## Problem Statement

Find the angle between the planes:

$x+y+z \sqrt{2}-3=0$

$x-y+z \sqrt{2}-1=0$
## Solution

The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:

$\overrightarrow{n_{1}}=\{1 ; 1 ; \sqrt{2}\}$

$\overrightarrow{n_{2}}=\{1 ;-1 ; \sqrt{2}\}$

The angle $\phi$ between the planes is determined by the formula:

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+1 \cdot(-1)+\sqrt{2} \cdot \sqrt{2}}{\sqrt{1^{2}+1^{2}+(\sqrt{2})^{2}} \cdot \sqrt{1^{2}+(-1)^{2}+(\sqrt{2})^{2}}}= \\
& =\frac{1-1+2}{\sqrt{1+1+2} \cdot \sqrt{1+1+2}}=\frac{2}{\sqrt{4} \cdot \sqrt{4}}=\frac{1}{2} \\
& \phi=\arccos \frac{1}{2}=\frac{\pi}{3}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry $10-30$