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

[chwala]

Homework Statement

Kindly see attached diagram.

Relevant Equations

3d graphs

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.

 

[Mark44]

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.

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.

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.

 

[chwala]

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.

 

[pbuk]

...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.

 

[chwala]

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! ...there is schlafil's Euclidean 4D and the other one formulated by Einstein. I will peruse through...cheers mate.

 

[chwala]

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;

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.

 

[Mark44]

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.

 

[pbuk]

Mostly correct, except that x = 0 and z = 2 are planes, not lines.

I think he meant the line (x = 0; z = 2).

 

[WWGD]

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?

 

[Mark44]

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.

 

[WWGD]

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.

 

[Mark44]

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...