Find parametric equations and symmetric equations for the line of intersection of the given planes.
x + y + z = 2, x + z = 0
X=1
y =
Z=
x=, y=
let z = t
then x+t = 0
x = -t
into the firs
-t + y +t = 2
y = 2
so
x = -t
y = 2
z = t for parametric
t = x/-1
t = z
symmetric:
x/-1 = z, y = 2
To find the parametric equations and symmetric equations for the line of intersection of the given planes, we can solve the system of equations formed by the two planes.
1) Adding the two equations, we obtain:
(x + y + z) + (x + z) = 2 + 0
2x + 2z = 2
2) Divide both sides of the equation by 2:
x + z = 1
3) We can consider x as a parameter, denoted by t, and solve for y and z:
x = t
z = 1 - x = 1 - t
Therefore, the parametric equations for the line of intersection are:
x = t
y = t
z = 1 - t
For the symmetric equations, we can write the equations in terms of the normal vectors of the planes:
The normal vector of the first plane is (1, 1, 1).
The normal vector of the second plane is (1, 0, 1).
The symmetric equations can be written as follows:
(x - X)/1 = (y - Y)/1 = (z - Z)/1
Substituting the known point (X, Y, Z) = (1, 0, 0), we have:
(x - 1)/1 = (y - 0)/1 = (z - 0)/1
Simplifying:
x - 1 = y = z
Therefore, the symmetric equations for the line of intersection are:
x - 1 = y = z
To find the parametric equations for the line of intersection, we can start by solving the system of equations formed by the given planes.
First, let's solve for y in terms of x and z from the first plane equation:
x + y + z = 2
Rearranging the equation, we have:
y = 2 - x - z
Now, substitute this expression for y into the second plane equation:
x + z = 0
Substituting y = 2 - x - z, we have:
x + z = 0
x + (2 - x - z) + z = 0
2 = 0
This is not a valid equation.
Since the planes do not intersect, there is no line of intersection. Therefore, there are no parametric equations or symmetric equations for the line of intersection.