## Problem Statement

Find the angle between the planes:

$2 x+2 y+z-1=0$

$x+z-1=0$
## Solution

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{2 \cdot 1+2 \cdot 0+1 \cdot 1}{\sqrt{2^{2}+2^{2}+1^{2}} \cdot \sqrt{1^{2}+0^{2}+1^{2}}}= \\
& =\frac{2+1}{\sqrt{4+4+1} \cdot \sqrt{1+0+1}}=\frac{3}{\sqrt{9} \cdot \sqrt{2}}=\frac{1}{\sqrt{2}} \\
& \phi=\arccos \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-11