## problem statement

Find the angle between the planes:

$x+2 y-1=0$

$x+y+6=0$
## Solution

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

$\overrightarrow{n_{1}}=\{1 ; 2 ; 0\}$

$\overrightarrow{n_{2}}=\{1 ; 1 ; 0\}$

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

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

$=\frac{1+2+0}{\sqrt{1+4+0} \cdot \sqrt{1+1+0}}=\frac{3}{\sqrt{5} \cdot \sqrt{2}}=\frac{3}{\sqrt{10}}$

$\phi=\arccos \frac{3}{\sqrt{10}} \approx 18^{\circ} 26^{\prime} 6^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-21