Question

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

Answer 1

`alpha=119.50^@`

Explanation:

To find the angle between two planes, one has to find the angle between normal vectors of these planes.

How to find a normal vector to a plane

Explanation of why the angle between two vectors, each one normal to a plane, gives the angle between the two involved planes

Normal Vectors
For the plane 1 , `x+2y-z+1=0`
`N vec 1= hat i +2*hat j-1*hat k`

For the plane 2 , `x-y+3z+4=0`
`N vec 2= hat i -hat j+3*hat k`

Angle between the 2 vectors
`N vec 1*N vec 2=|N vec 1|*|N vec 2|*cos alpha` => `cos alpha=(N vec 1*N vec 2)/(|N vec 1|*|N vec 2|)`
`N vec 1*N vec 2=1*1+2*(-1)+(-1)(3)=1-2-3=-4`
`|N vec 1|=sqrt(1^2+2^2+(-1)^2)=sqrt(1+4+1)=sqrt(6)`
`|N vec 2|=sqrt(1^2+(-1)^2+3^2)=sqrt(1+1+9)=sqrt(11)`
`cos alpha=-4/(sqrt(6)*sqrt(11)` => `cos alpha=-4/sqrt(66)` => `alpha=119.50^@`