## Task Condition

Find the angle between the planes

$x+2 y-2 z-7=0$

$x+y-35=0$
## Solution

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

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

$\overrightarrow{n_{2}}=\{1 ; 1 ; 0\}$

The angle $\phi_{\text{between the planes is determined by the formula: }}$

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+2 \cdot 1-2 \cdot 0}{\sqrt{1^{2}+2^{2}+(-2)^{2}} \cdot \sqrt{1^{2}+1^{2}+0^{2}}}=$

$$
\begin{aligned}
& =\frac{3}{\sqrt{9} \cdot \sqrt{2}}=\frac{3}{3 \cdot \sqrt{2}}=\frac{1}{\sqrt{2}} \\
& \phi=\arccos \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-31