## Task Condition

Find the angle between the planes

$3 y-z=0$

$2 y+z=0$
## Solution

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

$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{0 ; 3 ;-1\} \\
& \overrightarrow{n_{2}}=\{0 ; 2 ; 1\}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_9b2192c09ee753478b6bg-07.jpg?height=70&width=967&top_left_y=2269&top_left_x=156)

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{0 \cdot 0+3 \cdot 2+(-1) \cdot 1}{\sqrt{0^{2}+3^{2}+(-1)^{2}} \cdot \sqrt{0^{2}+2^{2}+1^{2}}}= \\
& =\frac{0+6-1}{\sqrt{0+9+1} \cdot \sqrt{0+4+1}}=\frac{5}{\sqrt{10} \cdot \sqrt{5}}=\frac{1}{\sqrt{2}}
\end{aligned}
$$

$\phi=\arccos \frac{1}{\sqrt{2}}=\frac{\pi}{4}$

## Problem Kuznetsov Analytic Geometry 10-7