## Task Condition

Find the angle between the planes:

$x-2 y+2 z+17=0$

$x-2 y-1=0$
## Solution

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

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+(-2) \cdot(-2)+2 \cdot 0}{\sqrt{1^{2}+(-2)^{2}+2^{2}} \cdot \sqrt{1^{2}+(-2)^{2}+0^{2}}}= \\
& =\frac{1+4+0}{\sqrt{1+4+4} \cdot \sqrt{1+4+0}}=\frac{5}{\sqrt{9} \cdot \sqrt{5}}=\sqrt{\frac{5}{9}} \\
& \phi=\arccos \sqrt{\frac{5}{9}} \approx 41^{\circ} 48^{\prime} 37^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytical Geometry $10-20$