Question

Em xin lời giản con b bài 1 với con b bài 2 ạ em cảm ơn ạ

Answer 1

thầy nguyễn việt lâm ko onl thì chịu:<

Answer 2

1b/ \(\overrightarrow{AB}=\left(1;1;3\right);\overrightarrow{u_{Oy}}=\left(0;1;0\right)\)

Vì \(\left(P_2\right)//AB//Oy\Rightarrow\overrightarrow{n_{\left(P_2\right)}}=\left[\overrightarrow{AB},\overrightarrow{u_{Oy}}\right]=\left(\left|\begin{matrix}1&3\\1&0\end{matrix}\right|,\left|\begin{matrix}3&1\\0&0\end{matrix}\right|,\left|\begin{matrix}1&1\\0&1\end{matrix}\right|\right)=\left(-3;0;1\right)\)

\(\Rightarrow\left(P_2\right):-3\left(x+3\right)+z-5=0\Leftrightarrow\left(P_2\right):3x-z+14=0\)

2b/

\(\overrightarrow{u_{Ox}}=\left(1;0;0\right);\overrightarrow{n_{\left(Q\right)}}=\left(3;2;5\right)\)

\(\Rightarrow\overrightarrow{n_{\left(\beta\right)}}=\left[\overrightarrow{u_{Ox}},\overrightarrow{n_{\left(Q\right)}}\right]=\left(0;-5;2\right)\)

\(d\left(O,\left(\beta\right)\right)=\dfrac{\left|d\right|}{\sqrt{25+4}}=\sqrt{29}\Rightarrow d=\pm29\)

\(\Rightarrow\left[{}\begin{matrix}\left(\beta\right):-5y+2z+29=0\\\left(\beta\right):-5y+2z-29=0\end{matrix}\right.\)