Proving Isometry Preserves Distance in R^3 with f(0)=0

[bookworm_07]

Ok I know that isometries preserve distance and in order for a fn to be an isometry || f(u) - f(v) || = || u - v ||
and in this question it asks to prove

prove that if an isometry satisfies f(0) = 0 then we have
f(u) x f(v) = +- f(u x v)
and what property of f determines the choice of sign

"x" is the cross product

Now i know that this space must be R^3 because its the cross product
and i know that f(0) = 0 because
|| f(v) - f(u) || = || f(0) - f(0) || = 0

I just don't know how to connect this knowledge to the cross product.
A push in the right direction would be awsome! I just need a start.
thank you very much

 

[AKG]

What do you mean that you "know f(0) = 0 because..."? You know f(0) = 0 because they tell you so, you didn't (can't) deduce that fact, since translation is an isometry for which that equation doesn't hold. Also, why in the world would you have:

||f(v) - f(u)|| = ||f(0) - f(0)||?

That appears to come out of nowhere. Moreover, it appears to have nothing to do with f(0) = 0, although you claim to use it as your reason for justifying it. Here's what you know:

||f(u) - f(v)|| = ||u - v|| because f is an isometry
f(0) = 0 because they tell you so
"|| f(v) - f(u) || = || f(0) - f(0) ||" is false in general

 

[bookworm_07]

oh ok, i was just trying anything really.
my biggest problem is using ||f(u) - f(v)|| = ||u - v||
to show that f(u) x f(v) = +- f(u x v)
i just don't know how to relate the two,
I don't want you to give me the answer AKG, i would much rather understand what i am suppose to do then get something free.