Найти угол плоскости x+2y+2z-3=0, 2x-y+2z+5=0

Решите задачу:

$\alpha :\; \; x+2y+2z-3=0\; \; \to; \; \vec {n_1}=(1,2,2)\\\\ \beta :\; \; 2x-y+2z+5=0\; \; \to \; \; \vec{n_2}=(2,-1,2)\\\\cos( \alpha \widehat {, }\beta )= \frac{\vec{n_1}\cdot \vec {n_2}}{|\vec{n_1}|\cdot |\vec {n_2}|} = \frac{2-2+4}{\sqrt{1+4+4}\cdot \sqrt{4+1+4}} = \frac{4}{3\cdot 3} =\frac{4}{9}\\\\( \alpha \widehat {, }\beta )=arccos \, \frac{4}{9}$