How do we find the parametric equation of the line of intersection of the planes x+y+z=1 and x-y-z=2 ?
If we let x=t(parameter) and try solving this eventually y and z terms get cancelled out.
How do we do this thrn? Thanks
add the two equations and you get
2x = 3
so how about
x = 3/2, y=t, z=t
or some variation of that
To find the parametric equation of the line of intersection of the planes x + y + z = 1 and x - y - z = 2, you can follow these steps:
1. Solve the system of equations formed by the two planes:
- Start by eliminating one variable between the equations. In this case, let's eliminate z. Add the two equations together to get rid of z:
(x + y + z) + (x - y - z) = 1 + 2
2x = 3
x = 3/2
- Substitute the value of x in either of the plane equations to find the value of y:
(3/2) + y + z = 1
y + z = -1/2
y = -1/2 - z
2. Now that we have the values of x and y in terms of z, we can parameterize the line using the variable t:
x = 3/2
y = -1/2 - z
z = t (parameter)
Combining these equations, we get the parametric equation of the line of intersection:
x = 3/2
y = -1/2 - t
z = t
So, the parametric equation of the line of intersection of the planes x + y + z = 1 and x - y - z = 2 is:
x = 3/2
y = -1/2 - t
z = t
Note: The parameterization assumes t can be any real number, which means it covers all points on the line of intersection.