Determine all planes orthogonal to (1,1,1)

AI Thread Summary
To determine all planes orthogonal to the vector (1,1,1) in R³, the point normal form of a plane can be utilized, leading to the equation (1,1,1)·(x-p)=0. This simplifies to the scalar equation x + y + z = D, where D can be any constant, indicating a family of planes. The normal vector (1,1,1) is perpendicular to all vectors in the plane, confirming the relationship between the normal vector and the plane's equation. The discussion emphasizes that the value of D can vary, allowing for multiple planes orthogonal to the given vector. Understanding this concept is crucial for solving similar problems in linear algebra.
Benny
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Hello, could someone please give me some help with the following question?

Q. Determine all planes (in R³) orthogonal to the vector (1,1,1).

This is how I started off but I am not really sure how I need to go about solving this problem. I begin(by somewhat assuming that the vector (1,1,1) is perpendicular to the relevant planes) by writing the point normal form of planes with the n = (1,1,1) so I get (1,1,1).(x-p)=0.

With x = (x,y,z) I get down to x + y + z = \left( {1,1,1} \right) \bullet \mathop p\limits_\~. With other questions I am given the point P so the dot product of the vectors 'n' and 'p' can be found. With this one the situation is different because I need to find all planes which are orthogonal to the vector (1,1,1). I thought about letting the vector p = (f,g,h) but that doesn't seem right. Could someone help me out with this one? Any help is appreciated.

Edit: My program for using Tex seems a little screwy at the moment so I had to fix part sof my post. X and p are supposed to denote vectors.
 
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If (x,y,z) is a general point on the plane and (x0,y0,z0) is a fixed point on the plane, then (x-x0)i+ (y-0j)+ (z-z0)k is a vector in the plane. If <1,1,1> is perpendicular to the plane, it is perpendicular to all vectors in the plane so <1,1,1>.<x-x0,y-y0,z-z0>= 0.

In fact, by the time you are expected to do a problem like this, you should already have learned that any plane can be written as Ax+ By+ Cz= D where <A, B, C> is a vector perpendicular to the plane
 
WHen a plane is orthogonal to a vector, what is that vector called??

THat gives you something about the family of planes and since we know that the scalar equation of a plane is Ax+By+Cz=D where (A,B,C) is the NORMAL vector to the plane and it doesn'd D is the intercept, but since in this case we want a family D can be anything!
 
Thanks for clarifying the point about D. I usually find the value of D by calculating n.p but since a family of planes is needed I can just denote p by by (f,g,h) or something similar.
 
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