# Question #b4645

Nov 1, 2016

$\frac{\pi}{3}$

#### Explanation:

Another representation for

${\Pi}_{1} \to 2 x - y + z = 3$ and
${\Pi}_{2} \to x + y + 2 z = 1$

is

${\Pi}_{1} \to \left\langle{\vec{n}}_{1} , p - {p}_{1}\right\rangle = 0$ and
${\Pi}_{2} \to \left\langle{\vec{n}}_{2} , p - {p}_{2}\right\rangle = 0$

with

$p = \left(x , y , z\right)$
${\vec{n}}_{1} = \left(2 , - 1 , 1\right) , {p}_{1} = \left(0 , 0 , 3\right)$ and
${\vec{n}}_{2} = \left(1 , 1 , 2\right) , {p}_{2} = \left(0 , 0 , \frac{1}{2}\right)$

where ${\vec{n}}_{1} , {\vec{n}}_{2}$ are the normal vectors to ${\Pi}_{1}$ and ${\Pi}_{2}$ respectively. So the dihedrical angle $\alpha$ between ${\Pi}_{1}$ and ${\Pi}_{2}$ is obtained by doing

$\left\langle{\vec{n}}_{1} , {\vec{n}}_{2}\right\rangle = \left\lVert {\vec{n}}_{1} \right\rVert \left\lVert {\vec{n}}_{2} \right\rVert \cos \alpha$

so

$\alpha = \arccos \left(\frac{\left\langle{\vec{n}}_{1} , {\vec{n}}_{2}\right\rangle}{\left\lVert {\vec{n}}_{1} \right\rVert \left\lVert {\vec{n}}_{2} \right\rVert}\right) = \frac{\pi}{3}$