Determine the intersection, if any, for the planes with equations
x + y – z + 12 =0 and 2x + 2y - 2z + 8 = 0.
x + y - z = -12
x + y - z = -4
never the twain shall meet. Sketch the graph in an x,y,z coordinate sketch
To find the intersection of two planes, we need to find the values of x, y, and z that satisfy both equations simultaneously.
First, let's write the equations of the planes in the standard form, which can be obtained by rearranging the terms:
Plane 1: x + y – z + 12 = 0
Plane 2: 2x + 2y - 2z + 8 = 0
Next, we can solve this system of equations using either the substitution method or the elimination method. I'll use the elimination method for this example.
To eliminate the variable x, we need to multiply the first equation by 2, which gives us:
2(x + y – z + 12) = 2(0)
2x + 2y - 2z + 24 = 0
Now, we can subtract the second equation from the modified first equation:
(2x + 2y - 2z + 24) - (2x + 2y - 2z + 8) = 0 - 0
2x + 2y - 2z - 2x - 2y + 2z + 24 - 8 = 0
0 = 16
This equation leads to 0 = 16, which is not true for any value of x, y, or z. Therefore, the planes do not intersect, and there is no common solution.
In summary, the planes with equations x + y – z + 12 = 0 and 2x + 2y - 2z + 8 = 0 do not intersect.