# Thread: Difference: Point inside and on a plane

1. 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.

2. If the point P is "inside the plane" I would assume that a line drawn from the origin to P would get to P before it intersects the plane.

3. Hi,
Thanks. Your answer is satisfactory for me. However, I have 2 Questions now.
i) Can we use the phrase "below the plane" instead of "inside the plane". Does it mean same?? Book uses the phrase "inside the plane" and so web material which i have gone through.
Plz guide me.

ii) Suppose A=1, B=1, C=0 & D=0 & we have to find that the plane defined by this normal passes through (1,1,0)?
1(1) + 1(1) + 0(0) +0 =0
2= 0 which is false so the plane does not pass through this point.
Is it correct??

Plz guide me. Thanks for your time.

Zulfi.

4. Zulfi,

1) when you say "below" the plane would need also to define which direction is "down". So you would need to choose which of the coordinates represents "up". Do you see the issue? With "inside" you can just work on distance from the origin.

2) yes. If A and B =1, then x+y = 0, so the point (1,1,0) cannot lie in the plane.

5. Hi,
Thanks. Though its not clear little bit but you have satisfied that we should use the term inside compared to below b/c we would need direction for below.

Now i have a question. Is there a concept of infinite plane in a 3D domain. What would be its equation?? For instance for x-y plane can we say that its an infinite plane although the example in above post makes it clear that it is not an infinite plane for all values of x & y & z=0??

Sorry for taking so much of your time. Thanks for helping me.

Zulfi.

6. Zulfi, all planes are infinite. x, y and z are allowed to be negative too!

7. Originally Posted by zak100
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.
The geometry is 3 dimensional. The first equation describes points on the plane, while the second describes points in space on one side.

There is an ambiguity about "inside" versus "outside". Multiply A,B,C,D by -1 and the plane doesn't change, but the second equation < becomes > for a given half space. Therefore what was "inside" before becomes "outside".

8. Hi,
Thanks for your answer. If A=1, B=1 & C=0, normal vector is (1, 1, 0). This is perpendicular to z-axis. But if A=B=C=1 , the normal vector is (1, 1, 1) what can we say about this normal vector? Is it perpendicular to any axis??
Thanks for helping me.

Zulfi.

9. Have you tried drawing it?

10. Hi,
I want to draw but dont know how to draw it. If you have time plz guide me. I would be thankful to you.

Zulfi.

11. Forgetting about the plane for a moment. If x+y =0, x=-y. Can you draw the line that represents this?

12. x plus y.png

Hi,
I have attached an image of x+y=0 . For my question, i feel that normal vector will point to 4th dimension.

Zulfi.

13. Nice diagram. Can you draw in a z axis and turn the line into a plane?

14. Hi,
Thanks for your response and update on this. I have added the z-axis. I dont know how the line acts like a plane. Plz guide me.

Zulfi.x plus y with z-axis.png

15. Zulfi, yes it's impossible, you won't be able to draw a plane from x+y+z=0, so get rid of the x+y=0 line. But you can with x+y+z=D. The intercept on each axis is at D. Draw on these three points and join them up to get the plane.

16. Hi.
Thanks. I have found the following link which tells me about the intercept:

So for eq:
x + y + z = D
x/D + y/D + z/D = 1

So D is the intercept on x, y and z axis. By following your suggestion, I have drawn the plane. Plz guide me.

intercept on x y and z-axis.png

Zulfi.

17. Hi, Nicely done! Can you see that the normal to this plane, coming out in the (1,1,1) direction won't be perpendicular to either the x,y or z axes, but will be normal to any line drawn in the plane?

18. Hi,
Originally Posted by Jilan
Hi, Nicely done! Can you see that the normal to this plane, coming out in the (1,1,1) direction won't be perpendicular to either the x,y or z axes, but will be normal to any line drawn in the plane?
Sorry, I cant see it.

Zulfi.

19. You could make an equilateral triangle out of a sheet of paper, put on the floor in the corner of the room and trying sticking a pencil through the middle of it......

20. Hi,