Trong không gian Oxyz cho ba điểm A(-2; 0; 0), B(0; – 2; 0), C(0;0; -2). Gọi D là điểm khác O sao cho DA, DB, DC đôi một vuông góc với nhau và I(a; b;

Đáp án:

\(S =  – 1.\)

Giải thích các bước giải:

\(A\left( { – 2;\,\,0;\,\,0} \right),\,\,B\left( {0; – 2;\,\,0} \right),\,\,\,C\left( {0;\,\,0; – 2} \right)\)

Gọi \(D\left( {{x_0};\,\,{y_0};\,\,{z_0}} \right) \Rightarrow \left\{ \begin{array}{l}\overrightarrow {DA}  = \left( { – 2 – {x_0};\,\, – {y_0}; – {z_0}} \right)\\\overrightarrow {DB}  = \left( { – {x_0};\, – 2 – {y_0}; – {z_0}} \right)\\\overrightarrow {DC}  = \left( { – {x_0}; – {y_0}; – 2 – {z_0}} \right)\end{array} \right..\)

Ta có: DA, DB, DC đôi một vuông góc nên ta có:

\[\begin{array}{l}\left\{ \begin{array}{l}\overrightarrow {DA} .\overrightarrow {DB}  = 0\\\overrightarrow {DA} .\overrightarrow {DC}  = 0\\\overrightarrow {DB} .\overrightarrow {DC}  = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_0}\left( {2 + {x_0}} \right) + {y_0}\left( {2 + {y_0}} \right) + z_0^2 = 0\\{x_0}\left( {2 + {x_0}} \right) + y_0^2 + {z_0}\left( {2 + {z_0}} \right) = 0\\x_0^2 + {y_0}\left( {{y_0} + 2} \right) + {z_0}\left( {{z_0} + 2} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x_0^2 + 2{x_0} + y_0^2 + 2{y_0} + z_0^2 = 0\\x_0^2 + 2{x_0} + y_0^2 + 2{z_0} + z_0^2 = 0\\x_0^2 + y_0^2 + 2{y_0} + 2{z_0} + z_0^2 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{y_0} = {z_0}\\{x_0} = {y_0}\\x_0^2 + 2{x_0} + x_0^2 + 2{x_0} + x_0^2 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x_0} = {y_0} = {z_0}\\3x_0^2 + 4{x_0} = 0\end{array} \right.\\ \Leftrightarrow {x_0} = {y_0} = {z_0} =  – \frac{4}{3}\,\,\,\,\left( {do\,\,\,D \ne O} \right).\\ \Rightarrow D\left( { – \frac{4}{3}; – \frac{4}{3}; – \frac{4}{3}} \right).\end{array}\]

\(I\left( {a;\,\,b;\,\,c} \right)\) là tâm mặt cầu ngoại tiếp tứ diện \(ABCD\)

\(\begin{array}{l} \Rightarrow I{A^2} = I{B^2} = I{C^2} = I{D^2}\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {2 + a} \right)^2} + {b^2} + {c^2} = {a^2} + {\left( {b + 2} \right)^2} + {c^2}\\{\left( {2 + a} \right)^2} + {b^2} + {c^2} = {a^2} + {b^2} + {\left( {c + 2} \right)^2}\\{\left( {2 + a} \right)^2} + {b^2} + {c^2} = {\left( {a + \frac{4}{3}} \right)^2} + {\left( {b + \frac{4}{3}} \right)^2} + {\left( {c + \frac{4}{3}} \right)^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4a = 4b\\4a = 4c\\4 + 4a = \frac{8}{3}a + \frac{8}{3}b + \frac{8}{3}c + \frac{{16}}{3}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = b = c\\4a =  – \frac{4}{3}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = b = c\\a =  – \frac{1}{3}\end{array} \right.\\ \Leftrightarrow a = b = c =  – \frac{1}{3} \Rightarrow I\left( { – \frac{1}{3}; – \frac{1}{3}; – \frac{1}{3}} \right)\\ \Rightarrow S = a + b + c = 3.\left( { – \frac{1}{3}} \right) =  – 1.\end{array}\)