Determine the equation of a plane, P3, that intersects the planes P1: x + y + 3z − 2 = 0 and P2: x − y + 2z = 0 in
a point;
a line.
To determine the equation of a plane that intersects two given planes in a point or a line, we'll need to follow a few steps.
1. If the planes P1 and P2 intersect, they will share a common line. We need to find the direction vector of this common line.
2. Once we have the direction vector, we can find a point on this common line by setting one of the variables to a specific value and solving the resulting system of equations.
3. Now that we have a point on the common line and the direction vector, we can write an equation for the line that passes through this point.
4. To find the equation of the plane that intersects both P1 and P2, we will need a normal vector for the plane. We can find this by taking the cross product of the normal vectors of P1 and P2.
5. With the normal vector and a point on the plane, we can write the equation of the plane in the form Ax + By + Cz + D = 0.
Now let's apply these steps to find the equation of plane P3 that intersects P1 and P2, either in a point or a line.
a) If the planes intersect in a point:
Step 1:
The normal vectors of P1 and P2 can be read directly from their equations:
P1: <1, 1, 3>
P2: <1, -1, 2>
The common line's direction vector can be found by taking the cross product of these two vectors:
<1, 1, 3> × <1, -1, 2> = <-1, -1, -2> = -1<1, 1, 2>
Step 2:
To find a point on the common line, we set one variable to a specific value in the system of equations for P1 and P2. Let's set z = 0:
P1: x + y - 2 = 0 --> x + y = 2 --> x = 2 - y
P2: x - y = 0 --> x = y
Substituting the value of x from P2 into P1:
y + y = 2
2y = 2
y = 1
x = 1
So, a point on the common line is (x, y, z) = (1, 1, 0).
Step 3:
Now that we have a point on the common line and its direction vector, we can write the equation of the line. Using parametric form, we get:
x = 1 - t
y = 1 - t
z = -2t
Step 4:
To find the normal vector for plane P3, we find the cross product of the normal vectors of P1 and P2:
<1, 1, 3> × <1, -1, 2> = <-1, -1, -2>
Step 5:
Using the point (1, 1, 0) and the normal vector <-1, -1, -2>, we can write the equation of plane P3 in the form Ax + By + Cz + D = 0:
-1(x - 1) - 1(y - 1) - 2z = 0
-x + 1 - y + 1 - 2z = 0
-x - y - 2z + 2 = 0
Therefore, the equation of plane P3 that intersects the planes P1 and P2 in a point is -x - y - 2z + 2 = 0.
b) If the planes intersect in a line:
In this case, the process is similar to the previous one until Step 3. So, let's pick up from there.
Step 3:
We found a point on the common line as (x, y, z) = (1, 1, 0). Now, we need to find a second point on this line to form a direction vector.
Again, we set z = 0 and solve for x and y in the system of equations for P1 and P2:
P1: x + y - 2 = 0 --> x + y = 2 --> x = 2 - y
P2: x - y + 2z = 0 --> x + y = 0 --> x = -y
Equating the two expressions of x:
2 - y = -y
2 = 0
This is not a valid equation, which means the two planes P1 and P2 do not intersect in a line.