# Orthogonal unit vectors also unit vectors?

If two vectors v, w are both unit vectors, then v+w and v-w will be orthogonal, but are v+w and v-w also unit vectors?

I would say no because the inner product of the two added, and the two subtracted would also have to be orthogonal.

<(v+w)+(v-w),(v+w)-(v-w)>

= <2v,2w>

and <2v,2w> must be greater than zero for the product to be defined in the first place.

*EDIT:
If it helps, the way I originally showed that they were orthogonal was to take
<v+w,v-w> = ||v||^2 - ||w||^2

if they are unit vectors then ||v|| = 1 and ||w|| = 1 so

<v+w,v-w> = 1 - 1 = 0 = orthogonal

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dextercioby
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What is $\left|| u-v\right|| ^2$ equal to ?

HallsofIvy
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If two vectors v, w are both unit vectors, then v+w and v-w will be orthogonal, but are v+w and v-w also unit vectors?

I would say no because the inner product of the two added, and the two subtracted would also have to be orthogonal.

<(v+w)+(v-w),(v+w)-(v-w)>

= <2v,2w>

and <2v,2w> must be greater than zero for the product to be defined in the first place.
??? Are you saying that an inner product can't be negative? In any case, I see no reason for looking at <(v+w)+ (v-w),(v+w)-(v-w)> . The question was asking about v+w and v-w separately. You should be looking at
||v+w||2= <v+w,v+w> and ||v-w||2= <v-w,v-w>.

Even more simply, what if v= w?

*EDIT:
If it helps, the way I originally showed that they were orthogonal was to take
<v+w,v-w> = ||v||^2 - ||w||^2

if they are unit vectors then ||v|| = 1 and ||w|| = 1 so

<v+w,v-w> = 1 - 1 = 0 = orthogonal[/QUOTE]

??? Are you saying that an inner product can't be negative? In any case, I see no reason for looking at <(v+w)+ (v-w),(v+w)-(v-w)> . The question was asking about v+w and v-w separately. You should be looking at
||v+w||2= <v+w,v+w> and ||v-w||2= <v-w,v-w>.

Even more simply, what if v= w?

Yes because if v = w, then <2v,2v> must be positive because 4<v,v> must obey positivity.

edit: whoops, I forgot the change the w at the end of my first post to v. This will work now, correct?

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I suppose that in the end dex's way would be the easiest and most straightforward (since it shows the norm is zero).

Dick