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

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1. Sep 13, 2015

### lucasLima

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?

2. Sep 13, 2015

### Staff: Mentor

If you write z with its components, what do you get as result?

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

3. Sep 13, 2015

### lucasLima

<[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?

4. Sep 13, 2015

### Staff: Mentor

Sure.

5. Sep 13, 2015

### lucasLima

That's it then! Thank you very much.

6. Sep 13, 2015

### Staff: Mentor

Well, it is the first part of the solution, yes.

7. Sep 13, 2015

### mathman

How about z=a(xXy), where X is cross product and a is any real number?
This gives you z=(2a,0,-2a).

8. Sep 13, 2015

### lucasLima

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. Sep 13, 2015

### Staff: Mentor

Correct (I wouldn't use x here as x is used as vector in the problem statement).

10. Sep 13, 2015

### Staff: Mentor

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>.