The normal vectors to the two planes are (3, 3, 2) and (2, -3, 2). The cross product of these will be the direction vector of the line of intersection, (12, -2, -15).
Using x=0, we can find a point on this line by solving the simultaneous equations that remain:
... 3y +2z = -2
... -3y +2z = 2
Adding these, we get
... 4z = 0
... z = 0
so the point we're looking for is (x, y, z) = (0, -2/3, 0). This gives rise to the parametric equations ...
By letting t=2/3, we can find a point on the line that has integer coefficients. That will be (x, y, z) = (8, -2, -10).
Then our parametric equations can be written as
The solution is:
x = - 4×t
y = -4/6 + 4/6 t for 0 ≤ t ≤ 1
To look for the intersection line, we have:
3×x + 3×y + 2×z + 2 = 0 or 2×z = - 3×x - 3×y - 2 (1)
2×x - 3×y + 2×z -2 = 0 or 2×z = - 2×x + 3×y + 2 (2)
Solving the two equation sysstem in terms of x and y. Equalitying (1) and (2)
- 3×x - 3×y - 2 = - 2×x + 3×y + 2
reordering - x - 6×y - 4 = 0
So we got the equation of the intersection line
If in that equation we look for the intersection points, we get:
x = 0 - 6×y - 4 = 0 y = -4/6 ( intersection point P with y-axis)
y = 0 -x - 4 = 0 x = -4 ( intersection point Q with x-axis)
Now that we got the whole information about the intersection line ( see attached drawing)
Looking at the attached file we can see that the segment PQ is
PQ = ( -4 ; 4/6) ( coordinates of Q minus coordinates of P )
We choose t as a parameter then:
When t = 0 x = 0 x = -4×t x = 0
When t = 1 x = -4 x = -4×t x = -4
Now for y
When t = 0 y = - 4/6 y = -4/6 + 4/6 t y = -4/6
When t = 1 y = 0 y = -4/6 + 4/6(1) y = 0
finally x = - 4×t
y = -4/6 + 4/6 t for 0 ≤ t ≤ 1
