## Task Condition

Find the angle between the planes:

$x+4 y-z+1=0$

$2 x+y+4 z-3=0$
## Solution

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

$\overrightarrow{n_{1}}=\{1 ; 4 ;-1\}$

$\overrightarrow{n_{2}}=\{2 ; 1 ; 4\}$

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

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

$=\frac{2+4+(-4)}{\sqrt{1+16+1} \cdot \sqrt{4+1+16}}=\frac{2}{\sqrt{18} \cdot \sqrt{21}}=\frac{2}{3 \sqrt{42}}$

$\phi=\arccos \frac{2}{3 \sqrt{42}} \approx 84^{\circ} 5^{\prime} 4^{\prime \prime}$

## Problem Kuznetsov Analytic Geometry 10-25