## Problem Statement

Find the angle between the planes:

\[
\begin{aligned}
& 3 x-2 y+3 z+23=0 \\
& y+z+5=0
\end{aligned}
\]
## Solution

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

$\overrightarrow{n_{1}}=\{3 ;-2 ; 3\}$

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{3 \cdot 0+(-2) \cdot 1+3 \cdot 1}{\sqrt{3^{2}+(-2)^{2}+3^{2}} \cdot \sqrt{0^{2}+1^{2}+1^{2}}}= \\
& =\frac{0+(-2)+3}{\sqrt{9+4+9} \cdot \sqrt{0+1+1}}=\frac{1}{\sqrt{22} \cdot \sqrt{2}}=\frac{1}{2 \sqrt{11}} \\
& \phi=\arccos \frac{1}{2 \sqrt{11}} \approx 81^{\circ} 19^{\prime} 45^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-18