## problem statement

Find the angle between the planes:

$$
\begin{aligned}
& 4 x-5 y+3 z-1=0 \\
& x-4 y-z+9=0
\end{aligned}
$$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{4 ;-5 ; 3\} \\
& \overrightarrow{n_{2}}=\{1 ;-4 ;-1\}
\end{aligned}
$$

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{4 \cdot 1+(-5) \cdot(-4)+3 \cdot(-1)}{\sqrt{4^{2}+(-5)^{2}+3^{2}} \cdot \sqrt{1^{2}+(-4)^{2}+(-1)^{2}}}= \\
& =\frac{4+20-3}{\sqrt{16+25+9} \cdot \sqrt{1+16+1}}=\frac{21}{\sqrt{50} \cdot \sqrt{18}}=\frac{21}{30}=0.7 \\
& \phi=\arccos 0.7 \approx 45^{\circ} 34^{\prime} 23^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry $10-3$