Parallel planes do not intersect.
so the slopes do not change, for planes this means the normal vector stays the same
the components of the normal vector are the coefficients of x,y,z respectively
n = <2,-1,3>
Given a point (x1,y1,z1) that plane passes through, the equation is
--> 2(x-x1) -(y-y1)+3(z-z1) = 0
since it passes through origin, the parallel plane is:
2x -y +3z = 0
For a perpendicular plane , there are many solutions as long as the 2 normal vectors are perpendicular
Let n1 be normal vector of a perpendicular plane
n1 = <a,b,c>
then
n*n1 = 0
2a -b +3c = 0
choose any (a,b,c) that satisfies the equation to obtain coefficients of perpendicular plane
Let a=1, b=-1, c=-1
--> (x-x1) -(y-y1)-(z-z1) = 0
since it passes through origin, a perpendicular plane is:
x -y -z = 0
