}, \tilde{\boldsymbol{v}} e \tilde{\boldsymbol{w}}.

Passo 1

O enunciado dá os vetores \boldsymbol{u}=(1,0,-2), \boldsymbol{v}=(1,3,1) e \boldsymbol{w}=(-1,1,1)
e as matrizes
U=\begin{pmatrix} x & 1 & 2\\\\ 0 & 3 & 1\\\\ 2 & 1 & 0 \end{pmatrix},\quad V=\begin{pmatrix} 1 & 1 & 2\\\\ 0 & y & 1\\\\ 1 & 3 & 2 \end{pmatrix},\quad W=\begin{pmatrix} 2 & 1 & 2\\\\ 1 & 3 & 1\\\\ 2 & 0 & z \end{pmatrix}.
Precisamos escolher x,y,z de modo que \tilde{\boldsymbol{u}}=U\boldsymbol{u}, \tilde{\boldsymbol{v}}=V\boldsymbol{v} e \tilde{\boldsymbol{w}}=W\boldsymbol{w}
sejam mutuamente ortogonais (produto interno zero) e depois normalizar cada um para obter o vetor unitário correspondente.

Passo 2

Calculemos primeiro cada vetor transformado:
Multiplicando U por \boldsymbol{u}:
\tilde{\boldsymbol{u}} =U\boldsymbol{u} =\begin{pmatrix} x & 1 & 2\\\\ 0 & 3 & 1\\\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix}1\\\ 0\\\ -2\end{pmatrix} =\begin{pmatrix} x\cdot1+1\cdot0+2\cdot(-2)\\\\ 0\cdot1+3\cdot0+1\cdot(-2)\\\\ 2\cdot1+1\cdot0+0\cdot(-2) \end{pmatrix} =\begin{pmatrix}x-4\\\ -2\\\ 2\end{pmatrix}.
Multiplicando V por \boldsymbol{v}:
\tilde{\boldsymbol{v}} =V\boldsymbol{v} =\begin{pmatrix} 1 & 1 & 2\\\\ 0 & y & 1\\\\ 1 & 3 & 2 \end{pmatrix} \begin{pmatrix}1\\\ 3\\\ 1\end{pmatrix} =\begin{pmatrix} 1+3+2\\\\ 0+3y+1\\\\ 1+9+2 \end{pmatrix} =\begin{pmatrix}6\\\ 3y+1\\\ 12\end{pmatrix}.
Multiplicando W por \boldsymbol{w}:
\tilde{\boldsymbol{w}} =W\boldsymbol{w} =\begin{pmatrix} 2 & 1 & 2\\\\ 1 & 3 & 1\\\\ 2 & 0 & z \end{pmatrix} \begin{pmatrix}-1\\\ 1\\\ 1\end{pmatrix} =\begin{pmatrix} -2+1+2\\\\ -1+3+1\\\\ -2+0+z \end{pmatrix} =\begin{pmatrix}1\\\ 3\z-2\end{pmatrix}.

Passo 3

Para que \tilde{\boldsymbol{u}},\tilde{\boldsymbol{v}},\tilde{\boldsymbol{w}} sejam ortogonais, impomos
\tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{v}}=0,\quad \tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{w}}=0,\quad \tilde{\boldsymbol{v}}\cdot\tilde{\boldsymbol{w}}=0.
Calculando cada um:
\tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{v}} =(x-4)\cdot6+(-2)(3y+1)+2\cdot12 =6x-24-6y-2+24 =6x-6y-2=0,
\tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{w}} =(x-4)\cdot1+(-2)\cdot3+2\,(z-2) =x-4-6+2z-4 =x+2z-14=0,
\tilde{\boldsymbol{v}}\cdot\tilde{\boldsymbol{w}} =6\cdot1+(3y+1)\cdot3+12\,(z-2) =6+9y+3+12z-24 =9y+12z-15=0.
Isso gera o sistema linear
\begin{cases} 6x-6y-2=0,\\\ x+2z-14=0,\\\\ 9y+12z-15=0, \end{cases}
ou simplificando,
\begin{cases} 3x-3y=1,\\\ x+2z=14,\\\\ 3y+4z=5. \end{cases}

Passo 4

Vamos aplicar a regra de Cramer a \displaystyle A=\begin{pmatrix}3&-3&0\\\ 1&0&2\\\ 0&3&4\end{pmatrix}, \boldsymbol{b}=(1,14,5)^T.
O determinante de A é
\det A =3\,(0\cdot4-2\cdot3)-(-3)\,(1\cdot4-2\cdot0) =3\,( -6)+3\,(4) =-18+12=-6.
Substituindo coluna a coluna, obtemos
D_x =\det\begin{pmatrix}1&-3&0\\\ 14&0&2\\\ 5&3&4\end{pmatrix} =-6+138=132, \quad D_y =\det\begin{pmatrix}3&1&0\\\ 1&14&2\\\ 0&5&4\end{pmatrix} =134, \quad D_z =\det\begin{pmatrix}3&-3&1\\\ 1&0&14\\\ 0&3&5\end{pmatrix} =-108.
Logo
x=\frac{D_x}{\det A}=\frac{132}{-6}=-22,\quad y=\frac{D_y}{\det A}=\frac{134}{-6}=-\tfrac{67}{3},\quad z=\frac{D_z}{\det A}=\frac{-108}{-6}=18.

Passo 5

Substituindo em cada vetor, fica
\tilde{\boldsymbol{u}} =(-22-4,\,-2,\,2)=(-26,-2,2),\quad \tilde{\boldsymbol{v}} =(6,\,3(-\tfrac{67}{3})+1,\,12)=(6,-66,12),\quad \tilde{\boldsymbol{w}} =(1,\,3,\,18-2)=(1,3,16).

Passo 6

Calculamos as normas:
\|\tilde{\boldsymbol{u}}\| =\sqrt{(-26)^2+(-2)^2+2^2} =\sqrt{676+4+4} =\sqrt{684} =6\sqrt{19},
\|\tilde{\boldsymbol{v}}\| =\sqrt{6^2+(-66)^2+12^2} =\sqrt{36+4356+144} =\sqrt{4536} =18\sqrt{14},
\|\tilde{\boldsymbol{w}}\| =\sqrt{1^2+3^2+16^2} =\sqrt{1+9+256} =\sqrt{266}.
Os vetores unitários são, portanto,
\hat{\boldsymbol{u}} =\frac{\tilde{\boldsymbol{u}}}{\|\tilde{\boldsymbol{u}}\|} =\frac{(-26,-2,2)}{6\sqrt{19}} =\bigl(-\tfrac{13}{3\sqrt{19}},\,-\tfrac{1}{3\sqrt{19}},\,\tfrac{1}{3\sqrt{19}}\bigr),
\hat{\boldsymbol{v}} =\frac{\tilde{\boldsymbol{v}}}{\|\tilde{\boldsymbol{v}}\|} =\frac{(6,-66,12)}{18\sqrt{14}} =\bigl(\tfrac{1}{3\sqrt{14}},\,-\tfrac{11}{3\sqrt{14}},\,\tfrac{2}{3\sqrt{14}}\bigr),
\hat{\boldsymbol{w}} =\frac{\tilde{\boldsymbol{w}}}{\|\tilde{\boldsymbol{w}}\|} =\frac{(1,3,16)}{\sqrt{266}}.

Resposta

x=-22,\quad y=-\tfrac{67}{3},\quad z=18.
Os vetores unitários são
\hat{\boldsymbol{u}} =\bigl(-\tfrac{13}{3\sqrt{19}},\,-\tfrac{1}{3\sqrt{19}},\,\tfrac{1}{3\sqrt{19}}\bigr),\quad \hat{\boldsymbol{v}} =\bigl(\tfrac{1}{3\sqrt{14}},\,-\tfrac{11}{3\sqrt{14}},\,\tfrac{2}{3\sqrt{14}}\bigr),\quad \hat{\boldsymbol{w}} =\frac{(1,3,16)}{\sqrt{266}}.

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