## Task Condition

Find the angle between the planes:

$x-3 y-2 z-8=0$

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

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

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

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

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

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

$=\frac{1-3+2}{\sqrt{1+9+4} \cdot \sqrt{1+1+1}}=\frac{0}{\sqrt{14} \cdot \sqrt{3}}=0$
$\phi=\arccos 0 \approx \frac{\pi}{2}$

## Problem Kuznetsov Analytic Geometry 10-17