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.
why should it multiply by 3
Well, in order to find the intersection of two planes, we need to solve the system of equations formed by their equations. To start, let's simplify the equations and put them in the form Ax + By + Cz + D = 0.
For the first plane, x + y - z + 12 = 0, we can rewrite it as x + y - z = -12.
For the second plane, 2x + 4y - 3z + 8 = 0, we can divide the entire equation by 2 to simplify it, giving us x + 2y - (3/2)z + 4 = 0.
Now we can write the system of equations formed by these two planes:
x + y - z = -12
x + 2y - (3/2)z = -4
To solve this system, we can use substitution or elimination method. Let's use the elimination method to eliminate x.
To do that, let's multiply the first equation by -1 and the second equation by 1:
-1(x + y - z) = -1(-12) => -x - y + z = 12
1(x + 2y - (3/2)z) = 1(-4) => x + 2y - (3/2)z = -4
Now we add both equations together:
(-x - y + z) + (x + 2y - (3/2)z) = 12 + (-4)
-z + y - (3/2)z = 8
Combining like terms, we get:
-(5/2)z + y = 8
To solve for y, we isolate it:
y = (5/2)z + 8
Now let's substitute this value of y into one of the original equations, such as the first one:
x + (5/2)z + 8 - z = -12
x + (3/2)z = -20
Now you have a system with two variables, x and z. To simplify it further, I'll leave it to you to solve it and find the values of x, y, and z. Good luck!
Remember, if at any point you need a laugh or a clown-like distraction, just let me know!
To find the intersection of the two planes, you can start by solving the system of equations formed by the given plane equations. Here are the steps to solve it:
Step 1: Write down the system of equations:
x + y - z + 12 = 0 Equation 1
2x + 4y - 3z + 8 = 0 Equation 2
Step 2: Choose any variable to eliminate first and set up the elimination:
In this case, let's eliminate the variable x by multiplying Equation 1 by 2 and Equation 2 by -1:
2(x + y - z + 12) = 2(0)
-(2x + 4y - 3z + 8) = -(0)
Which simplifies to:
2x + 2y - 2z + 24 = 0 Equation 3
-2x - 4y + 3z - 8 = 0 Equation 4
Step 3: Add Equation 3 and Equation 4 to eliminate the x variable:
(2x + 2y - 2z + 24) + (-2x - 4y + 3z - 8) = 0
2y + z + 16 = 0
Step 4: Solve for one variable in terms of the other:
Rearrange the equation to solve for z:
z = -2y - 16 Equation 5
Step 5: Substitute Equation 5 into one of the original equations to solve for another variable. Let's choose Equation 1:
x + y - (-2y - 16) + 12 = 0
x + y + 2y + 16 + 12 = 0
x + 3y + 28 = 0
Step 6: Rearrange the equation to solve for x:
x = -3y - 28 Equation 6
Step 7: You now have expressions for x, y, and z in terms of one variable (y). This represents the intersection of the two planes. We can express it as a parametric equation for the line of intersection:
x = -3y - 28
y = y
z = -2y - 16
So, the equation of the line of intersection is:
(x, y, z) = (-3y - 28, y, -2y - 16)
This means that for any value of y, you can plug it into the above equations to get the corresponding values of x, y, and z for a point on the line of intersection.
To find the intersection of the two planes, you need to solve the system of equations formed by their equations. Here are the steps you can follow:
Step 1: Simplify the equations (if necessary):
The given equations are already in a simplified form, so we don't need to do anything in this step.
Step 2: Find the normal vectors of the planes:
The normal vector of a plane is the coefficients of x, y, and z in its equation.
For the first plane: x + y - z + 12 = 0, the normal vector is (1, 1, -1).
For the second plane: 2x + 4y - 3z + 8 = 0, the normal vector is (2, 4, -3).
Step 3: Set up a system of equations:
To find the common solution to both planes, we can equate the equations in terms of their normal vectors:
1(x) + 1(y) - 1(z) = -12 ...(Equation 1)
2(x) + 4(y) - 3(z) = -8 ...(Equation 2)
Step 4: Solve the system of equations:
Using any method you prefer (substitution, elimination, or matrices), solve the system of equations formed by Equation 1 and Equation 2.
I'll use the substitution method as an example:
From Equation 1, we have:
x + y - z = -12 ...(Equation 3)
Let's solve Equation 2 in terms of x:
2x + 4y - 3z = -8
2x = 4y - 3z - 8
x = (4y - 3z - 8)/2
x = 2y - (3/2)z - 4 ...(Equation 4)
Now we can substitute Equation 4 into Equation 3:
2y - (3/2)z - 4 + y - z = -12
Combine like terms:
3y - (5/2)z = -8 ...(Equation 5)
Now we have a system of two equations (Equation 5 and Equation 2) with two variables (y and z). We can solve this system to find the values of y and z.
Step 5: Solve the system of equations (Equation 5 and Equation 2):
Again, you can use your preferred method (substitution, elimination, or matrices). I'll use elimination to demonstrate:
Multiply Equation 5 by 2 to eliminate the fractional coefficient:
6y - 5z = -16 ...(Equation 6)
Now, subtract Equation 2 from Equation 6:
(6y - 5z) - (2x + 4y - 3z) = -16 - (-8)
6y - 5z - 2x - 4y + 3z = -16 + 8
2y - 2z - 2x = -8 ...(Equation 7)
Now we have another system of two equations:
2y - 2z - 2x = -8 ...(Equation 7)
2x + 4y - 3z = -8 ...(Equation 2)
Solve this system to find the values of x, y, and z. Once you find these values, you will have the point of intersection (if any) of the two planes.
First of all you must understand what the intersection is
When you intersect two planes you get a straight line, unless the two planes are parallel
In this case they are not, since 1:1:-1 ≠ 2:4:-3
multiply the first by 3
3x+3x-3z = -36
subtract
2x+ 4y - 3z = -8
---------------
x - y = -28
x = y -28
pick any y, let y = 20
x = -8
in #1
-8 + 20-z = -12
z = 24 , so we have a point (-8 , 20 , 24)
another point: let y = 0
x = -28
-28 + 0 -z = -12
z = -16 , another point is (-28 , 0 , -16)
so the direction of our line is [20 , 20 , 40] or as a simplified vector [1,1,2]
using our first point of (-8,20,24), our line is
(x+8) = (y-2) = (z-24)/2
or as parametrics:
x = -28 + t
y = 20 + t
z = 24 + 2t