Question

Geometria Analitica
..............

Answer 1

Dados dois planos paralelos, encontrar a distância entre eles:

$\left\{ \!\begin{array}{l} \pi_1:~6x+4y-2z=28\\\\ \pi_2:~3x+2y-z=5 \end{array} \right.$


Da equação geral dos dois planos, tiramos que o vetor

$\right.\\\\\ \overrightarrow{\mathbf{n}}=(3,\,2,\,-1)$

é um vetor normal aos planos $\pi_1$ e $\pi_2.$

__________________________

Encontrando o versor de $\overrightarrow{\mathbf{n}}:$

( versor é um vetor com a mesma direção, cujo módulo é 1 )

$\overrightarrow{\mathbf{n}}^{\circ}=\dfrac{\overrightarrow{\mathbf{n}}}{\|\overrightarrow{\mathbf{n}}\|}\\\\\\ \overrightarrow{\mathbf{n}}^{\circ}=\dfrac{(3,\,2,\,-1)}{\sqrt{3^2+2^2+(-1)^2}}\\\\\\ \overrightarrow{\mathbf{n}}^{\circ}=\dfrac{(3,\,2,\,-1)}{\sqrt{9+4+1}}\\\\\\ \overrightarrow{\mathbf{n}}^{\circ}=\dfrac{(3,\,2,\,-1)}{\sqrt{14}}\\\\\\ \therefore~~\boxed{\begin{array}{c} \overrightarrow{\mathbf{n}}^{\circ}=\dfrac{1}{\sqrt{14}}\cdot (3,\,2,\,-1) \end{array}}$

________________________

O ponto $P(0,\,0,\,-14)$ pertence ao plano $\pi_1\,,$ pois suas coordenadas satisfazem a equação do plano.

Seja $Q(x_{_{Q}},\,y_{_{Q}},\,z_{_{Q}})$ um ponto do plano $\pi_2\,,$ de forma que

$\overrightarrow{PQ}=\lambda\overrightarrow{\mathbf{n}}^{\circ}~~~~~~\mathbf{(i)}$


A distância entre os planos é o valor absoluto de $\lambda\,,$ isto é, $\left|\lambda\right|.$

________________________

Então, pela equação $\mathbf{(i)}\,,$ temos

$Q-P=\lambda\overrightarrow{\mathbf{n}}^{\circ}\\\\ \big(x_{_{Q}}-0,\,y_{_{Q}}-0,\,z_{_{Q}}-(-14)\big)=\lambda\cdot \dfrac{1}{\sqrt{14}}\,(3,\,2,\,-1)\\\\\\ (x_{_{Q}},\,y_{_{Q}},\,z_{_{Q}}+14)=\dfrac{\lambda}{\sqrt{14}}\,(3,\,2,\,-1)\\\\\\ \left\{ \!\begin{array}{l} x_{_{Q}}=\dfrac{3}{\sqrt{14}}\,\lambda\\\\ y_{_{Q}}=\dfrac{2}{\sqrt{14}}\,\lambda\\\\ z_{_{Q}}=-\dfrac{1}{\sqrt{14}}\,\lambda-14 \end{array} \right.$


Substituindo as coordenadas de $Q$ na equação do plano $\pi_2:$

$3x_{_{Q}}+2y_{_{Q}}-z_{_{Q}}=5\\\\\\ 3\cdot \dfrac{3}{\sqrt{14}}\,\lambda+2\cdot \dfrac{2}{\sqrt{14}}\,\lambda-\left(-\dfrac{1}{\sqrt{14}}\,\lambda-14 \right )=5\\\\\\ \dfrac{9}{\sqrt{14}}\,\lambda+\dfrac{4}{\sqrt{14}}\,\lambda+\dfrac{1}{\sqrt{14}}\,\lambda+14=5\\\\\\ \dfrac{9+4+1}{\sqrt{14}}\,\lambda=5-14\\\\\\ \dfrac{14}{\sqrt{14}}\,\lambda=-9\\\\\\ \sqrt{14}\,\lambda=-9\\\\\\ \boxed{\begin{array}{c} \lambda=-\,\dfrac{9}{\sqrt{14}} \end{array}}$


Portanto, a distância entre os planos é

$\boxed{\begin{array}{c} \left|\lambda\right|=\dfrac{9}{\sqrt{14}}\mathrm{~u.c.} \end{array}}$