## problem statement

Find the angle between the planes:

$6 x+2 y-4 z+17=0$

$9 x+3 y-6 z-4=0$
## Solution

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

$\overrightarrow{n_{1}}=\{6 ; 2 ;-4\}$

$\overrightarrow{n_{2}}=\{9 ; 3 ;-6\}$

![](https://cdn.mathpix.com/cropped/2024_05_22_69978802cee1d8a5d318g-08.jpg?height=68&width=962&top_left_y=1256&top_left_x=164)

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{6 \cdot 9+2 \cdot 3+(-4) \cdot(-6)}{\sqrt{6^{2}+2^{2}+(-4)^{2}} \cdot \sqrt{9^{2}+3^{2}+(-6)^{2}}}=$

$=\frac{54+6+24}{\sqrt{36+4+16} \cdot \sqrt{81+9+36}}=\frac{84}{\sqrt{56} \cdot \sqrt{126}}=\frac{84}{84}=1$

$\phi=\arccos 1=0$

## Problem Kuznetsov Analytic Geometry $10-5$