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

In summary, the question is asking for all planes in R³ that are orthogonal to the vector (1,1,1). To solve this, we can use the point normal form of planes with the normal vector n = (1,1,1). This leads to the equation x + y + z = <1,1,1>⋅<x0,y0,z0>=0, where (x0,y0,z0) is a fixed point on the plane. We can also write any plane in R³ as Ax+ By+ Cz= D, where <A,B,C> is the normal vector to the plane. Therefore, the family of planes that are orthogonal to (1,1,1
  • #1
Benny
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0
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 [tex]x + y + z = \left( {1,1,1} \right) \bullet \mathop p\limits_\~[/tex]. 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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  • #2
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
 
  • #3
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!
 
  • #4
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.
 

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

What does it mean for a plane to be orthogonal to (1,1,1)?

A plane being orthogonal to (1,1,1) means that the plane is perpendicular to the vector (1,1,1) and any point on the plane will have a dot product of 0 with (1,1,1).

How do you determine all planes orthogonal to (1,1,1)?

To determine all planes orthogonal to (1,1,1), you can use the dot product formula: A∙B = |A||B|cosθ. Set A = (1,1,1) and B = (x,y,z), and solve for x, y, and z to get the equation of the plane. This will give you all possible planes orthogonal to (1,1,1).

Can there be more than one plane orthogonal to (1,1,1)?

Yes, there can be infinitely many planes orthogonal to (1,1,1). This is because for every point on the plane, there are infinite points that can form a perpendicular vector to (1,1,1).

What is the relationship between the normal vector of a plane and its orthogonality to (1,1,1)?

The normal vector of a plane is always orthogonal to the plane, so if the normal vector is (1,1,1), then the plane will be orthogonal to (1,1,1).

How can the concept of orthogonality to (1,1,1) be applied in real-world scenarios?

The concept of orthogonality to (1,1,1) is often used in physics and engineering, particularly in mechanics and structural design. It can also be applied in mathematical modeling and computer graphics, where orthogonal planes are used to create 3D objects and scenes.

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