Question

Dados os pontos A(8, 3, 8), B(2, 6, 4), C(-6, -4, 3) e D(3, x, -3).

a.) Determine o valor de x tal que os pontos acima sejam coplanares.

Answer 1

$\mathrm{A(8, 3, 8), B(2, 6, 4), C(-6, -4, 3)\ e\ D(3, x, -3)}\\\\ \textbf{Equa\c{c}\~ao do plano que cont\'em A, B e C:}\\\\ \mathrm{\vec{AB}=B-A=(2,6,4)-(8,3,8)=(-6,3,-4)}\\ \mathrm{\vec{AC}=C-A=(-6,-4,3)-(8,3,8)=(-14,-7,-5)}\\\\ \mathrm{\vec{AB}\times\vec{AC}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\-6&3&-4\\-14&-7&-5\end{array}\right|=(-43,26,84)}$

$\mathrm{*\ Seja\ P(x,y,z)\ um\ ponto\ contido\ no\ plano:}\\\\ \mathrm{\vec{AP}=P-A=(x,y,z)-(8,3,8)=(x-8,y-3,z-8)}\\\\ \mathrm{\vec{AP}(\vec{AB}\times\vec{AC})=0\ \to\ (x-8,y-3,z-8)(-43,26,84)=0}\\ \mathrm{-43(x-8)+26(y-3)+84(z-8)=0}\\ \mathrm{-43x+344+26y-78+84z-672=0}\\\\ \boxed{\mathbf{-43x+26y+84z-406=0}}\\\\ \textbf{Encontrando o valor de x em}\ \mathbf{D(3,x,-3):}\\\\ \mathrm{-43.3+26.x+84.(-3)-406=0}\\ \mathrm{-129+26x-252-406=0\ \to\ 26x-787=0}\\\\ \boxed{\mathbf{x=\dfrac{787}{26}}}$