What Does the Equation f(x, y, z) = 3x - y + 2z Represent in 3D Space?

[brendan]

Homework Statement

Describe the level surface

Homework Equations

f(x,y,z) = 3x - y + 2z

The Attempt at a Solution

f(x,y,z) = 3x - y + 2z

3x - y + 2z = w


f(-1,1,2) for w <= 0 <= is a plane

regards
Brendan

 

[yyat]

What do you mean by

f(-1,1,2) for w <= 0 <= is a plane

?

The solution may be easier then you think.

 

[brendan]

The only example I've been given is:

Describe the level surface of
w = x^2+(y-1)^2+(z-2)^2

we use
x^2+(y-1)^2+(z-2)^2= k where K is a constant

Which describes a surface which is a sphere of radius sqrt(k) and center (0,1,2)

 

[lanedance]

so as as you said a level surface is given by
F(x,y,z) = w for a constant w

so for a given k, what is the surface described by
f(x,y,z) = 3x - y + 2z = w

if you're unsure where to start, pick a w, say w = 0 and figure out what surface it is. Then look at what happens for different w. I find it also helps to look at the coordinate axis, say so what happens to z for different w values with x=y=0

as for your post

f(-1,1,2) for w <= 0 <= is a plane

has no real meaning, f(-1,1,2) represents a scalar value and I'm not sure what you mean by your comaprison sybols for w

however as a hint

is a plane

is close to the money

 

[brendan]

I know that the w is a constant so if w = 0

3x - y + 2z = w

This represents all the values of x y z which make the function = 0

thats where I got the f(-1,1,2) from because the function = 0 at those values.
But I know there are many vaules that would make the function = 0.

I know that there are no squares so its probabably not an ellipsoid or sphere.
I'm having a lot of trouble knowing how to graph it.

The example I've been given has a center of (0,1,2). I know that those figures make the function = 0 but why did the figures get chosen?

Brendan

 

[Mark44]

For each value of w, you can graph 3x - y + 2z = w.

For w = 0, the equation is
3x - y + 2z = 0
What does this look like. With one equation in three unknowns, you can pick any two and find the third.

Pick about 10 points and graph this surface.

If w = 5, the equation is
3x - y + 2z = 5
Do the same as above.

If w = -10, the equation is
3x - y + 2z = -10
Do the same as above.

Repeat until it clicks what the level surfaces look like.

 

[brendan]

Hi ,
I've been graphing the surface 3x - y + 2z = w and it looks like flat surfaces with center
x=0, y= 0, z = w/2 There doesn't seem to be any resitrictions on the value that z can take.

Now if w = 0 , 5 or -10 the su`rface is just centered on 0 , 2.5 or -5

To describe the function is it enough to say that it is a surface with no curvature centered on x=0, y= 0, z = w/2 ?

 

[Mark44]

These flat surfaces you finding are called "planes." They don't have centers. You can describe a plane completely by specifying three points that are on the plane, or by specifying a normal to the plane and one point on it.

To describe the function f(x, y, z), it's probably sufficient to say that the level surfaces are all planes.

 

[brendan]

Thanks about the method of graphing the function it helped a lot!

Would you say that when

"describing the function f(x, y, z), it's probably sufficient to say that the level surfaces are all planes."

would you mention the magnitude of the plan eg

The level surfaces of the function f(x, y, z) = w are all planes at w/2

regards
Brendan

 

[Mark44]

It doesn't make any sense to talk about the magnitude of a plane. And I don't know what "The level surfaces of the function f(x, y, z) = w are all planes at w/2" means.

 

[lanedance]

Thanks about the method of graphing the function it helped a lot!

Would you say that when

"describing the function f(x, y, z), it's probably sufficient to say that the level surfaces are all planes."

would you mention the magnitude of the plan eg

The level surfaces of the function f(x, y, z) = w are all planes at w/2

regards
Brendan

ummmm, I'm not sure what your last sentance means...

I would mention all the level surface represent parallel planes and try and specify the unique plane defined by a w value

As Mark said a plane can be described by either
- 3 points, or
- a normal and one point

so you could find the 3 points for each plane in terms of w (the axis corssings spring to mind)

or I think its easier to notice all the planes have the same noraml vector (can you find it?) and give a single axis crossing in terms of w

 

[brendan]

So the level surface represent parallel planes.

For f(x,y,z) 3x - y + 2z = 0

We give three points

P1(0,3,1.5)
P2(0,1,1/2)
P3(0,2,1)

All are point of a unique plane.

With Normal vector

P1->P2(0,-2,-1)
P1->P3(0,-1,-1/2)

P1->P2 X P1->P3

normal vector (0,0,0)

 

[Mark44]

So the level surface represent parallel planes.

Each level surface is a plane.

For f(x,y,z) 3x - y + 2z = 0

For f(x, y, z) = 3x -y + 2z, the level surface f(x, y, z) = 0 is a plane that contains the three points below.

We give three points

P1(0,3,1.5)
P2(0,1,1/2)
P3(0,2,1)

All are point of a unique plane.

With Normal vector

P1->P2(0,-2,-1)
P1->P3(0,-1,-1/2)

P1->P2 X P1->P3

normal vector (0,0,0)

Your vectors are calculated correctly, but you have unfortunately picked points that are on a line, so your cross product is zero. The normal vector to this level surface is not (0, 0, 0).

 

[brendan]

Okay I've picked three new points (I'm assuming they can be any points on the plane)

For f(x,y,z) 3x - y + 2z = 0

P1(1,2,-1/2)
P2(0,1,1/2)
P3(3,1,-4)



P1->P2(-1,-1,1)
P1->P3(2,-1,-3.5)

P1->P2 X P1->P3

normal vector (4.5,-1.5,3)

Is this right ?

Or am I using the wrong furmula for this problem ?

 

[Mark44]

This is correct. The most obvious normal to this plane is (3, -1, 2), which is a scalar multiple of the one you found (and vice versa). It is no coincidence that (3, -1, 2) is a normal to the family of planes 3x -y + 2z = k, where k is any real number.

 

[brendan]

Thanks a lot for your help I really appreciate it.

 

[HallsofIvy]

So the level surface represent parallel planes.

For f(x,y,z) 3x - y + 2z = 0

We give three points

P1(0,3,1.5)
P2(0,1,1/2)
P3(0,2,1)

All are point of a unique plane.

With Normal vector

P1->P2(0,-2,-1)
P1->P3(0,-1,-1/2)

P1->P2 X P1->P3

normal vector (0,0,0)

I'm not sure why you are focusing so much on w= 0. It should be immediately clear, from what you have already learned about planes, that f(x,y,z)= 3x - y + 2z = Constant is a plane with normal vector (3, -1, 2) for any Constant.