Prove Orthogonal Vectors: x ⊥ u and v implies x ⊥ u - v

[Dustinsfl]

If x is ⊥ u and v, then x is ⊥ u - v.

I know this is true because u - v is in the same place as u and v; therefore, x is orthogonal. How can this be written better?

 

[VeeEight]

Use the property of the dot product that a dot (b+c) = a dot b + a dot c

 

[Mark44]

If x is ⊥ u and v, then x is ⊥ u - v.

I know this is true because u - v is in the same place as u and v; therefore, x is orthogonal. How can this be written better?

"is in the same plane as u and v..."

Dot product?

 

[Dustinsfl]

I understand the dot product needs to be used to do this (VeeEight and Mark) but I don't know how to implement it correctly.

 

[VeeEight]

If two vectors are orthogonal, what is their dot product?

 

[Dustinsfl]

I got it <x,v>=0, <x,u>=0

<x,u-v>=<x,u>-<x,v>=0-0=0

 

[Mark44]

If x ⊥ u, how can you say the same thing using the dot product. You don't have to "implement" anything - just use the dot product.

 

[Mark44]

I got it <x,v>=0, <x,u>=0

<x,u-v>=<x,u>-<x,v>=0-0=0

OK, then what does this say about x and u - v?

 

[VeeEight]

I got it <x,v>=0, <x,u>=0

<x,u-v>=<x,u>-<x,v>=0-0=0

 

[Dustinsfl]

Use the property of the dot product that a dot (b+c) = a dot b + a dot c

I was thinking more of proof then doing.

 

[VeeEight]

I was thinking more of proof then doing.

 

[Mark44]

I second that.

 

[Dustinsfl]

I was thinking of a formal proof then just doing <x..> ... =0

 

[Mark44]

I was thinking of a formal proof then just doing <x..> ... =0

That is a formal proof. Don't overthink this stuff at the expense of understanding what's going on.

 

[ABarrios]

Use the linearity of the dot product ( , ) so,

(x,u)=(u,x)=0 and (x,v)=(v,x)=0,
So consider,
(u-v,x)=(u,x)-(v,x)=0 => (x,u-v)=0.