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Câu hỏi: Trong không gian $Oxyz$ cho hai mặt phẳng $\left( \alpha \right):x-2y+3z+1=0$ và $\left( \beta \right):2x-4y+6z+1=0$, khi đó:
A. $\left( \alpha \right)//\left( \beta \right)$.
B. $\left( \alpha \right)\equiv \left( \beta \right)$.
C. $\left( \alpha \right)\bot \left( \beta \right)$.
D. $\left( \alpha \right)$ cắt $\left( \beta \right)$.
A. $\left( \alpha \right)//\left( \beta \right)$.
B. $\left( \alpha \right)\equiv \left( \beta \right)$.
C. $\left( \alpha \right)\bot \left( \beta \right)$.
D. $\left( \alpha \right)$ cắt $\left( \beta \right)$.
Vectơ pháp tuyến $\overrightarrow{{{n}_{\alpha }}}=\left( 1;-2;3 \right)$ ; $\overrightarrow{{{n}_{\beta }}}=\left( 2;-4;6 \right)$
Ta có: $\left\{ \begin{aligned}
& \overrightarrow{{{n}_{\beta }}}=2\overrightarrow{{{n}_{\alpha }}} \\
& M(-1;0;0)\in \left( \alpha \right)\Rightarrow M\notin \left( \beta \right) \\
\end{aligned} \right.$
Chứng tỏ $\left( \alpha \right)//\left( \beta \right)$.
Ta có: $\left\{ \begin{aligned}
& \overrightarrow{{{n}_{\beta }}}=2\overrightarrow{{{n}_{\alpha }}} \\
& M(-1;0;0)\in \left( \alpha \right)\Rightarrow M\notin \left( \beta \right) \\
\end{aligned} \right.$
Chứng tỏ $\left( \alpha \right)//\left( \beta \right)$.
Đáp án A.
