Question

Determine o valor de "a" para que o vetor u = (a, -2a, 2a) seja seja um versor (vetor unitário). Podem me ajudar?

Answer 1

Resposta:

Explicação passo-a-passo:

Para ser unitário, módulo dele deve ser 1.

$Dado \ \vec{u}=(x,y,z)\\|\vec{u}|= \sqrt{x^2+y^2+z^2}\\\\Temos, \ no \ problema\\\vec{u}=(a,-2a,2a)\\|\vec{u}| \ para \ ser \ unitario\\\\\sqrt{a^2+(-2a)^2+(2a)^2} =1\\\\\sqrt{a^2+4a^2+4a^2}= 1\\\\\sqrt{9a^2} = 1\\\\(\sqrt{9a^2})^2=1^2\\\\9a^2=1\\\\a^2= \frac{1}{9}\\a= \± \sqrt{\frac{1}{9}} => a= \± \frac{1}{3}$

Answer 2

✅ Após resolver os cálculos, concluímos que o valor do parâmetro "a" que torna o vetor "u" em um versor é:

            $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf a = \pm \frac{1}{3} \:\:\:}}\end{gathered} }$

Seja o vetor dado:

          $\Large\displaystyle\begin{gathered}\vec{u} = (a, -2a, 2a) \end{gathered}$

Dizemos que um determinado vetor é versor - vetor unitário - se, e somente se, o módulo do referido vetor for igual a unidade, ou seja:

     $\Large\displaystyle\begin{gathered}\vec{u} = \hat{u}\:\:\:\Longleftrightarrow\:\:\:\|\vec{u}\| = 1 \end{gathered}$

Então, temos:

                                              $\Large\displaystyle\begin{gathered}\|\vec{u}\| = 1 \end{gathered}$

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

                  $\Large\displaystyle\text{ \begin{gathered} \sqrt{a^{2} + 4a^{2} + 4a^{2}} = 1\end{gathered} }$

                                           $\Large\displaystyle\text{ \begin{gathered} \sqrt{9a^{2}} = 1\end{gathered} }$

                                      $\Large\displaystyle\text{ \begin{gathered} (\sqrt[\!\diagup\!\!]{9a^{2}})^{\!\diagup\!\!\!\!2} = 1^{2} \end{gathered} }$

                                                $\Large\displaystyle\begin{gathered}9a^{2} = 1 \end{gathered}$

                                                   $\Large\displaystyle\begin{gathered}a^{2} = \frac{1}{9} \end{gathered}$

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

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

                                                     $\Large\displaystyle\begin{gathered}a = \pm\frac{1}{3} \end{gathered}$

✅ Portanto, o valor de "a" que torna o vetor "u" em um versor é:

                         $\Large\displaystyle\begin{gathered}a = \pm\frac{1}{3} \end{gathered}$      

Neste caso, teremos dois vetores unitários que são:

• Se a = 1/3:

        $\Large\displaystyle\begin{gathered}\hat{u}' = (a, -2a, 2a) \end{gathered}$

             $\Large\displaystyle\begin{gathered}= \Bigg[\frac{1}{3}, -2\cdot\Bigg(\frac{1}{3} \Bigg), 2\cdot\Bigg(\frac{1}{3} \Bigg) \Bigg] \end{gathered}$

             $\Large\displaystyle\begin{gathered}= \Bigg(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3} \Bigg) \end{gathered}$  

         $\Large\displaystyle\begin{gathered}\therefore\:\:\:\hat{u}' = \Bigg(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3} \Bigg) \end{gathered}$

• Se a = -1/3

        $\Large\displaystyle\begin{gathered}\hat{u}'' = (a, -2a, 2a) \end{gathered}$

             $\Large\displaystyle\begin{gathered}= \Bigg[\frac{-1}{3}, -2\cdot\Bigg(\frac{-1}{3} \Bigg), 2\cdot\Bigg(\frac{-1}{3} \Bigg) \Bigg] \end{gathered}$

             $\Large\displaystyle\begin{gathered}= \Bigg(\frac{-1}{3}, \frac{2}{3}, \frac{-2}{3} \Bigg) \end{gathered}$  

         $\Large\displaystyle\begin{gathered}\therefore\:\:\:\hat{u}'' = \Bigg(\frac{-1}{3}, \frac{2}{3}, \frac{-2}{3} \Bigg) \end{gathered}$

         

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Solução gráfica: