Question

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

Answer 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 ...

• x = 12t
• y = -2/3 -2t
• z = -15t

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

• x = 8 +12t
• y = -2 -2t
• z = -10 -15t