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

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SUMMARY

The discussion focuses on determining all planes in R³ that are orthogonal to the vector (1,1,1). The key approach involves using the point-normal form of a plane, expressed as (1,1,1)·(x-p)=0, leading to the equation x + y + z = D, where D can be any scalar value. The normal vector to the plane is (1,1,1), and the family of planes can be represented by varying D. Understanding the relationship between the normal vector and the plane's equation is crucial for solving this problem.

PREREQUISITES
  • Understanding of vector operations, specifically dot products.
  • Familiarity with the point-normal form of a plane.
  • Knowledge of scalar equations of planes in three-dimensional space.
  • Basic concepts of orthogonality in vector geometry.
NEXT STEPS
  • Study the point-normal form of planes in three-dimensional geometry.
  • Learn about the properties of normal vectors and their significance in defining planes.
  • Explore the concept of families of planes and how varying parameters affect their equations.
  • Investigate applications of orthogonal planes in physics and engineering contexts.
USEFUL FOR

Mathematics students, geometry enthusiasts, and anyone studying vector calculus or three-dimensional geometry will benefit from this discussion.

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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