## Task Condition

Find the angle between the planes:

$3 x-2 y-2 z-16=0$

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

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

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{3 \cdot 1+(-2) \cdot 1+(-2) \cdot(-3)}{\sqrt{3^{2}+(-2)^{2}+(-2)^{2}} \cdot \sqrt{1^{2}+1^{2}+(-3)^{2}}}= \\
& =\frac{3-2+6}{\sqrt{9+4+4} \cdot \sqrt{1+1+9}}=\frac{7}{\sqrt{17} \cdot \sqrt{11}}=\frac{7}{\sqrt{187}} \\
& \phi=\arccos \frac{7}{\sqrt{187}} \approx 59^{\circ} 12^{\prime} 37^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-13