Given two planes:
p1: x+y-z=4
p2: 3x-y+5z=3
The corresponding normal vectors are:
n1=<1,1,-1>
n2=<3,-1,5>
(they are simply the corresponding coefficients of x,y,z)
The line of intersection of p1 and p2 has a direction vector perpendicular to both p1 and p2, thus equal the cross product of n1 and n2,
L1=n1 x n2
=
i j k
1 1 -1
3 -1 5
=<5-1, -3-5, -1-3>
=<4,-8,-4>
Since the length of the direction vector does not matter, we simplify the vector to
L1=<1,-2,-1>
The required line must pass through the point (-2,3,3).
The line therefore has the equation
L: (-2,3,3)+t(1,-2,-1)
or alternatively,
L: x=-2+t, y=3-2t, z=3-t
