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Câu hỏi: Trong không gian $Oxyz,$ cho hai mặt phẳng $\left( P \right):x+y=0$ và $\left( Q \right):x-2y+2z-1=0.$ Cosin góc giữa hai mặt phẳng $\left( P \right)$ và $\left( Q \right)$ là
A. $-\dfrac{\sqrt{3}}{2}.$
B. $\dfrac{\sqrt{2}}{6}.$
C. $-\dfrac{\sqrt{2}}{6}.$
D. $\dfrac{\sqrt{3}}{2}.$
A. $-\dfrac{\sqrt{3}}{2}.$
B. $\dfrac{\sqrt{2}}{6}.$
C. $-\dfrac{\sqrt{2}}{6}.$
D. $\dfrac{\sqrt{3}}{2}.$
Ta có $\overrightarrow{{{n}_{P}}}=\left( 1;1;0 \right),\overrightarrow{{{n}_{Q}}}=\left( 1;-2;2 \right)$
$\Rightarrow \cos \left( \left( P \right);\left( Q \right) \right)=\dfrac{1.1+1.\left( -2 \right)+0.2}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{0}^{2}}}.\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{2}^{2}}}}=-\dfrac{\sqrt{2}}{6}.$
$\Rightarrow \cos \left( \left( P \right);\left( Q \right) \right)=\dfrac{1.1+1.\left( -2 \right)+0.2}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{0}^{2}}}.\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{2}^{2}}}}=-\dfrac{\sqrt{2}}{6}.$
Đáp án C.