Question

Calcule o ângulo entre os planos π1 : − y + 1 = 0 e π2 : y + z + 2 = 0.

Answer 1

✅ Após resolver os cálculos, concluímos que o ângulo entre os referidos planos é:

                    $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \theta = 135^{\circ}\:\:\:}}\end{gathered} }$

Sejam os planos:

                 $\Large\begin{cases} \pi_{1}: -y + 1 = 0\\\pi_{2} : y + z + 2 = 0\end{cases}$

Para calcular o ângulo entre os planos devemos calcular o ângulo entre seus vetores normais. Então, devemos:

• Recuperar os vetores normais de ambos os planos.

        Sabemos que as componentes do vetor normal de um plano são os coeficientes de cada uma de suas variáveis. Então, temos:

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

• Calcular o ângulo entre os vetores normais.

           $\Large\displaystyle\text{ \begin{gathered} \theta = \textrm{ang}(\vec{n_{1}},\,\vec{n_{2}})\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{\vec{n_{1}}\cdot\vec{n_{2}}}{\parallel\vec{n_{1}}\parallel\cdot\parallel\vec{n_{2}}\parallel}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{(0,-1, 0)\cdot(0,1,1)}{\sqrt{0^{2} + (-1)^{2} + 0^{2}}\cdot\sqrt{0^{2}+ 1^{2} + 1^{2}}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{0\cdot0 + (-1)\cdot1 + 0\cdot1}{\sqrt{1} \cdot\sqrt{2}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{0 - 1 + 0}{\sqrt{1\cdot2}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(-\frac{1}{\sqrt{2}}\bigg)\end{gathered} }$

              $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(-\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\bigg)\end{gathered} }$

              $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(-\frac{\sqrt{2}}{(\sqrt[\!\diagup\!]{2})^{\!\diagup\!\!\!\!2}}\bigg)\end{gathered} }$

              $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(-\frac{\sqrt{2}}{2}\bigg)\end{gathered} }$

              $\Large\displaystyle\begin{gathered} = 135^{\circ}\end{gathered}$

Portanto, o ângulo entre os planos é:

          $\Large\displaystyle\begin{gathered} \theta = 135^{\circ}\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$