Determine the intersection, if any, of the planes with equations x + y - z + 12 = 0 and 2x + 4y - 3z + 8 = 0.
I have no clue where to even begin. If someone could help and give me the steps to complete it, that would be greatly appreciated. Do I have to first simplify the equations? of find the normal vector for them? I am really stuck.
done, look back please
To find the intersection of two planes, you need to solve the system of equations formed by their equations. In this case, you have the following equations:
1) x + y - z + 12 = 0
2) 2x + 4y - 3z + 8 = 0
One way to solve this system is by eliminating one variable at a time using substitution or elimination method. Let's use the elimination method:
Step 1: Multiply equation 1 by 2 to make the coefficients of x in both equations equal:
2(x + y - z + 12) = 0
2x + 2y - 2z + 24 = 0
Step 2: Subtract equation 2 from the modified equation 1 to eliminate x:
(2x + 2y - 2z + 24) - (2x + 4y - 3z + 8) = 0
2y - z + 16 = 0
So, you now have a new equation: 2y - z + 16 = 0.
Step 3: Solve the new equation for one variable. Let's isolate z:
-z = -2y - 16
z = 2y + 16
Now you have z in terms of y.
Step 4: Substitute the expression for z in terms of y into either of the original equations. Let's use equation 1:
x + y - (2y + 16) + 12 = 0
x - y - 4 = 0
So, you now have x in terms of y as well.
The intersection of the planes is the set of points (x, y, z) that satisfy both equations. In other words, it is the set of values of x, y, and z that simultaneously make both equations true.
To summarize, you have obtained the following system of equations representing the intersection of the planes:
x - y - 4 = 0
z = 2y + 16
You can solve this system further if needed to find specific values of x, y, and z that satisfy both equations.