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Câu hỏi: Trong không gian Oxyz, cho đường thẳng $\Delta $ : $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{-1}$ và mặt phẳng $\left( \alpha \right)$ : $x-y+2z=0$. Góc giữa đường thẳng $\Delta $ và mặt phẳng $\left( \alpha \right)$ bằng
A. 30°.
B. 60°.
C. 150°.
D. 120°.
A. 30°.
B. 60°.
C. 150°.
D. 120°.
Ta có $\left\{ \begin{aligned}
& \overrightarrow{{{u}_{\Delta }}}=\left( 1;2;-1 \right) \\
& \overrightarrow{{{n}_{\left( \alpha \right)}}}=\left( 1;-1;2 \right) \\
\end{aligned} \right.\to \sin \left( \Delta ;\left( \alpha \right) \right)=\dfrac{\left| \overrightarrow{{{u}_{\Delta }}}.\overrightarrow{{{n}_{\left( \alpha \right)}}} \right|}{\left| \overrightarrow{{{u}_{\Delta }}} \right|\left| \overrightarrow{{{n}_{\left( \alpha \right)}}} \right|}=\dfrac{\left| 1-2-2 \right|}{\sqrt{6}\sqrt{6}}=\dfrac{1}{2}\Rightarrow \left( \Delta ;\left( \alpha \right) \right)=30{}^\circ $.
& \overrightarrow{{{u}_{\Delta }}}=\left( 1;2;-1 \right) \\
& \overrightarrow{{{n}_{\left( \alpha \right)}}}=\left( 1;-1;2 \right) \\
\end{aligned} \right.\to \sin \left( \Delta ;\left( \alpha \right) \right)=\dfrac{\left| \overrightarrow{{{u}_{\Delta }}}.\overrightarrow{{{n}_{\left( \alpha \right)}}} \right|}{\left| \overrightarrow{{{u}_{\Delta }}} \right|\left| \overrightarrow{{{n}_{\left( \alpha \right)}}} \right|}=\dfrac{\left| 1-2-2 \right|}{\sqrt{6}\sqrt{6}}=\dfrac{1}{2}\Rightarrow \left( \Delta ;\left( \alpha \right) \right)=30{}^\circ $.
Đáp án A.