## Task Condition

Find the angle between the planes:

$2 x-z+5=0$

$2 x+3 y-7=0$
## Solution

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

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

$\overrightarrow{n_{2}}=\{2 ; 3 ; 0\}$

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{2 \cdot 2+0 \cdot 3+(-1) \cdot 0}{\sqrt{2^{2}+0^{2}+(-1)^{2}} \cdot \sqrt{2^{2}+3^{2}+0^{2}}}= \\
& =\frac{4+0+0}{\sqrt{4+0+1} \cdot \sqrt{4+9+0}}=\frac{4}{\sqrt{5} \cdot \sqrt{13}}=\frac{4}{\sqrt{65}} \\
& \phi=\arccos \frac{4}{\sqrt{65}} \approx 60^{0} 15^{\prime} 18^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-22