Viết pt mặt cầu (S) đi qua 4 đỉnh của tứ diện A(1,0,1), B(3,4,-2), C(4,-1,1), D(3,0,3). 06/12/2021 Bởi Adalyn Viết pt mặt cầu (S) đi qua 4 đỉnh của tứ diện A(1,0,1), B(3,4,-2), C(4,-1,1), D(3,0,3).
Đáp án: \[{\left( {x – \frac{{115}}{{34}}} \right)^2} + {\left( {y – \frac{{73}}{{34}}} \right)^2} + {\left( {z – \frac{{21}}{{34}}} \right)^2} = \frac{{12059}}{{1156}}\] Giải thích các bước giải: Gọi tâm của mặt cầu ngoại tiếp tứ diện là \(I\left( {a;b;c} \right)\). Ta có: \(\begin{array}{l}I\left( {a;b;c} \right);\,\,\,\,A\left( {1;0;1} \right);\,\,\,\,B\left( {3;4; – 2} \right);\,\,\,\,C\left( {4; – 1;1} \right);\,\,\,\,D\left( {3;0;3} \right)\\ \Rightarrow \left\{ \begin{array}{l}\overrightarrow {AI} = \left( {a – 1;\,\,b;\,\,c – 1} \right)\\\overrightarrow {BI} = \left( {a – 3;\,\,b – 4;\,\,c + 2} \right)\\\overrightarrow {CI} = \left( {a – 4;\,\,b + 1;\,\,c – 1} \right)\\\overrightarrow {DI} = \left( {a – 3;\,\,b;\,\,\,c – 3} \right)\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}AI = \sqrt {{{\left( {a – 1} \right)}^2} + {b^2} + {{\left( {c – 1} \right)}^2}} \\BI = \sqrt {{{\left( {a – 3} \right)}^2} + {{\left( {b – 4} \right)}^2} + {{\left( {c + 2} \right)}^2}} \\CI = \sqrt {{{\left( {a – 4} \right)}^2} + {{\left( {b + 1} \right)}^2} + {{\left( {c – 1} \right)}^2}} \\DI = \sqrt {{{\left( {a – 3} \right)}^2} + {b^2} + {{\left( {c – 3} \right)}^2}} \end{array} \right.\\R = IA = IB = IC = ID\\ \Rightarrow \left\{ \begin{array}{l}IA = IB\\IA = IC\\IA = ID\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {a – 1} \right)^2} + {b^2} + {\left( {c – 1} \right)^2} = {\left( {a – 3} \right)^2} + {\left( {b – 4} \right)^2} + {\left( {c + 2} \right)^2}\\{\left( {a – 1} \right)^2} + {b^2} + {\left( {c – 1} \right)^2} = {\left( {a – 4} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c – 1} \right)^2}\\{\left( {a – 1} \right)^2} + {b^2} + {\left( {c – 1} \right)^2} = {\left( {a – 3} \right)^2} + {b^2} + {\left( {c – 3} \right)^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{a^2} + {b^2} + {c^2} – 2a – 2c + 2 = {a^2} + {b^2} + {c^2} – 6a – 8b + 4c + 29\\{a^2} + {b^2} + {c^2} – 2a – 2c + 2 = {a^2} + {b^2} + {c^2} – 8a + 2b – 2c + 18\\{a^2} + {b^2} + {c^2} – 2a – 2c + 2 = {a^2} + {b^2} + {c^2} – 6a – 6c + 18\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4a + 8b – 6c = 27\\6a – 2b = 16\\4a + 4c = 16\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{{115}}{{34}}\\b = \frac{{73}}{{34}}\\c = \frac{{21}}{{34}}\end{array} \right.\\ \Rightarrow R = IA = IB = IC = ID = \frac{{\sqrt {12059} }}{{34}}\\ \end{array}\) Vậy phương trình mặt cầu cần tìm là: \({\left( {x – \frac{{115}}{{34}}} \right)^2} + {\left( {y – \frac{{73}}{{34}}} \right)^2} + {\left( {z – \frac{{21}}{{34}}} \right)^2} = \frac{{12059}}{{1156}}\) Bình luận
Đáp án:
\[{\left( {x – \frac{{115}}{{34}}} \right)^2} + {\left( {y – \frac{{73}}{{34}}} \right)^2} + {\left( {z – \frac{{21}}{{34}}} \right)^2} = \frac{{12059}}{{1156}}\]
Giải thích các bước giải:
Gọi tâm của mặt cầu ngoại tiếp tứ diện là \(I\left( {a;b;c} \right)\). Ta có:
\(\begin{array}{l}
I\left( {a;b;c} \right);\,\,\,\,A\left( {1;0;1} \right);\,\,\,\,B\left( {3;4; – 2} \right);\,\,\,\,C\left( {4; – 1;1} \right);\,\,\,\,D\left( {3;0;3} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AI} = \left( {a – 1;\,\,b;\,\,c – 1} \right)\\
\overrightarrow {BI} = \left( {a – 3;\,\,b – 4;\,\,c + 2} \right)\\
\overrightarrow {CI} = \left( {a – 4;\,\,b + 1;\,\,c – 1} \right)\\
\overrightarrow {DI} = \left( {a – 3;\,\,b;\,\,\,c – 3} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
AI = \sqrt {{{\left( {a – 1} \right)}^2} + {b^2} + {{\left( {c – 1} \right)}^2}} \\
BI = \sqrt {{{\left( {a – 3} \right)}^2} + {{\left( {b – 4} \right)}^2} + {{\left( {c + 2} \right)}^2}} \\
CI = \sqrt {{{\left( {a – 4} \right)}^2} + {{\left( {b + 1} \right)}^2} + {{\left( {c – 1} \right)}^2}} \\
DI = \sqrt {{{\left( {a – 3} \right)}^2} + {b^2} + {{\left( {c – 3} \right)}^2}}
\end{array} \right.\\
R = IA = IB = IC = ID\\
\Rightarrow \left\{ \begin{array}{l}
IA = IB\\
IA = IC\\
IA = ID
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {a – 1} \right)^2} + {b^2} + {\left( {c – 1} \right)^2} = {\left( {a – 3} \right)^2} + {\left( {b – 4} \right)^2} + {\left( {c + 2} \right)^2}\\
{\left( {a – 1} \right)^2} + {b^2} + {\left( {c – 1} \right)^2} = {\left( {a – 4} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c – 1} \right)^2}\\
{\left( {a – 1} \right)^2} + {b^2} + {\left( {c – 1} \right)^2} = {\left( {a – 3} \right)^2} + {b^2} + {\left( {c – 3} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{a^2} + {b^2} + {c^2} – 2a – 2c + 2 = {a^2} + {b^2} + {c^2} – 6a – 8b + 4c + 29\\
{a^2} + {b^2} + {c^2} – 2a – 2c + 2 = {a^2} + {b^2} + {c^2} – 8a + 2b – 2c + 18\\
{a^2} + {b^2} + {c^2} – 2a – 2c + 2 = {a^2} + {b^2} + {c^2} – 6a – 6c + 18
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4a + 8b – 6c = 27\\
6a – 2b = 16\\
4a + 4c = 16
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = \frac{{115}}{{34}}\\
b = \frac{{73}}{{34}}\\
c = \frac{{21}}{{34}}
\end{array} \right.\\
\Rightarrow R = IA = IB = IC = ID = \frac{{\sqrt {12059} }}{{34}}\\
\end{array}\)
Vậy phương trình mặt cầu cần tìm là:
\({\left( {x – \frac{{115}}{{34}}} \right)^2} + {\left( {y – \frac{{73}}{{34}}} \right)^2} + {\left( {z – \frac{{21}}{{34}}} \right)^2} = \frac{{12059}}{{1156}}\)