## Problem Statement

Find the angle between the planes:

\[
\begin{aligned}
& 3 x+y+z-4=0 \\
& y+z+5=0
\end{aligned}
\]
## Solution

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

$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{3 ; 1 ; 1\} \\
& \overrightarrow{n_{2}}=\{0 ; 1 ; 1\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{3 \cdot 0+1 \cdot 1+1 \cdot 1}{\sqrt{3^{2}+1^{2}+1^{2}} \cdot \sqrt{0^{2}+1^{2}+1^{2}}}= \\
& =\frac{0+1+1}{\sqrt{9+1+1} \cdot \sqrt{0+1+1}}=\frac{2}{\sqrt{11} \cdot \sqrt{2}}=\sqrt{\frac{2}{11}} \\
& \phi=\arccos \sqrt{\frac{2}{11}} \approx 64^{\circ} 45^{\prime} 38^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-12