## Problem Statement

Find the angle between the planes:

\[
\begin{aligned}
& x+y+3z-7=0 \\
& y+z-1=0
\end{aligned}
\]
## Solution

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

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

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

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

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 0+1 \cdot 1+3 \cdot 1}{\sqrt{1^{2}+1^{2}+3^{2}} \cdot \sqrt{0^{2}+1^{2}+1^{2}}}=$

$=\frac{0+1+3}{\sqrt{1+1+9} \cdot \sqrt{0+1+1}}=\frac{4}{\sqrt{11} \cdot \sqrt{2}}=2 \sqrt{\frac{2}{11}}$

$\phi=\arccos 2 \sqrt{\frac{2}{11}} \approx 31^{0} 28^{\prime} 56^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-19