Cho 4 điểm $A(1; 1; 1) ; B(1; 2; 1); C(1; 1; 2)$ và $D(2; 2; 1)$
1, Chứng tỏ rằng $A, B, C, D$ không đồng phẳng. Tính thể tích tứ diện $ABCD$
2, Lập phương trình mặt cầu ngoại tiếp tứ diện $ABCD$
2,
Giả sử mặt cầu (S) có dạng:
$(S): x^2+y^2+z^2-2ax-2by-2cz+d=0$ điều kiện: $a^2+b^2+c^2-d>0$
Điểm A, B, C, D $\in$ (S) ta được:
$\begin{cases}3-2a-2b-2c+d=0 \\ 6-2a-4b-2c+d=0\\6-2a-2b-4c+d=0\\9-4a-4b-2c+d=0 \end{cases} \Leftrightarrow  \begin{cases}a=b=c=\frac{3}{2} \\ d=6 \end{cases}$ (thỏa mãn điều kiện)
Vậy phương trình mặt cầu (S) có dạng:$(S): x^2+y^2+z^2-3x-3y-3z+6=0$
1,
Ta có: $\overrightarrow{AB}(0; 1; 0), \overrightarrow{AC}(0; 0; 1); \overrightarrow{AD}(1; 1; 0)$, suy ra:
$[\overrightarrow{AB}. \overrightarrow{AC}].\overrightarrow{AD}=(1; 0; 0)(1;1;0)=1\neq 0$
$\Rightarrow \overrightarrow{AB}, \overrightarrow{AC}; \overrightarrow{AD} $ không đồng phẳng
$\Leftrightarrow $ A, B, C, D không đồng phẳng
Ta có: $V_{ABCD}=\frac{1}{6}|[\overrightarrow{AB}. \overrightarrow{AC}].\overrightarrow{AD}|=\frac{1}{6}  $  

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