Question

Ache u de norma √5, ortogonal a v = (2, 1, -1) e tal que (u, (1, 1, 1), (0, 1, -1)) seja LD.

Answer 1

Seja $\overrightarrow{\mathbf{u}}=(a,\,b,\,c)$ o vetor procurado.


$\bullet~~$ Queremos que a norma de $\overrightarrow{\mathbf{u}}$ seja $\sqrt{5}:$

$\|\overrightarrow{\mathbf{u}}\|=\sqrt{5}\\\\ \|(a,\,b,\,c)\|=\sqrt{5}\\\\ \sqrt{a^2+b^2+c^2}=\sqrt{5}\\\\ a^2+b^2+c^2=5~~~~~~\mathbf{(i)}$


$\bullet~~$ Queremos que $\overrightarrow{\mathbf{u}}$ seja ortogonal a $\overrightarrow{\mathbf{v}}=(2,\,1,\,-1):$

(Dois vetores são ortogonais se e só se o produto escalar entre eles for zero )

$\overrightarrow{\mathbf{u}}\cdot \overrightarrow{\mathbf{v}}=0\\\\ (a,\,b,\,c)\cdot (2,\,1,\,-1)=0\\\\ 2a+b-c=0~~~~~~\mathbf{(ii)}$


$\bullet~~$ Queremos que o conjunto

$A=\{\overrightarrow{\mathbf{u}};\,(1,\,1,\,1);\,(0,\,1,\,-1)\}\\\\ =\{(a,\,b,\,c);\,(1,\,1,\,1);\,(0,\,1,\,-1)\}$


seja um conjunto de vetores linearmente dependentes. Isto acontecerá somente se

$\det\!\left[ \begin{array}{ccc} a&b&c\\ 1&1&1\\ 0&1&-1 \end{array} \right]=0\\\\\\ (1\cdot (-1)-1\cdot 1)a-(1\cdot (-1)-0\cdot 1)b+(1\cdot 1-0\cdot 1)c=0\\\\ (-1-1)a-(-1-0)b+(1-0)c=0\\\\ -2a+b+c=0~~~~~~\mathbf{(iii)}$

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Basta agora resolver o sistema formado pelas equações $\mathbf{(i)},\,\mathbf{(ii)}$ e $\mathbf{(iii)}:$

$\left\{ \!\begin{array}{rc} a^2+b^2+c^2=5&~~~~\mathbf{(i)}\\\\ 2a+b-c=0&~~~~\mathbf{(ii)}\\\\ -2a+b+c=0&~~~~\mathbf{(iii)} \end{array} \right.$


Somando as equações $\mathbf{(ii)}$ e $\mathbf{(iii)}$ membro a membro, obtemos

$2b=0~~\Rightarrow~~\boxed{\begin{array}{c} b=0 \end{array}}$


Então o sistema se reduz a

$\left\{ \!\begin{array}{rc} a^2+c^2=5&~~~~\mathbf{(iv)}\\\\ 2a-c=0&~~~~\mathbf{(v)} \end{array} \right.$


Isolando $c$ na equação $\mathbf{(v)}$ e substituindo em $\mathbf{(iv)},$ obtemos

$c=2a\\\\\\ a^2+(2a)^2=5\\\\ a^2+4a^2=5\\\\ 5a^2=5\\\\ a^2=1\\\\ a=\pm 1\\\\ \boxed{\begin{array}{c} a=-1~~\text{ ou }~~a=1 \end{array}}$


$\bullet~~$ Para $a=-1~~\Rightarrow~~\boxed{\begin{array}{c}c=-2 \end{array}}$

$\bullet~~$ Para $a=1~~\Rightarrow~~\boxed{\begin{array}{c}c=2 \end{array}}$

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Logo, temos duas possibilidades para o vetor $\overrightarrow{\mathbf{u}}:$

$\boxed{\begin{array}{c} \overrightarrow{\mathbf{u}}=(-1,\,0,\,-2)~~\text{ ou }~~\overrightarrow{\mathbf{u}}=(1,\,0,\,2) \end{array}}$