?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 wf(-1,1,2) for w <= 0 <= is a plane
is close to the moneyis a plane
ummmm, I'm not sure what your last sentance means...Thanks about the method of graphing the function it helped alot!
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
Each level surface is a plane.So the level surface represent parallel planes.
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.For f(x,y,z) 3x - y + 2z = 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).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.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)