## Task Condition

Find the angle between the planes:

$x+2 y+2 z-3=0$

$2 x-y+2 z+5=0$
## Solution

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

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 2+2 \cdot(-1)+2 \cdot 2}{\sqrt{1^{2}+2^{2}+2^{2}} \cdot \sqrt{2^{2}+(-1)^{2}+2^{2}}}= \\
& =\frac{2-2+4}{\sqrt{1+4+4} \cdot \sqrt{4+1+4}}=\frac{4}{\sqrt{9} \cdot \sqrt{9}}=\frac{4}{9} \\
& \phi=\arccos \frac{4}{9} \approx 63^{0} 36^{\prime} 44^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-15