## Task Condition

Find the angle between the planes

$6 x+3 y-2 z=0$

$x+2 y+6 z-12=0$
## Solution

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

$\overrightarrow{n_{1}}=\{6 ; 3 ;-2\}$

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

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

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

$$
\begin{aligned}
& =\frac{6+6-12}{\sqrt{36+9+4} \cdot \sqrt{1+4+36}}=\frac{0}{\sqrt{49} \cdot \sqrt{41}}=0 \\
& \phi=\arccos 0=\frac{\pi}{2}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry $10-8$