## Task Condition

Find the angle between the planes:

$x-y+7 z-1=0$

$2 x-2 y-5=0$
## Solution

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

$\overrightarrow{n_{1}}=\{1 ;-1 ; 7\}$

$\overrightarrow{n_{2}}=\{2 ;-2 ; 0\}$

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 2+(-1) \cdot(-2)+7 \cdot 0}{\sqrt{1^{2}+(-1)^{2}+7^{2}} \cdot \sqrt{2^{2}+(-2)^{2}+0^{2}}}= \\
& =\frac{2+2+0}{\sqrt{1+1+49} \cdot \sqrt{4+4+0}}=\frac{4}{\sqrt{51} \cdot \sqrt{8}}=\frac{2}{\sqrt{102}} \\
& \phi=\arccos \frac{2}{\sqrt{102}} \approx 78^{\circ} 34^{\prime} 42^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-28