## Task Condition

Find the angle between the planes:

$4 x+3 z-2=0$

$x+2 y+2 z+5=0$
## Solution

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

$\overrightarrow{n_{1}}=\{4 ; 0 ; 3\}$

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

![](https://cdn.mathpix.com/cropped/2024_05_22_5214a658a43dd63b9438g-08.jpg?height=68&width=971&top_left_y=606&top_left_x=154)

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{4 \cdot 1+0 \cdot 2+3 \cdot 2}{\sqrt{4^{2}+0^{2}+3^{2}} \cdot \sqrt{1^{2}+2^{2}+2^{2}}}=$

$=\frac{4+0+6}{\sqrt{16+0+9} \cdot \sqrt{1+4+4}}=\frac{10}{\sqrt{25} \cdot \sqrt{9}}=\frac{2}{3}$

$\phi=\arccos \frac{2}{3} \approx 48^{0} 11^{\prime} 23^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-24