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Câu hỏi: Trong không gian với hệ trục tọa độ $Oxyz$, cho hai mặt phẳng $\left( P \right):2x-y-\text{z}-3=0$ và $\left( Q \right):x-\text{z}-2=0.$ Góc giữa hai mặt phẳng $\left( P \right)$ và $\left( Q \right)$ bằng
A. $30{}^\circ $.
B. $45{}^\circ $.
C. $60{}^\circ $.
D. $90{}^\circ $.
A. $30{}^\circ $.
B. $45{}^\circ $.
C. $60{}^\circ $.
D. $90{}^\circ $.
Ta có $\left( P \right):2x-y-\text{z}-3=0\Rightarrow $ VTPT $\overrightarrow{{{n}_{1}}}=\left( 2;-1;-1 \right)$.
$\left( Q \right):x-\text{z}-2=0\Rightarrow $ VTPT $\overrightarrow{{{n}_{2}}}=\left( 1;0;-1 \right)$.
Khi đó $\cos \left( \left( P \right),\left( Q \right) \right)=\dfrac{\left| \overrightarrow{{{n}_{1}}}.\overrightarrow{{{n}_{2}}} \right|}{\left| \overrightarrow{{{n}_{1}}} \right|.\left| \overrightarrow{{{n}_{2}}} \right|}=\dfrac{\left| 2.1+0.\left( -1 \right)+\left( -1 \right).\left( -1 \right) \right|}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}}.\sqrt{{{1}^{2}}+{{0}^{2}}+{{\left( -1 \right)}^{2}}}}=\dfrac{\sqrt{3}}{2}$.
Do đó $\left( \left( P \right),\left( Q \right) \right)=30{}^\circ $.
$\left( Q \right):x-\text{z}-2=0\Rightarrow $ VTPT $\overrightarrow{{{n}_{2}}}=\left( 1;0;-1 \right)$.
Khi đó $\cos \left( \left( P \right),\left( Q \right) \right)=\dfrac{\left| \overrightarrow{{{n}_{1}}}.\overrightarrow{{{n}_{2}}} \right|}{\left| \overrightarrow{{{n}_{1}}} \right|.\left| \overrightarrow{{{n}_{2}}} \right|}=\dfrac{\left| 2.1+0.\left( -1 \right)+\left( -1 \right).\left( -1 \right) \right|}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}}.\sqrt{{{1}^{2}}+{{0}^{2}}+{{\left( -1 \right)}^{2}}}}=\dfrac{\sqrt{3}}{2}$.
Do đó $\left( \left( P \right),\left( Q \right) \right)=30{}^\circ $.
Đáp án A.