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Câu hỏi: Trong không gian $Oxyz$, cho hai mặt phẳng $\left( P \right):x-2y+2z-1=0$ và $\left( Q \right):2x+2y-z-3=0$. Gọi $\alpha $ là góc giữa hai mặt phẳng $\left( P \right)$ và $\left( Q \right)$. Tính $\cos \alpha $.
A. $-\dfrac{4}{9}$.
B. $\dfrac{4}{9}$.
C. $\dfrac{2}{3}$.
D. $-\dfrac{2}{3}$.
A. $-\dfrac{4}{9}$.
B. $\dfrac{4}{9}$.
C. $\dfrac{2}{3}$.
D. $-\dfrac{2}{3}$.
Ta có vtpt $\overrightarrow{{{n}_{P}}}=\left( 1;-2;2 \right)$ và vtpt $\overrightarrow{{{n}_{Q}}}=\left( 2;2;-1 \right)$.
Suy ra $\cos \alpha =\dfrac{\left| \overrightarrow{{{n}_{P}}}.\overrightarrow{{{n}_{Q}}} \right|}{\left| \overrightarrow{{{n}_{P}}} \right|.\left| \overrightarrow{{{n}_{Q}}} \right|}=\dfrac{\left| 1.2+\left( -2 \right).2+2.\left( -1 \right) \right|}{\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{2}^{2}}}.\sqrt{{{2}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}}=\dfrac{4}{9}$.
Suy ra $\cos \alpha =\dfrac{\left| \overrightarrow{{{n}_{P}}}.\overrightarrow{{{n}_{Q}}} \right|}{\left| \overrightarrow{{{n}_{P}}} \right|.\left| \overrightarrow{{{n}_{Q}}} \right|}=\dfrac{\left| 1.2+\left( -2 \right).2+2.\left( -1 \right) \right|}{\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{2}^{2}}}.\sqrt{{{2}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}}=\dfrac{4}{9}$.
Đáp án B.