## problem statement

Find the angle between the planes:

$2 x-y+5 z+16=0$

$x+2 y+3 z+8=0$
## Solution

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

$\overrightarrow{n_{1}}=\{2 ;-1 ; 5\}$

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

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

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{2 \cdot 1+(-1) \cdot 2+5 \cdot 3}{\sqrt{2^{2}+(-1)^{2}+5^{2}} \cdot \sqrt{1^{2}+2^{2}+3^{2}}}=$

$=\frac{2-2+15}{\sqrt{4+1+25} \cdot \sqrt{1+4+9}}=\frac{15}{\sqrt{30} \cdot \sqrt{14}}=\sqrt{\frac{15}{28}}$

$\phi=\arccos \sqrt{\frac{15}{28}} \approx 42^{0} 57^{\prime} 7^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-10