Chọn véc tơ pháp tuyến của mặt phẳng(MNE) là \(\overrightarrow{n}=\left(1;2;2\right) \)
Ta có
\(+) \cos \left(\left(MNE\right),\left(Oxy\right)\right)=\left|\cos \left(\overrightarrow{n},\overrightarrow{k}\right)\right|=\frac{\left|\overrightarrow{n}.\overrightarrow{k}\right|}{\left|\overrightarrow{n}\right|\left|\overrightarrow{k}\right|} =\frac{2}{3} \Rightarrow \widehat{\left(\left(MNE\right),\left(Oxy\right)\right)}\approx 48^{0} 11' \)
\(+) \cos \left(\left(MNE\right),\left(Oxz\right)\right)=\left|\cos \left(\overrightarrow{n},\overrightarrow{j}\right)\right|=\frac{\left|\overrightarrow{n}.\overrightarrow{j}\right|}{\left|\overrightarrow{n}\right|\left|\overrightarrow{j}\right|} =\frac{2}{3} \Rightarrow \widehat{\left(\left(MNE\right),\left(Oxy\right)\right)}\approx 48^{0} 11' \)
\(+) \cos \left(\left(MNE\right),\left(Oyz\right)\right)=\left|\cos \left(\overrightarrow{n},\overrightarrow{i}\right)\right|=\frac{\left|\overrightarrow{n}.\overrightarrow{i}\right|}{\left|\overrightarrow{n}\right|\left|\overrightarrow{i}\right|} =\frac{1}{3} \Rightarrow \widehat{\left(\left(MNE\right),\left(Oyz\right)\right)}\approx 70^{0} 31' \)