Question

Encontre um vetor unitário de sentido contrário e na mesma direção do vetor (1,2,3)

Answer 1

$\large\begin{array}{l} \textsf{Queremos encontrar um vetor }\overrightarrow{\mathsf{u}},\textsf{ de modo que}\\\\ \overrightarrow{\mathsf{u}}\mathsf{=t\cdot \overrightarrow{\mathsf{v}}}\\\\ \overrightarrow{\mathsf{u}}\mathsf{=t\cdot (1,\,2,\,3)}\\\\ \textsf{para algum }\mathsf{t<0.}\\\\\\ \textsf{Como }\overrightarrow{\mathsf{u}}\textsf{ \'e unit\'ario,}\\\\ \mathsf{\|\overrightarrow{\mathsf{u}}\|=1}\end{array}$

$\large\begin{array}{l}\mathsf{\|t\cdot (1,\,2,\,3)\|=1}\\\\ \mathsf{|t|\cdot \|(1,\,2,\,3)\|=1}\\\\\mathsf{|t|\cdot \sqrt{1^2+2^2+3^2}=1}\\\\ \mathsf{|t|=\dfrac{1}{\sqrt{1^2+2^2+3^2}}}\\\\ \mathsf{|t|=\dfrac{1}{\sqrt{1+4+9}}}\\\\ \mathsf{|t|=\dfrac{1}{\sqrt{14}}}\\\\ \mathsf{t=\pm \dfrac{1}{\sqrt{14}}} \end{array}$


$\large\begin{array}{l}\textsf{Como queremos }\mathsf{t<0,}\textsf{ ent\~ao } \mathsf{t=-\,\dfrac{1}{\sqrt{14}},}\textsf{ de forma que}\\\\ \overrightarrow{\mathsf{u}}\mathsf{=-\,\dfrac{1}{\sqrt{14}}\cdot (1,\,2,\,3)}\\\\\\ \boxed{\begin{array}{c}\mathsf{\overrightarrow{\mathsf{u}}=\left(-\,\dfrac{1}{\sqrt{14}},\,-\,\dfrac{2}{\sqrt{14}},\,-\,\dfrac{3}{\sqrt{14}}\right)} \end{array}} \end{array}$


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

Answer 2

✅ Após resolver todos os cálculos, concluímos que o simétrico ou oposto do vetor unitário de "v" é:

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S(\hat{v}) = \Bigg(- \frac{\sqrt{14}}{14}, - \frac{2\sqrt{14}}{14}, - \frac{3\sqrt{14}}{14} \Bigg)\:\:\:}}\end{gathered} }$

Seja o vetor:

                   $\Large\displaystyle\begin{gathered}\vec{v} = (1, 2, 3) \end{gathered}$

Se estamos querendo encontrar um vetor unitário de sentido contrário e na mesma direção do vetor "v", então devemos encontra o vetor simétrico ou oposto do vetor unitário de "v". Para isso, devemos utilizar a seguinte fórmula:

             $\Large\displaystyle\text{ \begin{gathered}S(\hat{v}) = (-1)\cdot\frac{\vec{v}}{\|\vec{v}\|} \end{gathered} }$

Então temos:

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

                  $\Large\displaystyle\text{ \begin{gathered}= (-1)\cdot\frac{(1, 2, 3)}{\sqrt{1 + 4 + 9}} \end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}= (-1)\cdot\frac{(1, 2, 3)}{\sqrt{14}} \end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}= (-1)\cdot\Bigg(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \Bigg) \end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}= (-1)\cdot\Bigg(\frac{\sqrt{14}}{14}, \frac{2\sqrt{14}}{14}, \frac{3\sqrt{14}}{14} \Bigg) \end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}= \Bigg(-\frac{\sqrt{14}}{14}, - \frac{2\sqrt{14}}{14}, -\frac{3\sqrt{14}}{14} \Bigg) \end{gathered} }$    

✅ Portanto, o vetor procurado é:

           $\Large\displaystyle\text{ \begin{gathered}S(\hat{v}) = \Bigg(-\frac{\sqrt{14}}{14}, - \frac{2\sqrt{14}}{14}, -\frac{3\sqrt{14}}{14} \Bigg) \end{gathered} }$    

Saiba mais:

1. https://brainly.com.br/tarefa/7115567
2. https://brainly.com.br/tarefa/7292789