Question

determine a equaçao geral do plano que contém os pontos A(1,2,4), B(-2,-1,5) e C(7,5,-1).

Answer 1

Bom dia Lary!

Solução!


$A(1,2,4)\\\\\ B(-2,-1,5)\\\\\ C(7,5,-1)$


Vamos transformar os pontos em vetores!



$B-A= \overrightarrow{AB}=(-2,-1,5)+(-1,-2,-4)\Rightarrow \overrightarrow{AB}=(-3,-3,1)\\\\\\\ C-A= \overrightarrow{AC}=(7,5,-1)+(-1,-2,-4)\Rightarrow \overrightarrow{AC}=(6,3,-5)$


Sendo o vetor normal do plano determinado pelo produto de  $\overrightarrow{AB} \times \overrightarrow{AC}$



$\overrightarrow{AB} \times \overrightarrow{AC}=\begin{vmatrix} i & j&k \\ -3 &-3&1 \\ 6&3&-5 \end{vmatrix}\\\\\\\\ \overrightarrow{AB} \times \overrightarrow{AC}=\begin{vmatrix} i & j&k&i&j \\ -3 &-3&1&-3&-3 \\ 6&3&-5&6&3\end{vmatrix}\\\\\\\ Multiplicando~~as~~diagonais~~do~~determinante!\\\\\\\\ (15i+6j-9k)+(18k-3i-15j)\\\\\\\ (15i-3i,6j-15j,9k,-15k)\\\\\\ (12i,-9j,-6k)$




Equação geral do plano!


$a(x- x_{A})+b(y -y_{A} )+(z- z_{A})=0$


Pegando qualquer ponto pertencente ao plano!


$A(1,2,4)\\\\\\ (12i,-9j,-6k)=(a=12~~b=-9~~c=-6)\\\\\\\ Agora~~e~~ so\´~~ substituir~~ na ~~equac\~ao ~~do~~ plano!$

$12(x- 1)-9(y -2 )-6(z- 4)=0 \\\\\\\ 12x-12-9y+18-6z+24=0\\\\\\ 12x-9y-6z-12+18+24=0\\\\\\ 12x-9y-6z+30=0\\\\\\\\\ \boxed{Resposta: Eq~~do~~plano~~ \Rightarrow~~12x-9y-6z+30=0}$




Bom dia!
Bons estudos!

Answer 2

✅ Tendo resolvido todos os cálculos, concluímos que a equação geral do plano "π" que passa pelos pontos "A", "B" e "C" é:

  $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \pi = 4x - 3y + 3z - 10 = 0\:\:\:}}\end{gathered} }$

Sejam os pontos:

                   $\Large\begin{cases} A(1, 2, 4)\\B(-2, -1, 5)\\C(7, 5, -1)\end{cases}$

Para determinar a equação geral do plano, além de utilizarmos a seguinte equação:

$\Large\displaystyle\begin{gathered} \bf(I)\end{gathered}$      $\large\displaystyle\begin{gathered} X_{n}\cdot X + Y_{n}\cdot Y + Z_{n}\cdot Z = X_{n}\cdot X_{A} + Y_{n}\cdot Y_{A} + Z_{n}\cdot Z_{A}\end{gathered}$

Devemos ter o vetor normal "n" ao plano e um ponto qualquer pertencente ao plano-  que, neste caso, será o ponto "A" - ou seja, precisamos dos seguintes itens:

                      $\Large\begin{cases} \vec{n} = (X_{n}, Y_{n}, Z_{n})\\A(X_{A}, Y_{A}, Z_{A})\end{cases}$

A partir de agora devemos:

• Calcular os vetores diretores do plano:

             $\Large\displaystyle\begin{gathered} \vec{u} = \overrightarrow{AB}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = B - A\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = (-2, -1, 5) - (1, 2, 4)\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = (-2 - 1, -1 - 2, 5 - 4)\end{gathered}$

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

             $\Large\displaystyle\begin{gathered} \therefore\:\:\:\vec{u} = (-3, -3, 1)\end{gathered}$

              $\Large\displaystyle\begin{gathered} \vec{v} = \overrightarrow{AC}\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = C - A\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = (7, 5, -1) - (1, 2, 4)\end{gathered}$

                   $\Large\displaystyle\begin{gathered} = (7 - 1, 5 - 2, -1 -4)\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = (6, 3, -5)\end{gathered}$

            $\Large\displaystyle\begin{gathered} \therefore\:\:\:\vec{v} = (6, 3, -5)\end{gathered}$

• Calcular o vetor normal "n".

        Para calcular o vetor normal "n" devemos calcular o produto vetorial entre os vetores diretores, ou seja:

               $\Large\displaystyle\begin{gathered} \vec{n} = \vec{u}\wedge\vec{v}\end{gathered}$

                    $\Large\displaystyle\begin{gathered} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\-3 & -3 & 1\\6 & 3 & -5\end{vmatrix}\end{gathered}$

                    $\Large\displaystyle\begin{gathered} = \begin{vmatrix}-3 & 1\\ 3 & -5\end{vmatrix}\vec{i} - \begin{vmatrix}-3 & 1\\ 6 & -5\end{vmatrix}\vec{j} + \begin{vmatrix}-3 & -3\\ 6 & 3\end{vmatrix}\vec{k}\end{gathered}$

                    $\Large\displaystyle\begin{gathered} = (15 - 3)\vec{i} - (15 - 6)\vec{j} + (-9 + 18)\vec{k}\end{gathered}$

                    $\Large\displaystyle\begin{gathered} = 12\vec{i} - 9\vec{j} + 9\vec{k}\end{gathered}$

                    $\Large\displaystyle\begin{gathered} = (12, -9, 9)\end{gathered}$

              $\Large\displaystyle\begin{gathered} \therefore\:\:\:\vec{n} = (12, -9, 9)\end{gathered}$

• Montar a equação geral do plano "π".

        Para isso devemos substituir tanto as coordenadas do vetor normal "n" quanto as coordenadas do ponto "A" na equação "I", ou seja:

         $\Large\displaystyle\begin{gathered} 12\cdot x + (-9)\cdot y + 9\cdot z = 12\cdot1 + (-9)\cdot2 + 9\cdot4\end{gathered}$  

                            $\Large\displaystyle\begin{gathered} 12x - 9y + 9z = 12 - 18 + 36\end{gathered}$  

                            $\Large\displaystyle\begin{gathered} 12x - 9y + 9z = 30\end{gathered}$

                 $\Large\displaystyle\begin{gathered} (\div3)\:12x - 9y + 9z = (\div3) \:30\end{gathered}$

                               $\Large\displaystyle\begin{gathered} 4x - 3y + 3z = 10\end{gathered}$

                    $\Large\displaystyle\begin{gathered} 4x - 3y + 3z - 10 = 0\end{gathered}$

✅ Portanto, a equação geral do plano procurado é:

             $\Large\displaystyle\begin{gathered}\pi: 4x - 3y + 3z - 10 = 0 \end{gathered}$

               

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Solução gráfica: