Final answer:
The distance between the parallel planes z = 2y - 2x and 3z = -7 - 6x + 6y is 1/3 units.
Step-by-step explanation:
To find the distance between two parallel planes given by z = 2y - 2x and 3z = -7 - 6x + 6y, we first need to rewrite the second equation in terms of z only. This gives us z = (-7/3) - 2x + 2y. Both planes have the same normal vector (unit vector normal to the plane) which is represented by the coefficients of x and y, that is, [-2, 2, 1].
To find the distance (d) between the two parallel planes, we can use the distance formula for parallel planes: d = |c2 - c1| / ||N||, where c1 and c2 are the constant terms from the plane equations (without x, y, and z) and N is the normal vector.
Here c1 = -2 and c2 = -7/3, and the magnitude of the normal vector N is ||N|| = √((-2)^2 + 2^2 + 1^2) = 3. Substituting the values in the formula, we get d = |(-7/3) + 2| / 3 = 1/3. Therefore, the distance between the planes is 1/3 units.