Solving the nature of intersection between 3 planes

[sawdee]

Stuck on the last 3 systems on my worksheet, if someone could give me more than just the answer but also teach me how, it would be much appreciated!

"Determine the nature of the intersection if it exists) between the following sets of planes. If it is a line, find the equation of it. If it is a point, determine it.

a.) x + y - z + 3 = 0

-4x + y + 4z - 7 = 0

-2x + 3y + 2z - 2 = 0

b.) 2x - 3y + 4z - 1 = 0

x - y - z + 1 = 0

-x + 2y -z + 2 = 0

c.) 2x - y + 2z + 1 = 0

-4x + 2y -4z -2 = 0

6x - 3y + 6x + 1 = 0

 

[MarkFL]

Can you tell us what we require in a system of 3 planes for the intersection to be a point, and what we require for the intersection to be a point?

 

[sawdee]

Can you tell us what we require in a system of 3 planes for the intersection to be a point, and what we require for the intersection to be a point?

I've actually managed to solve for the first two systems, however I'm stumped on the third one. I'll do my best to write it out here:

2x-y+2x =-1
-4x+2y-4z =2
6x-3y+6z =-1

I managed to bring this down to:

2x-y+2z =-1
0x+0y+0z = 0
6x-3y+6z =-1

This would imply that the system has infinitely many solution, however I kept going

2x-y+2z=1
0x+0y+0z =0
0x-0y-0z = 2

This final step is equal to a number, so that confuses me as that means there are no solutions.

I know if I got a value equal to a point, I can substitute that back into two other equations. However this question gives me both 0=0 and 0=2, so I am beyond confused.

 

[MarkFL]

In problem c), we are given the 3X3 system:

$$2x-y+2z+1=0\tag{1}$$

$$-4x+2y-4z-2=0\tag{2}$$

$$6x-3y+6z+1=0\tag{3}$$

We should observe that (2) can be obtained by multiplying (1) by -2, and so we really have only two independent equations:

$$2x-y+2z+1=0$$

$$6x-3y+6z+1=0$$

Next, we should observe that these two planes are parallel, and distinct, and so there will be no intersection between them. :)

 

[sawdee]

In problem c), we are given the 3X3 system:

$$2x-y+2z+1=0\tag{1}$$

$$-4x+2y-4z-2=0\tag{2}$$

$$6x-3y+6z+1=0\tag{3}$$

We should observe that (2) can be obtained by multiplying (1) by -2, and so we really have only two independent equations:

$$2x-y+2z+1=0$$

$$6x-3y+6z+1=0$$

Next, we should observe that these two planes are parallel, and distinct, and so there will be no intersection between them. :)

Ok! so the no solution trumps over the infinitely many solutions?

 

[MarkFL]

Ok! so the no solution trumps over the infinitely many solutions?

If you look at the first two equations, you will get infinitely many solutions, because as we saw, they are the same equations, just in slightly different forms.

Then, if we look at the third equation and either of the first two, we will find no solutions because they are parallel, but distinct, and so they have no intersection. Since we are interested in where all 3 planes intersect, there is none, and so we say the system is inconsistent, or no solution.

 

[sawdee]

Awesome thank you so much!