## Task Condition

Find the angle between the planes:

$x-3 y+5=0$

$2 x-y+5 z-16=0$
## Solution

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

$\overrightarrow{n_{1}}=\{1 ;-3 ; 0\}$

$\overrightarrow{n_{2}}=\{2 ;-1 ; 5\}$

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 2+(-3) \cdot(-1)+0 \cdot 5}{\sqrt{1^{2}+(-3)^{2}+0^{2}} \cdot \sqrt{2^{2}+(-1)^{2}+5^{2}}}= \\
& =\frac{2+3+0}{\sqrt{1+9+0} \cdot \sqrt{4+1+25}}=\frac{5}{\sqrt{10} \cdot \sqrt{30}}=\frac{5}{10 \sqrt{3}}=\frac{\sqrt{3}}{6} \\
& \phi=\arccos \frac{\sqrt{3}}{6} \approx 73^{0} 13^{\prime} 17^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry $10-1$