## Task Condition

Find the angle between the planes:
$5 x+3 y+z-18=0$

$2 y+z-9=0$
## Solution

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

$\overrightarrow{n_{1}}=\{5 ; 3 ; 1\}$

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{5 \cdot 0+3 \cdot 2+1 \cdot 1}{\sqrt{5^{2}+3^{2}+1^{2}} \cdot \sqrt{0^{2}+2^{2}+1^{2}}}= \\
& =\frac{0+6+1}{\sqrt{25+9+1} \cdot \sqrt{0+4+1}}=\frac{7}{\sqrt{35} \cdot \sqrt{5}}=\frac{\sqrt{7}}{5} \\
& \phi=\arccos \frac{\sqrt{7}}{5} \approx 58^{0} 3^{\prime} 7^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-23