## Task Condition

Find the angle between the planes

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

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

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

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+(-\sqrt{2}) \cdot \sqrt{2}+1 \cdot(-1)}{\sqrt{1^{2}+(-\sqrt{2})^{2}+1^{2}} \cdot \sqrt{1^{2}+(\sqrt{2})^{2}+(-1)^{2}}}= \\
& =\frac{1-2-1}{\sqrt{1+2+1} \cdot \sqrt{1+2+1}}=\frac{-2}{\sqrt{4} \cdot \sqrt{4}}=-\frac{1}{2} \\
& \phi=\arccos -\frac{1}{2}=120^{\circ}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-6