## Problem Statement

Find the angle between the planes:

$$
\begin{aligned}
& 3 x-y+2 z+15=0 \\
& 5 x+9 y-3 z-1=0
\end{aligned}
$$
## Solution

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

$\overrightarrow{n_{1}}=\{3 ;-1 ; 2\}$

$\overrightarrow{n_{2}}=\{5 ; 9 ;-3\}$

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

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{3 \cdot 5+(-1) \cdot 9+2 \cdot(-3)}{\sqrt{3^{2}+(-1)^{2}+2^{2}} \cdot \sqrt{5^{2}+9^{2}+(-3)^{2}}}=$

$=\frac{15-9-6}{\sqrt{9+1+4} \cdot \sqrt{25+81+9}}=\frac{0}{\sqrt{14} \cdot \sqrt{115}}=0$
$\phi=\arccos 0=\frac{\pi}{2}$

## Problem Kuznetsov Analytic Geometry 10-4