Đáp án:
\(\begin{array}{l}
a)\,\,AB = \sqrt {122} ;\,\,AC = \sqrt {53} ;\,\,BC = 5.\\
b)\,\,\,\overrightarrow u = \left( {1; - 40} \right)\\
c)\,\,{m_a} = \frac{{5\sqrt {13} }}{2};\,\,{m_b} = \frac{{\sqrt {241} }}{2};\,\,{m_c} = \sqrt 3 \\
d)\,\,D\left( {1;\,\,24} \right),\,\,\,E\left( { - 3; - 14} \right).
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}A\left( {0;\,\,5} \right),\,\,B\left( {1; - 6} \right),\,\,C\left( { - 2; - 2} \right)\\a)\,\,\,\left\{ \begin{array}{l}\overrightarrow {AB} = \left( {1; - 11} \right) \Rightarrow AB = \sqrt {1 + {{11}^2}} = \sqrt {122} \\\overrightarrow {AC} = \left( { - 2; - 7} \right) \Rightarrow AC = \sqrt {{2^2} + {7^2}} = \sqrt {53} \\\overrightarrow {BC} = \left( { - 3;\,\,4} \right) \Rightarrow BC = \sqrt {{3^2} + {4^2}} = 5.\end{array} \right.\\b)\,\,\overrightarrow u = \overrightarrow {AB} + 3\overrightarrow {AC} - 2\overrightarrow {BC} \\ = \left( {1; - 11} \right) + 3\left( { - 2; - 7} \right) - 2\left( { - 3;4} \right)\\ = \left( {1 - 6 + 6;\,\, - 11 - 3.7 - 2.4} \right)\\ = \left( {1;\,\, - 40} \right).\end{array}\)
- c) Áp dụng công thức tính độ dài đường trung tuyến ta có:
\(\begin{array}{l}{m_a} = \sqrt {\frac{{2A{B^2} + 2A{C^2} - B{C^2}}}{4}} = \sqrt {\frac{{2.122 + 2.53 - {5^2}}}{4}} = \frac{{5\sqrt {13} }}{2}.\\{m_b} = \sqrt {\frac{{2A{B^2} + 2B{C^2} - A{C^2}}}{4}} = \sqrt {\frac{{2.122 + 2.25 - 53}}{4}} = \frac{{\sqrt {241} }}{2}\\{m_c} = \sqrt {\frac{{2B{C^2} + 2A{C^2} - A{B^2}}}{4}} = \sqrt {\frac{{2.25 + 2.53 - 122}}{4}} = \sqrt 3 .\end{array}\)
- d) Gọi \(D\left( {a;\,\,b} \right)\)
Ta có \(A\) là trọng tâm \(\Delta BCD \Rightarrow \left\{ \begin{array}{l}a = 3{x_A} - {x_B} - {x_C}\\b = 3{y_A} - {y_B} - {y_C}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 24\end{array} \right. \Rightarrow D\left( {1;\,\,24} \right).\)
Gọi \(E\left( {u;\,\,v} \right).\) Tứ giác \(ABCE\) là hình bình hành \( \Leftrightarrow \overrightarrow {AB} = \overrightarrow {EC} \)
\( \Leftrightarrow \left( {1;\,\,11} \right) = \left( { - 2 - u; - 3 - v} \right) \Leftrightarrow \left\{ \begin{array}{l} - 2 - u = 1\\ - 3 - v = 11\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}u = - 3\\v = - 14\end{array} \right. \Rightarrow E\left( { - 3; - 14} \right).\)
