Question

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

Answer 1

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=(1,2,-1)` is a normal vector to plane 1
and `w=(1,-1,3)` 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 `absv, absw` are the module of vectors.

`absv=sqrt(1^2+4^2+1^2)=sqrt6`
`absw=sqrt(1^2+1^2+3^2)=sqrt11`

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

`arccos(-0.4923659)=119º29´46´´`