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
In order to answer any question, reasonable people use
the information they're given, together with what they know
about the situation.
You've given us no information, and we know nothing about
the situation, so we basically don't stand a chance.
Answer:
8:3
Step-by-step explanation: