Finding all vectors <x,z>=<y,z>=0

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Discussion Overview

The discussion revolves around finding all vectors \( z \) such that the inner products \( \langle x, z \rangle = \langle y, z \rangle = 0 \) hold true. Participants explore the implications of these conditions in the context of vector algebra and inner products, with a focus on the geometric interpretation of perpendicularity in Euclidean space.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a series of equations leading to the conclusion that \( x - y = [0, 2, 0] \) and seeks further assistance.
  • Another participant prompts for the components of \( z \) and notes that the "=0" part has not been utilized yet.
  • There is a discussion about assuming the inner product in Euclidean space is defined as \( \langle x, y \rangle = x_1 y_1 + x_2 y_2 + x_3 y_3 \).
  • A participant suggests a specific form for \( z \) as \( z = a(x \times y) \), leading to \( z = (2a, 0, -2a) \).
  • Another participant derives that \( z \) must satisfy \( z_2 = 0 \) and \( z \) can be expressed in the form \( (x, 0, -x) \).
  • One participant cautions against using \( x \) in a different context, as it is already defined as a vector in the problem statement.

Areas of Agreement / Disagreement

Participants express various approaches to the problem, with no consensus on a single solution or method. Different interpretations of the conditions and the form of \( z \) are presented, indicating multiple competing views.

Contextual Notes

Some participants note the importance of the inner product notation and its implications for the geometric interpretation of the vectors involved. There is also mention of the need to clarify the components of \( z \) and how they relate to the conditions set by \( x \) and \( y \).

lucasLima
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5iZ64h0.png


Hi Guys, that's what i got

<x,z>=<y,z>
<x,z>-<y,z>=0
<x,z>+<-y,z>=0
<x-y,z>=0

x-y = [0,2,0]

<2*[0,1,0],Z>=0
2<[0,1,0],z> = 0
<[0,1,0],z>=0

So 'im stuck at that. Any ideas?
 
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lucasLima said:
<[0,1,0],z>=0
If you write z with its components, what do you get as result?

Unrelated:
You didn't use the "=0" part yet.
 
mfb said:
If you write z with its components, what do you get as result?

Unrelated:
You didn't use the "=0" part yet.

<[0,1,0],[z1,z2,z3]>=0
Can i assume that the inner product is <x,y>=x1*y1+x2*y2+x3*y3 because it's a Euclidean Space?
 
That's it then! Thank you very much.
 
Well, it is the first part of the solution, yes.
 
How about z=a(xXy), where X is cross product and a is any real number?
This gives you z=(2a,0,-2a).
 
mfb said:
Well, it is the first part of the solution, yes.

0*z1+1*z2+0*z3=0
z2=0

<x,z>=0
z1+z3=0
z2=-z3

z is all vectors in the form (x,0,-x)
 
Correct (I wouldn't use x here as x is used as vector in the problem statement).
 
  • #10
lucasLima said:
5iZ64h0.png


Hi Guys, that's what i got

<x,z>=<y,z>
<x,z>-<y,z>=0
<x,z>+<-y,z>=0
<x-y,z>=0

x-y = [0,2,0]

<2*[0,1,0],Z>=0
2<[0,1,0],z> = 0
<[0,1,0],z>=0

So 'im stuck at that. Any ideas?
One thing I don't see mentioned in this thread is that the notation <x, z> represents the inner product of x and z, I believe. If z is an arbitrary vector with z = <z1, z2, z3>, then <x, z> = 0 means that x and z are perpendicular. Also, <x, z> = ##x_1z_1 + x_2z_2 + x_3z_3 = z_1 + z_2 + z_3##, and similarly for <y, z>.
 

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