Final answer:
Parametric equations for the line of intersection are x = (6 + 3t) / 2, y = -3t/4 + 3/2, and z = t. The parameter t represents any value that y can take, and by substituting into the plane equations, these satisfy both plane equations.
Step-by-step explanation:
To find the parametric equations for the line of intersection of the planes x + 2y - 3z = 3 and x - y - z = 3, we need to solve for three variables (x, y, z) given these two equations. With only two equations, we can express one variable in terms of a parameter t.
First, let's solve the system of equations to express x in terms of y and z. By adding the two equations together, we get:
Let y be the parameter t. Now, we can express x and z in terms of t as follows:
- x = (6 + 4z - t) / 2
- z is free to be any value, so let's let z = t
- x = (6 + 4t - t) / 2 = (6 + 3t) / 2
Now we substitute these expressions into our original equations to eliminate z and solve for y:
- Substituting x and z into the first plane equation, we get (6 + 3t)/2 + 2y - 3(t) = 3, which simplifies to y = (3t - 6) / -4 = -3t/4 + 3/2.
Therefore, the parametric equations for the line are:
- x = (6 + 3t) / 2
- y = -3t/4 + 3/2
- z = t
These equations can be used to get any point on the line for a given value of t. To verify that these parametric equations are correct, substitute them back into the original plane equations to ensure that they satisfy both equations.