Question

Determine um vetor unitário que é simultaneamente ortogonal aos vetores u=(1,1,0) e v=(2,-1,3).

Answer 1

Um vetor "w" ortogonal a outros dois vetores "u" e "v" pode ser determinado pelo produto vetorial entre "u" e "v", acompanhe:

$\vec{w}~=~\vec{u}\times\vec{v}\\\\\\\vec{w}~=~\left|\begin{array}{ccc}i&j&k\\u_i&u_j&u_k\\v_i&v_j&v_k\end{array}\right|\\\\\\\vec{w}~=~\left|\begin{array}{ccc}i&j&k\\1&1&0\\2&-1&3\end{array}\right|\\\\\\\vec{w}~=~(~i.1.3+1.(-1).k+2.j.0~)~-~(~k.1.2+0.(-1).i+3.j.1~)\\\\\\\vec{w}~=~(~3i-k+0~)~-~(~2k+0+3j~)\\\\\\\vec{w}~=~3i-k-2k-3j\\\\\\\boxed{\vec{w}~=~3i-3j-3k}$

Vamos agora verificar se esse vetor é unitário:

$|\vec{w}|~=~\sqrt{w_i^{~2}+w_j^{~2}+w_k^{~2}}\\\\\\|\vec{w}|~=~\sqrt{3^2+(-3)^2+(-3)^2}\\\\\\|\vec{w}|~=~\sqrt{9+9+9}\\\\\\\boxed{|\vec{w}|~=~3\sqrt{3}}$

Como w não é unitário, vamos normaliza-lo dividindo pelo seu modulo:

$\hat{w}~=~\frac{\vec{w}}{|\vec{w}|}\\\\\\\hat{w}~=~\frac{3i~-~3j~-~3k}{3\sqrt{3}}\\\\\\\hat{w}~=~\frac{1}{\sqrt{3}}\,i~-~\frac{1}{\sqrt{3}}\,j~-~\frac{1}{\sqrt{3}}\,k\\\\\\Podemos~racionalizar~os~coeficientes\\\\\\\boxed{\hat{w}~=~\frac{\sqrt{3}}{3}\,i~-~\frac{\sqrt{3}}{3}\,j~-~\frac{\sqrt{3}}{3}\,k}~~ou~~\boxed{\hat{w}~=~\left(\dfrac{\sqrt{3}}{3}~,\,-\dfrac{\sqrt{3}}{3}~,\,-\dfrac{\sqrt{3}}{3}\right)}$

$\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$

Answer 2

✅ Após ter resolvido os cálculos, concluímos que o vetor unitário procurado é:

         $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \hat{w} = \Bigg(\frac{\sqrt{3}}{3}, \frac{-\sqrt{3}}{3}, \frac{-\sqrt{3}}{3} \Bigg) \:\:\:}}\end{gathered} }$

Sejam  os vetores:

                $\Large\begin{cases}\vec{u} = (1, 1, 0)\\\vec{v} = (2, -1, 3) \end{cases}$

Dizemos que um vetor é unitário se, e somente se, o seu módulo for igual a "1", ou seja:

   $\Large\displaystyle\begin{gathered}\hat{w} = Unit\acute{a}rio\:\:\:\Longleftrightarrow\:\:\:\|\hat{w}\| = 1 \end{gathered}$

E isto só irá acontecer quando o vetor "w" for normalizado, ou seja:

                       $\Large\displaystyle\text{ \begin{gathered}\hat{w} = \frac{\vec{w}}{\|\vec{w}\|} \end{gathered} }$

Como estamos querendo encontrar o vetor unitário que é simultaneamente ortogonal aos vetores supracitados então, devemos encontrar o vetor unitário do produto vetorial entre os dois vetores citados. Neste caso, temos:

      $\Large\displaystyle\text{ \begin{gathered}\vec{w} = \vec{u}\wedge\vec{v} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k}\\x_{\vec{u}} & y_{\vec{u}} & z_{\vec{u}}\\x_{\vec{v}} & y_{\vec{v}} & z_{\vec{v}} \end{vmatrix}, \:\:\:com\:\vec{u}\nparallel\vec{v}\end{gathered} }$

• Calculando o produto vetorial:

              $\Large\displaystyle\begin{gathered}\vec{w} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k}\\1 & 1 & 0\\2 & -1 & 3 \end{vmatrix} \end{gathered}$

                    $\Large\displaystyle\begin{gathered}= \begin{vmatrix}1 & 0\\-1 & 3 \end{vmatrix}\vec{i} - \begin{vmatrix}1 & 0\\2 & 3 \end{vmatrix}\vec{j} + \begin{vmatrix}1 & 1\\2 & -1 \end{vmatrix}\vec{k} \end{gathered}$

                    $\Large\displaystyle\begin{gathered}= (3 + 0)\vec{i} - (3 - 0)\vec{j} + (-1 - 2)\vec{k} \end{gathered}$

                    $\Large\displaystyle\begin{gathered}= 3\vec{i} - 3\vec{j} - 3\vec{k} \end{gathered}$

                   $\Large\displaystyle\begin{gathered}= (3, -3, -3) \end{gathered}$

           $\Large\displaystyle\begin{gathered}\therefore\:\:\:\vec{w} = (3, -3, -3) \end{gathered}$

• Calculando o vetor unitário do vetor "w":

             $\Large\displaystyle\text{ \begin{gathered}\hat{w} = \frac{\vec{w}}{\|\vec{w}\|} \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}= \frac{(3, -3, -3)}{\sqrt{3^{2} + (-3)^{2} + (-3)^{2}}} \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}= \frac{(3, -3, -3)}{\sqrt{9 + 9 + 9}} \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}= \frac{(3, -3, -3)}{\sqrt{27}} \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}= \frac{(3, -3, -3)}{\sqrt{3^{2}\cdot3}} \end{gathered} }$

                    $\Large\displaystyle\text{ \begin{gathered}= \frac{(3, -3, -3)}{3\sqrt{3}} \end{gathered} }$

                    $\Large\displaystyle\text{ \begin{gathered}= \Bigg(\frac{3}{3\sqrt{3}}, \frac{-3}{3\sqrt{3}}, \frac{-3}{3\sqrt{3}} \Bigg) \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}= \Bigg(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \Bigg) \end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}= \Bigg(\frac{\sqrt{3}}{3}, \frac{-\sqrt{3}}{3}, \frac{-\sqrt{3}}{3} \Bigg) \end{gathered} }$

✅ Portanto, o vetor unitário procurado é:

            $\Large\displaystyle\text{ \begin{gathered}\hat{w} = \Bigg(\frac{\sqrt{3}}{3}, \frac{-\sqrt{3}}{3}, \frac{-\sqrt{3}}{3} \Bigg) \end{gathered} }$

Prova:

                                                                        $\Large\displaystyle\begin{gathered}\|\hat{w}\| = 1 \end{gathered}$

  $\Large\displaystyle\text{ \begin{gathered}\sqrt{\Bigg(\frac{\sqrt{3}}{3} \Bigg)^{2} + \Bigg(\frac{-\sqrt{3}}{3} \Bigg)^{2} + \Bigg(\frac{-\sqrt{3}}{3} \Bigg)^{2}} = 1\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered}\sqrt{\frac{(\sqrt{3})^{2}}{3^{2}} + \frac{(-\sqrt{3})^{2}}{3^{2}} + \frac{(-\sqrt{3})^{2}}{3^{2}} } = 1\end{gathered} }$

                                                      $\Large\displaystyle\text{ \begin{gathered}\sqrt{\frac{3}{9} + \frac{3}{9} + \frac{3}{9} } = 1\end{gathered} }$

                                                                         $\Large\displaystyle\text{ \begin{gathered} \sqrt{\frac{9}{9} } = 1\end{gathered} }$

                                                                           $\Large\displaystyle\begin{gathered}\sqrt{1} = 1 \end{gathered}$

                                                                               $\Large\displaystyle\begin{gathered} 1 = 1\end{gathered}$

           

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