Finding the volume of the tetrahedron with vertices (0,0,0), (2,0,0), (0,2,0), (0,0,2). I get 8; answer is 4/3.

In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors A, B and C then the area of the object is $A\cdot (B\times C)$. Let V be the volume we are trying to find.

$\begin{array}{rl}{x}^{2}& =6-{y}^{2}-{z}^{2}\\ A& =(2,0,0)-(0,0,0)=(2,0,0)\\ B& =(0,2,0)-(0,0,0)=(0,2,0)\\ C& =(0,0,2)-(0,0,0)=(0,0,2)\\ V& =\left|\begin{array}{ccc}{a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\\ {c}_{1}& {c}_{2}& {c}_{3}\end{array}\right|=\left|\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right|\\ & =2\left|\begin{array}{cc}2& 0\\ 0& 2\end{array}\right|=2(4-0)\\ & =8\end{array}$

However, the book gets $\frac{4}{3}$

In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors A, B and C then the area of the object is $A\cdot (B\times C)$. Let V be the volume we are trying to find.

$\begin{array}{rl}{x}^{2}& =6-{y}^{2}-{z}^{2}\\ A& =(2,0,0)-(0,0,0)=(2,0,0)\\ B& =(0,2,0)-(0,0,0)=(0,2,0)\\ C& =(0,0,2)-(0,0,0)=(0,0,2)\\ V& =\left|\begin{array}{ccc}{a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\\ {c}_{1}& {c}_{2}& {c}_{3}\end{array}\right|=\left|\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right|\\ & =2\left|\begin{array}{cc}2& 0\\ 0& 2\end{array}\right|=2(4-0)\\ & =8\end{array}$

However, the book gets $\frac{4}{3}$