## Problem Statement

Find the angle between the planes:

$x-3 y+z-1=0$

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

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

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

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

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

$$
\begin{aligned}
& \cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+(-3) \cdot 0+1 \cdot 1}{\sqrt{1^{2}+(-3)^{2}+1^{2}} \cdot \sqrt{1^{2}+0^{2}+1^{2}}}= \\
& =\frac{1+0+1}{\sqrt{1+9+1} \cdot \sqrt{1+0+1}}=\frac{2}{\sqrt{11} \cdot \sqrt{2}}=\sqrt{\frac{2}{11}} \\
& \phi=\arccos \sqrt{\frac{2}{11}} \approx 64^{\circ} 45^{\prime} 38^{\prime \prime}
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 10-2