## Task Condition

Find the angle between the planes:

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

$16 x+12 y-15 z-1=0$
## Solution

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

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

$\overrightarrow{n_{2}}=\{16 ; 12 ;-15\}$

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

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 16+2 \cdot 12+2 \cdot(-15)}{\sqrt{1^{2}+2^{2}+2^{2}} \cdot \sqrt{16^{2}+12^{2}+(-15)^{2}}}=$

$=\frac{16+24-30}{\sqrt{1+4+4} \cdot \sqrt{256+144+225}}=\frac{10}{\sqrt{9} \cdot \sqrt{625}}=\frac{10}{75}=\frac{2}{15}$

$\phi=\arccos \frac{2}{15} \approx 82^{0} 20^{\prime} 16^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-9