## Task Condition

Find the angle between the planes:

$$
3 x+2 y-3 z-1=0
$$

$x+y+z-7=0$
## Solution

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

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{3 \cdot 1+2 \cdot 1-3 \cdot 1}{\sqrt{3^{2}+2^{2}+(-3)^{2}} \cdot \sqrt{1^{2}+1^{2}+1^{2}}}= \\
& =\frac{2}{\sqrt{9+4+9} \cdot \sqrt{1+1+1}}=\frac{2}{\sqrt{22} \cdot \sqrt{3}}=\frac{2}{\sqrt{66}} \\
& \phi=\arccos \frac{2}{\sqrt{66}} \approx 75^{\circ} 44^{\prime} 54^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-16