Question

Três vértices de um tetraedo de volume 2 u.v. são A(-3,0,-6), B(-8,3,8) e C(6,-3,-6).

Determine o quarto vértice D sabendo que ele pertence ao eixo Oy. Basta informar apenas uma das respostas possíveis:
D'= ( ), ( ), ( )

Answer 1

Olá

A(-3,0,-6)
B(-8,3,8)
C(6,-3,-6)
D(0,y,0)                <- O ponto se encontra no eixo Oy

V. tetraedro = 2
V. Paralelepípedo = 6*V. tetraedro   -> V paralelepípedo = 12 


Criando os 3 vetores

AB = B-A = (-5,3,14)
AC = C-A = (9,-3,0)
AD = D-A = (3,y,6)


Calculando o determinante e igualando a 12.

$[(\vec{AB},\vec{AC},\vec{AD})] \left[\begin{array}{ccc}-5&3&14\\9&-3&0\\3&y&6\end{array}\right] =12\\\\\\\\\mathsf{\underbrace{(\mathsf{90+0+126y})}_{diag.~principal}~-\underbrace{(\mathsf{162+0-126})}_{diag.~secund\'aria}}=12\\\\\\\\\mathsf{90+126y-36=12}\\\\\mathsf{126y+54=12}\\\\\mathsf{y= \frac{-42}{126} }\\\\\boxed{\mathsf{y=- \frac{1}{3} }}$


D(0,y,0)
D(0,-1/3,0)                   ←      RESPOSTA



Confirmando o resultado

Agora vamos calcular o volume do tetraedro para vermos se resultará em 2.


AB = B-A = (-5,3,14)
AC = C-A = (9,-3,0)
AD = D-A = (3,-1/3,6)

$(\vec{AB},\vec{AC},\vec{AD})] \left[\begin{array}{ccc}-5&3&14\\9&-3&0\\3&- \frac{1}{3} &6\end{array}\right] \\\\\\\\=\mathsf{\underbrace{(\mathsf{90+0-42})}_{diag.~principal}~-\underbrace{(\mathsf{162+0-126})}_{diag.~secund\'aria}}\\\\\\\mathsf{=48-36}\\\\\mathsf{=12}$

$\mathsf{V_{tetraedro}= \frac{1}{6}V_{paralelepipedo} }\\\\\\\mathsf{V_{tetraedro}= \frac{12}{6} }\\\\\\\boxed{\mathsf{V_{tetraedro}= 2}}\qquad \checkmark$