Vector.. find the symmetric equation of the line intersection of the two planes...

find the symmetric equation of the line intersection of the two planes:
x + y + z = 1 and x - 2y + 3z = 1

The line is parallel to each plane, hence perpendicular to each plane's normal. So take the direction vector to be the cross product of the normals.

n1 = <1, 1, 1>, n2 = <1, -2, 3> ==> v = n1 x n2 = <5, -2, -3>.

Find some point on the line. If z = 0, solve the system

x + y = 1
x - 2y = 1 ==> x = 1, y = 0.

So the line contains (1, 0, 0) and is parallel to <5, -2, -3>. The symmetric form is

(x - 1)/5 = y/-2 = z/-3.

say the first plane is S1, and the second is S2
find the normal of the vectors
N1 = <1,1,1> and N2 = <1,-2,3> the cross product of these two will give us a vector that is perpendicular to both normals and parallel to both planes (vector along the line of intersection)
so i solved V = N1xN2= <5,-2,-3>
we also need a point on the line. so a sample point could be the point of intersection of the line with the xz plane where y=0
so S1: x+z=1 , S2: x+3z=1. >>> z=0, x=1. we have P(1,0,0)
so our parametric equation of line: is
x=1+5t
y=-2t
z=-3t
solving for t and setting them equal to eachother gives us the symmetric form.
(x-1)/5 = y/-2 = z/-3