## problem statement

Find the angle between the planes:

$2 x-6 y+14 z-1=0$

$5 x-15 y+35 z-3=0$
## Solution

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

$\overrightarrow{n_{1}}=\{2 ;-6 ; 14\}$

$\overrightarrow{n_{2}}=\{5 ;-15 ; 35\}$

The angle $\phi_{\text{between the planes is determined by the formula: }}$

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{2 \cdot 5+(-6) \cdot(-15)+14 \cdot 35}{\sqrt{2^{2}+(-6)^{2}+14^{2}} \cdot \sqrt{5^{2}+(-15)^{2}+35^{2}}}= \\
& =\frac{10+90+490}{\sqrt{4+36+196} \cdot \sqrt{25+225+1225}}=\frac{590}{\sqrt{236} \cdot \sqrt{1475}}=\frac{590}{590}=1 \\
& \phi=\arccos 1=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-27