Hi,
I have found that for a point (x, y, z) and with parameters A, B, C, D, if:
Ax + By + Cz + D= 0
it is on a plane but if
Ax + By + Cz + D < 0,
it is inside a plane.

Kindly guide me what is difference between a point inside and on a plane??


Zulfi.

 

You question makes no sense. A plane in 3-space extends to infinity in all directions. Considering a plane as a set of points, "on" means the same thing as "belongs to the set of points" that make up the plane.

The notion of inside and outside, imagines a closed surface, consisting of another infinite set of points, as a boundary where the terms inside and outside have a precise mathematical description.

Consider the equation of a sphere or radius r with a center at the origin (0,0,0). A point P is on the sphere if the distance from P to the origin is exactly r. It is inside the boundary of the sphere if the distance from P to the origin is less than r. It is outside the boundary of the sphere is the distance from P to the origin is greater than r.

You might be able to establish which side of the plane a point is on. Is that what you're trying to get at?

 

Either a point is (x,y,z), which are just 3 numbers and x,y and z are the coodinates of the point.

or you are referring to a function where x, y , z are variables.

So which is it?

 

If I recall correctly, there are 3 possible locations for the point
Ax + By + Cz + D > 0 point is above the plane
Ax + By + Cz + D = 0 point is on the plane
Ax + By + Cz + D < 0 point is below the plane.

 

If I recall correctly, there are 3 possible locations for the point
Ax + By + Cz + D > 0 point is above the plane
Ax + By + Cz + D = 0 point is on the plane
Ax + By + Cz + D < 0 point is below the plane.

can't see how this works.

Firstly what do you mean by above or below?

Now your formulae should work for any specific value so consider the point

x=1; y=1; z=0

This lies on the xy plane.

Now set A = B = 1 ; C = 0 and D=-2 and substitue for the coordinates of the point
in the expression Ax+By+Cz+D
(1*1) + (1*1) + (0*0) + (-2) = 0

But this is a plane displace by -2 from the xy plane on which the point lies so the point does not lie on the plane

x+y-2=0, since it lies on the plane x+y=0

Yet by your formula it should lie on that plane.

Oh and welcome to AllAboutCircuits

 

can't see how this works.

Firstly what do you mean by above or below?

Now your formulae should work for any specific value so consider the point

x=1; y=1; z=0

This lies on the xy plane.

Now set A = B = 1 ; C = 0 and D=-2 and substitue for the coordinates of the point
in the expression Ax+By+Cz+D
(1*1) + (1*1) + (0*0) + (-2) = 0

But this is a plane displace by -2 from the xy plane on which the point lies so the point does not lie on the plane

x+y-2=0, since it lies on the plane x+y=0

Yet by your formula it should lie on that plane.

Oh and welcome to AllAboutCircuits

The falacy in your argument is that you don't get to choose things independently. A plane divides 3-dimensional space into two half spaces. One of the half-spaces is above the plane and one of the half-spaces is below the plane. Choosing the xy plane fixes A,B,C, and D. You can't just assign it an arbitrary value of -2.

 

But I didn't choose the xy plane.

There are three coordinate axes so the point (x,y,z) requires 3 numbers not 4.

I specified them.

And that point is most definitely on the xy plane. I could have chosen differently, the end result wouldhave been the same.

The proposition was whether the point xyz lies on a particular plane or not and the condition given required the calculation of the three coordinates along with 4 coefficients (of the plane) to equal zero.

This test should work for any point and any plane.

My apologies, I rushed my reply to get it in whilst you guys were online. So it was very scruffy.

Consider the point (1,1,0) and the planes defined by A=1, B=1, C=0, D=0 and A=1, B=1, C=0, D=-2

Apply the test

Plane1):- (1*1) + (1*1) + (0*0) + (0) which is not equal to zero

Pane2):- (1*1) + (1*1) + (0*0) + (-2) = 0


So on which plane does the point (1,1,0) lie?

Well it lies on the intersection of both planes.

I was wrong initially, they are not parallel.

Plane1 is indeed the xy plane. This is otherwise known as the plane z=0,
edit : which means that it is not the plane A=1, B=1, C=0, D=0 The point does not lie on this plane and the test is accurate.

Plane 2 is the plane x+y = -2. This plane is parallel to the z axis and perpendicular to the plane z=0

 

Hi,
Thanks for your replies. Book uses the terms "inside" , "outside" and "on" the plane instead of above and below the plane.

Plane1):- (1*1) + (1*1) + (0*0) + (0)

This tests says that the point is outside the plane. But if the value of the test is less than zero then it means that the point is "inside the plane" which is not same as "on the plane".

I like this method of testing points but as Papabravo said that the normal vector is fixed. I have a feeling he is right.

One of the half-spaces is above the plane and one of the half-spaces is below the plane.

What about points on the plane?? Which half is used??

Kindly guide me about the difference between the terms inside and on the plane.

Zulfi.

 

The three coefficients of X, Y, and Z define a normal direction to the plane. A point could be considered inside or outside of the plane depending on whether or not it is in the direction of that normal, or in the direction of its opposite.

 

Hi,
Thanks for your reply and welcome to this great helping forum.

A point could be considered inside or outside of the plane depending on whether or not it is in the direction of that normal, or in the direction of its opposite.

How to find the direction of the point?
book says that :
Ax + By + Cz + D= 0 (Point on the plane)
Ax + By + Cz + D< 0 (Point inside the plane)
Ax + By + Cz + D> 0 (Point outside the plane)

You can find this in the slides also for which i have provided the link or by using google search. Plz provide reference of what you are saying.

Zulfi.