Trong không gian Oxyz, cho hai điểm A(0;2;-2),B(2;2;-4)

* Hướng dẫn giải

Đáp án A

Ta có$\overrightarrow{OA}=\left(0;2;-2\right),\overrightarrow{OB}=\left(2;2;-4\right)$ , .

Phương trình mặt phẳng (OAB) là $x+y+z=0$

  $I\in \left(OAB\right)\Rightarrow a+b+c=0$

$\overrightarrow{AI}=\left(a;b-2;c+2\right),\overrightarrow{BI}=\left(a-2;b-2;c+4\right),\overrightarrow{OI}=\left(a;b;c\right)$

Ta có hệ $\left\{\begin{array}{l}AI=BI\\ AI=OI\end{array}\right.\iff \left\{\begin{array}{l}{a}^2+{\left(c+2\right)}^2={\left(a-2\right)}^2+{\left(c+4\right)}^2\\ {\left(b-2\right)}^2+{\left(c+2\right)}^2=b^2+c^2\end{array}\right.\iff \left\{\begin{array}{l}a-c=4\\ -b+c=-2\end{array}\right.$

Từ (1) và (2), suy ra $\left\{\begin{array}{l}a-c=4\\ -b+c=-2\\ a+b+c=0\end{array}\right.\iff \left\{\begin{array}{l}a-c=4\\ -b+c=-2\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=2\\ b=0\\ c=-2\end{array}\right.$

Vậy $I\left(2;0;-2\right)\Rightarrow T=a^2+b^2+c^2=8$ .