## Task Condition

Find the angle between the planes:

$2 y+z-9=0$

$x-y+2 z-1=0$
## Solution

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

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

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

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

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{0 \cdot 1+2 \cdot(-1)+1 \cdot 2}{\sqrt{0^{2}+2^{2}+1^{2}} \cdot \sqrt{1^{2}+(-1)^{2}+2^{2}}}=$
$=\frac{0-2+2}{\sqrt{0+4+1} \cdot \sqrt{1+1+4}}=\frac{0}{\sqrt{5} \cdot \sqrt{6}}=0$
$\phi=\arccos 0=\frac{\pi}{2}$

Problem Kuznetsov Analytic Geometry 10-26