## Problem Statement

Find the angle between the planes:

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

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

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

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

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

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{3 \cdot 2+(-1) \cdot 1+0 \cdot 0}{\sqrt{3^{2}+(-1)^{2}+0^{2}} \cdot \sqrt{2^{2}+1^{2}+0^{2}}}=$

$=\frac{6-1+0}{\sqrt{9+1+0} \cdot \sqrt{4+1+0}}=\frac{5}{\sqrt{10} \cdot \sqrt{5}}=\frac{1}{\sqrt{2}}$

$\phi=\arccos \frac{1}{\sqrt{2}}=\frac{\pi}{4}$

## Problem Kuznetsov Analytic Geometry 10-29