What is the role of 'd' in defining a plane's normal form?

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The discussion focuses on understanding the role of 'd' in the normal form of a plane equation, represented as nx + ny + nz + d = 0. The variable 'd' indicates the distance from the origin of the coordinate system to the plane along the normal vector. It is derived from the equation by rearranging the vector equation of the plane, which involves a point on the plane and its normal vector. Specifically, 'd' can be calculated as d = -(ax_o + by_o + cz_o), where (x_o, y_o, z_o) is a point on the plane. The conversation clarifies that 'd' represents the distance from the origin to the intersection point of a line normal to the plane.
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Hi guys,

I am sort of new here. So I am not pretty sure if I am to post this question in here.

I am a software programmer and I need to write a class for defining a plane. I came across the plane in its normal form nx+ny+nz+d=0

I need to feed in the values of the plane from another part of my program.

I can understand that nx,ny and nz are the normals of the plane. So where does the d come from. How exactly do you arrive at the value of d;

May sound very basic but then it would be nice if some one could help me out
 
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The equation should read nxx+nyy+nzz+d=0. d essentially defines how far the plane is from the origin of the coordinate system.
 
If I may rephrase my question:

If the orign of my co ordinate system is (0,0,0) then d is the distance between (0,0,0) and which point on the plane ? Or am I totally misunderstanding this?? Can you please explain
 
Ok, let us first try to come up with the vector equation of the plane, and then we will switch to cartesian coordinates, and you will probbably see how the d comes into play.

A plane is generally uniqely determined by a point call it P_o(x_o,y_o,z_o) and a vector normal on the plane n=<a,b,c>

Now, let P(x,y,z) be any other point in the plane, then its position vector would be:

r=<x,y,z>

while let

r_o=<x_o,y_o,z_o> be the position vector to the point P_o.

Now, if you draw a picture you will se that the following relation holds:

(r-r_o)*n=0

"*" holds for the dot product. Notice that (r-ro) and n are normal vectors.

Now, switching to the coordinate representation of the above vectors we get:

<x-x_o,y-y_o,z-z_o>*<a,b,c>=0=>a(x-x_o)+b(y-y_o)+c(z-z_o)=0

After rearranging the stuff in there we get:

ax+by+cz-(ax_o+by_o+cz_o)=0

So,

d=-(ax_o+by_o+cz_o)
 
oh..
many thanks for explaining stuff to me.. I got confused after looking at many websites none of which gave me what d is .

Thanks anyways
 
sundar0206 said:
If I may rephrase my question:

If the orign of my co ordinate system is (0,0,0) then d is the distance between (0,0,0) and which point on the plane ? Or am I totally misunderstanding this?? Can you please explain
Take a line normal to the plane starting at the origin. This line will hit the plane at a distance d from the origin. The hit point will have coordinates (-dnx,-dny,-dnz)
 
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