Question

Determine o ângulo entre os planos π1:y+z+3=0 e π2:x+y−4=0.


a) π/2


b) π/3


c) π/6


d) π/4


e) 0

Answer 1

✅ Após resolver os cálculos, concluímos que o ângulo entre os referidos planos é numericamente igual ao ângulo entre os vetores normais dos referidos planos, ou seja:

            $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \theta = \frac{\pi}{3}\,rad\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

        $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\:B\:\:\:}}\end{gathered} }$

Sejam as equações dos planos:

           $\Large\begin{cases} \pi_{1} : y + z + 3 = 0\\\pi_{2} : x + y - 4 = 0\end{cases}$

Para resolver esta questão devemos:

• Recuperar os vetores normais dos respectivos planos:

        Sabendo que a equação geral do plano no R³ pode ser dado por:

               $\Large\displaystyle\begin{gathered} ax + by + cz + d = 0\end{gathered}$

        Sabemos também que o vetor normal "n" de um plano é:

                       $\Large\displaystyle\begin{gathered} \vec{n} = (a, b, c)\end{gathered}$

         Desta forma, temos os seguintes vetores normais:

                      $\Large\displaystyle\text{ \begin{gathered} \vec{n_{1}} = (0, 1, 1)\end{gathered} }$

                      $\Large\displaystyle\text{ \begin{gathered} \vec{n_{2}} = (1, 1, 0)\end{gathered} }$

• Calcular o ângulo entre os vetores normais dos planos. Para isso, devemos fazer:

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

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

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

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

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

                 $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{1}{\sqrt{2\cdot2}}\bigg)\end{gathered} }$

                 $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{1}{\sqrt{4}}\bigg)\end{gathered} }$

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

                 $\Large\displaystyle\begin{gathered} = 60^{\circ}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = \frac{\pi}{3}\,\textrm{rad}\end{gathered}$

✅ Portanto, o ângulo entre os planos é:

             $\Large\displaystyle\begin{gathered} \theta = \frac{\pi}{3}\,\textrm{rad}\end{gathered}$

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

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