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

In summary: So when you have <x, z> = <y, z>, you can equate the components and get a solution.In summary, the conversation discusses the inner product and its applications in solving equations in Euclidean space. The participants mention using the inner product to find perpendicular vectors and solving equations by equating the components. They also briefly touch on the use of cross product and a real number in finding a solution.
  • #1
lucasLima
17
0
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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  • #2
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.
 
  • #3
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?
 
  • #5
That's it then! Thank you very much.
 
  • #6
Well, it is the first part of the solution, yes.
 
  • #7
How about z=a(xXy), where X is cross product and a is any real number?
This gives you z=(2a,0,-2a).
 
  • #8
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)
 
  • #9
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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