Determine um vetor unitário ortogonal simultaneamente aos vectores u = (1, −5, −1) e v = (0, 2, 1).... Pergunta de ideia deOpsy

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Determine um vetor unitário ortogonal simultaneamente aos vectores
u = (1, −5, −1) e v = (0, 2, 1)

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$\large\begin{array}{l} \textsf{Determinar um vetor }\overrightarrow{\mathsf{w}}\mathsf{(a,\,b,\,c),}\textsf{ de modo que} \end{array}$

•   $\large\begin{array}{l} \overrightarrow{\mathsf{w}}\textsf{ \'e simultaneamente ortogonal aos vetores}\end{array}$

$\large\begin{array}{l} \overrightarrow{\mathsf{u}}\mathsf{(1,\,-5,\,-1)}\textsf{ e }\overrightarrow{\mathsf{v}}\mathsf{(0,\,2,\,1);}\end{array}$

•   $\large\begin{array}{l} \overrightarrow{\mathsf{w}}\textsf{ \'e unit\'ario, isto \'e, sua norma vale 1.} \end{array}$

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$\large\begin{array}{l} \textsf{Dois vetores s\~ao ortogonais se e somente se o produto}\\\textsf{escalar entre eles \'e igual a zero. Logo, devemos ter}\\\\ \left\{\! \begin{array}{l} \overrightarrow{\mathsf{u}}\cdot \overrightarrow{\mathsf{w}}=\mathsf{0}\\\\ \overrightarrow{\mathsf{v}}\cdot \overrightarrow{\mathsf{w}}=\mathsf{0}\\\\ \|\overrightarrow{\mathsf{w}}\|=\mathsf{1} \end{array} \right. \end{array}$

$\large\begin{array}{l} \left\{\! \begin{array}{l} \mathsf{(1,\,-5,\,-1)\cdot (a,\,b,\,c)=0}\\\\ \mathsf{(0,\,2,\,1)\cdot (a,\,b,\,c)=0}\\\\ \mathsf{\|(a,\,b,\,c)\|=1} \end{array} \right. \end{array}$

$\large\begin{array}{l} \left\{\! \begin{array}{lc} \mathsf{a-5b-c=0}&\quad\mathsf{(i)}\\\\ \mathsf{0a+2b+c=0}&\quad\mathsf{(ii)}\\\\ \mathsf{a^2+b^2+c^2=1}&\quad\mathsf{(iii)} \end{array} \right. \end{array}$

$\large\begin{array}{l} \textsf{Multiplique a equa\c{c}\~ao (i) por 2, e a equa\c{c}\~ao (ii) por 5.}\\\textsf{Depois some as equa\c{c}\~oes membro a membro:}\\\\ \mathsf{2a-10b-2c=0}\\\\ \mathsf{0a+10b+5c=0}\\\\\\ \mathsf{2a+3c=0}\\\\ \mathsf{2a=-3c}\\\\ \mathsf{a=-\,\dfrac{3}{2}\,c}\end{array}$

$\large\begin{array}{l} \textsf{Da equa\c{c}\~ao (ii), tiramos tamb\'em que}\\\\ \mathsf{2b=-c}\\\\ \mathsf{b=-\,\dfrac{1}{2}\,c} \end{array}$

$\large\begin{array}{l} \textsf{Substituindo tudo na equa\c{c}\~ao (iii), ficamos com}\\\\ \mathsf{\Big(-\,\dfrac{3}{2}\,c\Big)^2+\Big(-\dfrac{1}{2}\,c\Big)^2+c^2=1}\\\\ \mathsf{\dfrac{9}{4}\,c^2+\dfrac{1}{4}\,c^2+c^2=1}\\\\ \mathsf{9c^2+c^2+4c^2=4}\\\\ \mathsf{14c^2=4}\\\\ \mathsf{c^2=\dfrac{4}{14}} \end{array}$

$\large\begin{array}{l} \mathsf{c=\pm\,\sqrt{\dfrac{4}{14}}}\\\\ \mathsf{c=\pm\,\dfrac{2}{\sqrt{14}}} \end{array}$

•   $\large\begin{array}{l} \textsf{Para }\mathsf{c=\dfrac{2}{\sqrt{14}},}\textsf{ obtemos} \end{array}$

$\large\begin{array}{l} \mathsf{a=-\dfrac{3}{2}\cdot \dfrac{2}{\sqrt{14}}}\\\\ \mathsf{a=-\,\dfrac{3}{\sqrt{14}}}\qquad\quad\checkmark\\\\\\ \mathsf{b=-\,\dfrac{1}{2}\cdot \dfrac{2}{\sqrt{14}}}\\\\ \mathsf{b=-\,\dfrac{1}{\sqrt{14}}}\qquad\quad\checkmark \end{array}$

$\large\begin{array}{l} \textsf{Encontramos uma possibilidade para o vetor }\overrightarrow{\mathsf{w}}:\\\\ \boxed{\begin{array}{c}\overrightarrow{\mathsf{w}}=\mathsf{\dfrac{1}{\sqrt{14}}\,(-3,\,-1,\,2)} \end{array}}\qquad\quad\checkmark \end{array}$

•   $\large\begin{array}{l} \textsf{Para }\mathsf{c=-\,\dfrac{2}{\sqrt{14}},}\textsf{ obtemos outro vetor de forma an\'aloga:} \end{array}$

$\large\begin{array}{l} \vdots\\\\ \boxed{\begin{array}{c}\overrightarrow{\mathsf{w}}=\mathsf{\dfrac{1}{\sqrt{14}}\,(3,\,1,\,-2)} \end{array}}\qquad\quad\checkmark\\\\\\ \textsf{(s\'o inverteram-se os sinais das coordenadas)} \end{array}$

$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$

Tags: vetor ortogonal unitário produto escalar norma sistema de equações álgebra linear geometria analítica

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