# How do you find the angle between the planes x + 2y - z + 1 = 0 and x - y + 3z + 4 = 0?

Jun 1, 2018

See below

#### Explanation:

Angle between two planes is given by the angle of their normal vectors (caracteristic). The caracteristic vectors are given by coefficients of x, y and z in equation plane

In our case $v = \left(1 , 2 , - 1\right)$ is a normal vector to plane 1
and $w = \left(1 , - 1 , 3\right)$ is the normal vector to plane 2

By other hand we know that dot product vector is given by

v·w=v_1w_1+v_2w_2+v_3w_3 where ${v}_{i} , {w}_{i}$ are components of both vectors. Then in our case

v·w=1·1+2·(-1)+(-1)·3=1-2-3=-4

The dot product also can write as v·w=absv·absw·costheta (1) where $\theta$ is angle between both vectors and $\left\mid v \right\mid , \left\mid w \right\mid$ are the module of vectors.

$\left\mid v \right\mid = \sqrt{{1}^{2} + {4}^{2} + {1}^{2}} = \sqrt{6}$
$\left\mid w \right\mid = \sqrt{{1}^{2} + {1}^{2} + {3}^{2}} = \sqrt{11}$

From (1) costheta=(v·w)/(absv·absw)=-4/(sqrt6·sqrt11)=-0.4923659

arccos(-0.4923659)=119º29´46´´