To find the parametric equations for the line of intersection of the planes x+y+z=3 and x-y+2z=2, you need to solve the system of equations simultaneously.
First, let's solve the system by using the method of elimination:
Given equations:
1) x + y + z = 3
2) x - y + 2z = 2
To eliminate x, let's subtract equation (2) from equation (1):
(x + y + z) - (x - y + 2z) = 3 - 2
x + y + z - x + y - 2z = 1
2y - z = 1
Now, we have two equations:
1) x + y + z = 3
2) 2y - z = 1
To eliminate z, let's multiply equation (1) by 2 and add it to equation (2):
2(x + y + z) + 2y - z = 6 + 1
2x + 3y = 7
Now, we have two equations:
1) x + y + z = 3
2) 2x + 3y = 7
To further eliminate x, let's multiply equation (1) by 2 and subtract it from equation (2):
2(2x + 3y) - (x + y + z) = 14 - 3
4x + 6y - x - y - z = 11
3x + 5y - z = 11
Now, we have two equations:
1) x + y + z = 3
2) 3x + 5y - z = 11
To eliminate z, let's add equation (1) to equation (2):
(x + y + z) + (3x + 5y - z) = 3 + 11
4x + 6y = 14
Now, we have one equation:
4x + 6y = 14
To find the parameterization, solve this equation for x and y:
4x + 6y = 14
2x + 3y = 7
Let's set y as the parameter t:
2x + 3t = 7
2x = 7 - 3t
x = 7/2 - (3/2)t
Now, we can substitute this value of x into the equation 1) x + y + z = 3:
(7/2 - (3/2)t) + y + z = 3
y + z = 3 - (7/2) + (3/2)t
y + z = 6/2 - 7/2 + (3/2)t
y + z = (-1/2) + (3/2)t
Let's substitute y + z as another parameter s:
s = (-1/2) + (3/2)t
Now, we can express y and z in terms of the parameter s:
y = s - z
y = [(-1/2) + (3/2)t] - z
y = (-1/2) + (3/2)t -z
Substitute z as another parameter w:
y = (-1/2) + (3/2)t - w
Finally, we can express the line of intersection as parametric equations:
x = 7/2 - (3/2)t
y = (-1/2) + (3/2)t - w
z = w
These are the correct parametric equations for the line of intersection of the given planes.