Describe the set of points that satisfy ##(x-4)(z-2)=0##

In summary, in 3D, if x-4 and z-2 are both zero, then there exists a solution on the line x=4; z=2. If x-4 is not zero, then there exists a solution on the line x=-4; z=2. If z-2 is not zero, then there exists a solution on the line z=2; x=4.
  • #1
chwala
Gold Member
2,730
384
Homework Statement
Kindly see attached diagram.
Relevant Equations
3d graphs
1689304792445.png


1689304837966.png


We are told that ##(x-4)(z-2)=0## The ##0## refers to what variable? The ##z=2## i think the sketch on diagram may be misleading (not to scale). Why have the ##x## crossing the axis on negative side and same applies to ##z##...
To put this into context, we may have ##(x,y,z)=(4,y,z)## where ##y## and ##z## may take any values and ##(x,y,z)=(x,y,2)## where ##x## and ##y## may take any values.
 
Last edited:
Physics news on Phys.org
  • #2
chwala said:
Homework Statement: Kindly see attached diagram.
Relevant Equations: 3d graphs

View attachment 329227

View attachment 329228

We are told that ##(x-4)(z-2)=0## The ##0## refers to what variable?
Zero is just the value of the expression (x - 4)(z - 2) for specific values of x and z. It doesn't refer to any variable at all.
chwala said:
The ##z=2## i think the sketch on diagram may be misleading (not to scale). Why have the ##x## crossing the axis on negative side
That's not the negative side of the x-axis. The positive side of the x and y axes extends out from the origin. The marks on these axes show positive numbers on the positive sides of these axes. This is the standard way of doing things for 3D graphs.
chwala said:
and same applies to ##z##...
To put this into context, we may have ##(x,y,z)=(4,y,z)## where ##y## and ##z## may take any values and ##(x,y,z)=(x,y,2)## where ##x## and ##y## may take any values.
And these two sets of points define the two planes shown in the graph. I agree that the plane x = 2 doesn't appear to be in the right place, according to the tick marks at 6 and 8. The graph should intersect the x-axis a lot closer to the 6 tick mark. Since there are no tick marks on the z-axis, it's impossible to say whether the plane z = 2 is located correctly.
 
  • Like
Likes Math100 and chwala
  • #3
Mark44 said:
Zero is just the value of the expression (x - 4)(z - 2) for specific values of x and z. It doesn't refer to any variable at all.
That's not the negative side of the x-axis. The positive side of the x and y axes extends out from the origin. The marks on these axes show positive numbers on the positive sides of these axes. This is the standard way of doing things for 3D graphs.And these two sets of points define the two planes shown in the graph. I agree that the plane x = 2 doesn't appear to be in the right place, according to the tick marks at 6 and 8. The graph should intersect the x-axis a lot closer to the 6 tick mark. Since there are no tick marks on the z-axis, it's impossible to say whether the plane z = 2 is located correctly.
...am reading this now...looks like it does not work the same way as 2D- straight lines.
 
  • #4
chwala said:
...am reading this now...looks like it does not work the same way as 2D- straight lines.
It works in a perfectly analogous way in any number of dimensions greater than 1.

In 2D we have ## (x-4)(y-2) = 0 ## which implies that either ## x-4 = 0 ## or ## y-2 = 0 ##. ## x=4 ## and ## y=2 ## define lines in 2D, and so all solutions lie on at least one of these lines. Exactly one solution lies on both of these lines: this is the solution at the point ## x = 4; y = 2 ##.

In 3D we have ## (x-4)(z-2) = 0 ## which implies that either ## x-4 = 0 ## or ## z-2 = 0 ##. ## x=4 ## and ## z=2 ## each define planes in 3D, and so all solutions lie in at least one of these planes. An uncountable inifinity of solutions lie in both of these planes: these are the solutions on the line ## x = 4; z = 2 ##.

See if you can guess what happens in 4D.
 
Last edited:
  • Like
Likes Math100, chwala and Mark44
  • #5
pbuk said:
It works in a perfectly analogous way in any number of dimensions greater than 1.

In 2D we have ## (x-4)(y-2) = 0 ## which implies that either ## x-4 = 0 ## or ## y-2 = 0 ##. ## x-4 ## and ## y-2 ## define lines in 2D, and so all solutions lie on at least one of these lines. Exactly one solution lies on both of these lines: this is the solution at the point ## x = 4; y = 2 ##.

In 3D we have ## (x-4)(z-2) = 0 ## which implies that either ## x-4 = 0 ## or ## z-2 = 0 ##. ## x-4 ## and ## z-2 ## define planes in 3D, and so all solutions lie in at least one of these planes. An uncountable inifinity of solutions lie in both of these planes: these are the solutions on the line ## x = 4; z = 2 ##.

See if you can guess what happens in 4D.
...4D would be a whole new world altogether! :smile:...there is schlafil's Euclidean 4D and the other one formulated by Einstein. I will peruse through...cheers mate.
 
Last edited:
  • #6
This is just but a continuation of the problem; same chapter that is,

i hope i can still post it here rather than start a new thread;

1689348978387.png


My attempt;
##x## and ##z## co ordinates form a circle in the ##xz## plane with radius= 4,centred at ##(0,2)##. There is no restriction on ##y## thus the result is a circular cylinder of ##r=4##, centred on the line ##x=0## and ##z=2##. The cylinder extends indefinetely on the ##y## axis.
 
  • Like
Likes pbuk
  • #7
chwala said:
This is just but a continuation of the problem; same chapter that is,

i hope i can still post it here rather than start a new thread;

View attachment 329252

My attempt;
##x## and ##z## co ordinates form a circle in the ##xz## plane with radius= 4,centred at ##(0,2)##. There is no restriction on ##y## thus the result is a circular cylinder of ##r=4##, centred on the line ##x=0## and ##z=2##. The cylinder extends indefinetely on the ##y## axis.
Mostly correct, except that x = 0 and z = 2 are planes, not lines. The central axis of the cylinder goes through the point (0, 0, 2) and lies on the line through this point that is perpendicular to the xz plane.
 
  • #8
Mark44 said:
Mostly correct, except that x = 0 and z = 2 are planes, not lines.
I think he meant the line (x = 0; z = 2).
 
  • Like
Likes chwala
  • #9
pbuk said:
I think he meant the line (x = 0; z = 2).
Isn't the set x=0; z=2 , an infinite rectangular strip, varying along y?
 
  • #10
WWGD said:
Isn't the set x=0; z=2 , an infinite rectangular strip, varying along y?
How do you conclude that these equations represent an infinitely long rectangular strip? Each equation represents a plane. The intersection of these planes is a line through the point (0, 0, 2) that is perpendicular to the xz plane.
 
  • #11
Mark44 said:
How do you conclude that these equations represent an infinitely long rectangular strip? Each equation represents a plane. The intersection of these planes is a line through the point (0, 0, 2) that is perpendicular to the xz plane.
I thought it was the region bounded by these, i.e., the region ## \{ (0,y,2); -\infty < y < \infty \} ##. But maybe I misread something.
 
  • #12
WWGD said:
I thought it was the region bounded by these, i.e., the region ## \{ (0,y,2); -\infty < y < \infty \} ##. But maybe I misread something.
Yes, that's the region, but it's a line. You could think of it as a rectangular region, but one that is very thin...:oldbiggrin:
 
  • Haha
Likes pbuk

FAQ: Describe the set of points that satisfy ##(x-4)(z-2)=0##

What does the equation ##(x-4)(z-2)=0## represent geometrically?

The equation ##(x-4)(z-2)=0## represents the union of two planes in three-dimensional space. Specifically, it describes the set of points where either ##x=4## or ##z=2##. This means the solution set includes all points on the plane where ##x=4## and all points on the plane where ##z=2##.

How can we visualize the set of points that satisfy ##(x-4)(z-2)=0##?

To visualize this, imagine a three-dimensional Cartesian coordinate system. The plane ##x=4## is a vertical plane parallel to the yz-plane, intersecting the x-axis at x=4. The plane ##z=2## is a horizontal plane parallel to the xy-plane, intersecting the z-axis at z=2. The set of points satisfying the equation is the combination of these two planes.

Are there any points that belong to both planes described by the equation ##(x-4)(z-2)=0##?

Yes, the points that belong to both planes are those that satisfy both ##x=4## and ##z=2## simultaneously. Therefore, the line of intersection of these two planes is given by the set of points (4, y, 2) for any real number y. This line is parallel to the y-axis and passes through the point (4, 0, 2).

How do we write the solution set of the equation ##(x-4)(z-2)=0## in set notation?

The solution set can be written as the union of two sets: the set of points where ##x=4## and the set of points where ##z=2##. In set notation, this is expressed as:\[ \{ (x, y, z) \in \mathbb{R}^3 \mid x=4 \} \cup \{ (x, y, z) \in \mathbb{R}^3 \mid z=2 \}. \]

What is the significance of the individual factors in the equation ##(x-4)(z-2)=0##?

The individual factors in the equation, ##(x-4)## and ##(z-2)##, represent conditions that must be satisfied for the equation to hold true. The factor ##(x-4)## equals zero when ##x=4##, defining the vertical plane. Similarly, the factor ##(z-2)## equals zero when ##z=2##, defining the horizontal plane. The equation as a whole describes the union of these two geometric conditions

Back
Top