Find a set of scalar parametric equations for the line formed by the two intersecting planes. 3x+3y+2z+2=0 2x−3y+2z−2=0

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

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