Question

Find the angle of intersection of the plane −3x−5y+3z=3−3x−5y+3z=3 with the plane −5x+3y−5z=−4−5x+3y−5z=−4. Answer in radians: and in degrees:

Answer 1

`-3x-5y+3z=3` has normal vector (-3, -5, 3)

`-5x+3y-5z=-4` has normal vector (-5, 3, -5)

Compute the angle between these normal vectors:

`(-3,-5,3)\cdot(-5,3,-5)=\|(-3,-5,3)\|\|(-5,3,-5)\|\cos\theta\implies\theta\approx1.873\,\mathrm{rad}\approx107.326^\circ`