## Problem Statement

Find the angle between the planes:

$$
\begin{aligned}
& 2 x+2 y+z+9=0 \\
& x-y+3 z-1=0
\end{aligned}
$$
## Solution

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

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

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

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

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

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

$\phi=\arccos \frac{1}{\sqrt{11}} \approx 72^{\circ} 27^{\prime} 6^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-14