Find an equation of the plane containing the line of intersection of x+y+z=1 and
x−y+2z=2, and perpendicular to the x-y plane.
To find the equation of the plane containing the line of intersection of two given planes and perpendicular to another plane, we can follow these steps:
Step 1: Find the line of intersection of the two given planes.
Step 2: Find a vector that is parallel to the line of intersection.
Step 3: Find a vector that is perpendicular to the desired plane.
Step 4: Use the normal vector and a point on the line to get the equation of the plane.
Let's go through the steps one by one:
Step 1: Find the line of intersection of x+y+z=1 and x−y+2z=2.
To find the line of intersection, we need to solve the system of equations formed by the two planes:
x+y+z=1
x−y+2z=2
By solving the system, we get:
x = 1
y = 0
z = 0
So, the line of intersection is represented by the parametric equations:
x = 1
y = 0
z = 0 + t
Where t is a parameter.
Step 2: Find a vector that is parallel to the line of intersection.
Since the line of intersection has a parallel vector along the line, we can choose the coefficients of x, y, and z in the parametric equation as the components of the parallel vector.
A vector parallel to the line of intersection is:
V1 = <1, 0, 1>
Step 3: Find a vector that is perpendicular to the desired plane, which is the x-y plane.
Since the desired plane is perpendicular to the x-y plane, any vector in the x-y plane will be perpendicular to it.
A vector in the x-y plane is:
V2 = <1, 1, 0>
Step 4: Use the normal vector and a point on the line to get the equation of the plane.
To find the equation of the plane, we need a normal vector perpendicular to the plane. We can find the normal vector by taking the cross product of the two vectors found in steps 2 and 3.
The cross product of V1 and V2 gives the normal vector:
N = V1 x V2 = <1, 0, 1> x <1, 1, 0> = <-1, 1, 1>
Now, we have the normal vector N = <-1, 1, 1> and we can use a point on the line of intersection, which is (1, 0, 0), to find the equation of the plane.
The equation of the plane is given by:
-1(x - 1) + 1y + 1(z - 0) = 0
Which simplifies to:
-x + y + z + 1 = 0
Therefore, the equation of the plane containing the line of intersection of x+y+z=1 and x−y+2z=2, and perpendicular to the x-y plane is -x + y + z + 1 = 0.