Question

Determinar o ângulo entre os planos
π1:x+y+2z-7=0 e π2:4x+5y+3z+2=0

Answer 1

El ángulo entre los planos es lo mismo que el ángulo entre sus vectores normales, y estos son:
PARA $\Pi_1$ : $(1,1,2)$
PARA: $\Pi_2: (4,5,3)$

Si el ángulo es $\theta$ entonces:

$\cos \theta =\dfrac{(1,1,2)\cdot(4,5,3)}{\|(1,1,2)\|\cdot \|(4,5,3)\|}\\ \\ \cos \theta =\dfrac{15}{\sqrt{6}\cdot\sqrt{50}}\\ \\ \cos \theta =\dfrac{\sqrt{3}}{2}\\ \\ \boxed{\theta =\dfrac{\pi}{6}\mbox{ rad}}$

Answer 2

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

                   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \theta = 30^{\circ}\:\:\:}}\end{gathered} }$

Sejam os planos:

           $\Large\begin{cases} \pi_{1}: x + y + 2z - 7 = 0\\\pi_{2} : 4x + 5y + 3z + 2 = 0\end{cases}$

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:

                          $\Large\begin{cases} \vec{n_{1}} = (1, 1, 2)\\\vec{n_{2}} = (4, 5, 3)\end{cases}$

• Calcular o ângulo entre os vetores normais.

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

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

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{(1,1, 2)\cdot(4,5,3)}{\sqrt{1^{2} + 1^{2} + 2^{2}}\cdot\sqrt{4^{2}+ 5^{2} + 3^{2}}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{1\cdot4 + 1\cdot5 + 2\cdot3}{\sqrt{6} \cdot\sqrt{50}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{4 + 5 + 6}{\sqrt{6\cdot50}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{15}{\sqrt{300}}\bigg)\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{15}{\sqrt{10^{2}\cdot3}}\bigg)\end{gathered} }$

                $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{15}{10\sqrt{3}}\bigg)\end{gathered} }$

                $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{15}{10\sqrt{3}}\cdot\frac{10\sqrt{3}}{10\sqrt{3}}\bigg)\end{gathered} }$

                $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{150\sqrt{3}}{300}\bigg)\end{gathered} }$

                $\Large\displaystyle\text{ \begin{gathered} = \arccos\bigg(\frac{\sqrt{3}}{2}\bigg)\end{gathered} }$

               $\Large\displaystyle\begin{gathered} = 30^{\circ}\end{gathered}$

Portanto, o ângulo entre os planos é:

          $\Large\displaystyle\begin{gathered} \theta = 30^{\circ}\end{gathered}$

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

Saiba mais:

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3. https://brainly.com.br/tarefa/50224080

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