Matéria: Matemática
Autor: altmikaelmarlon
Perguntado 3 anos atrás

sabendo que o plano pi terminado pelo ponto a(2,-1,3) b(3,2,-4) e c (2,2,5) assinale a alternativa que representa à equação geral de pi

Respostas

respondido por: solkarped

✅ Após resolver os cálculos, concluímos que a equação geral do plano que contém os três referidos pontos é:

$\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf \pi: 27x - 2y + 3z - 65 = 0 \:\:\:}}\end{gathered}$}$

Sejam os pontos:

$\Large\begin{cases} A(2, -1, 3)\\B(3, 2, -4)\\C(2, 2, 5)\end{cases}$

Para resolver esta questão devemos:

Calcular o vetor diretor "u":

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

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

$\Large\displaystyle\text{$\begin{gathered} = (3, 2, -4) - (2, -1, 3)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = (3 - 2, 2 - (-1), -4 - 3)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = (1, 3, -7)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:\vec{u} = (1, 3, -7)\end{gathered}$}$

Calcular o vetor diretor "v":

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

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

$\Large\displaystyle\text{$\begin{gathered} = (2, 2, 5) - (2, -1, 3)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = (2-2,2 - (-1), 5 - 3)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = (0, 3, 2)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:\vec{v} = (0, 3, 2)\end{gathered}$}$

Calcular o vetor "w" que será o produto vetorial entre "u" e "v":

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

$\Large\displaystyle\text{$\begin{gathered} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\1 & 3 & -7\\0 & 3 & 2\end{vmatrix}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = \begin{vmatrix} 3 & -7\\3 & 2\end{vmatrix}\cdot\vec{i} - \begin{vmatrix}1 & -7\\0 & 2 \end{vmatrix}\cdot\vec{j} + \begin{vmatrix}1 & 3\\0 & 3 \end{vmatrix}\cdot\vec{k}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = (6 + 21)\vec{i} - (2 + 0)\vec{j} + (3 - 0)\vec{k}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = 27\vec{i} - 2\vec{j} + 3\vec{k}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} = (27, -2, 3)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:\vec{w} = (27, -2, 3)\end{gathered}$}$

Montar a equação geral da reta: Para isso, devemos utilizar a seguinte fórmula:

$\large\displaystyle\text{$\begin{gathered} X_{w}\cdot X + Y_{w}\cdot Y + Z_{w}\cdot Z = X_{w}\cdot X_{A} + Y_{w}\cdot Y_{A} + Z_{w}\cdot Z_{A}\end{gathered}$}$