Find the acute angle between the planes $2 x-y-3 z+5=0$ and $x+y-2=0$.

#
Vectors $\overrightarrow{n_{1}}=(2 ;-1 ;-3)$ and $\overrightarrow{n_{2}}=(1 ; 1 ; 0)$ are respectively perpendicular to the given planes. If $\alpha-$ is the acute angle between the planes, and $\phi-$ is the angle between vectors $\overrightarrow{n_{\mathbf{I}}}$ and $\overrightarrow{n_{2}}$, then

$$
\begin{gathered}
\cos \phi=\frac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{m_{1}}\right| \cdot\left|\overrightarrow{a_{2}}\right|}=\frac{2 \cdot 1-1 \cdot 1-3 \cdot 0}{\sqrt{2^{2}+1^{2}+3^{2}} \sqrt{1^{2}+1^{2}+0^{2}}}=\frac{1}{2 \sqrt{7}} \\
\cos \alpha=|\cos \phi|=\frac{1}{2 \sqrt{7}}
\end{gathered}
$$

## Answer

 $\arccos \frac{1}{2 \sqrt{7}}$.