# Thread: Let U=... V=... Find ||...||

1. ## Let U=... V=... Find ||...||

Let
U=
|-1|
|2 |
|1 |

V=
|3 |
|1 |
|-1|

Find:

|| [1/(||U-V||)](u-v)||

Starters?

2. You can compute that expression without doing any numerical computations, if you use some of the properties of norms, such as || av || = |a| ||v||. Does this give you some ideas?

3. What can you say about the length of $\displaystyle \frac{\vec{v}}{||\vec{v}||}$?

4. ## Need more help

I don't get it...please explain more

Thanks

5. You would agree that

$\displaystyle \dfrac{1}{\|u-v\|}$ is a scalar, right? And it's positive?

So

$\displaystyle \left\|\dfrac{1}{\|u-v\|}(u-v)\right\|=\left\|\dfrac{1}{\|u-v\|}\right\|\,\|u-v\|=\dots?$

6. If $\displaystyle v$ is a nonzero vector then $\displaystyle \left\| {\dfrac{v} {{\left\| v \right\|}}} \right\| = 1$
That is all there is to it.

7. So does it end up being = 1? because you times the thing up.?

8. Well, I don't know what you mean by "times the thing up". You have a/a, where a is not zero. That's always 1.

9. Why don't you know the basic properties of the norm function?
If $\displaystyle \alpha$ is a scalar and $\displaystyle v$ is a vector then $\displaystyle \left\| {\alpha v} \right\| = \left| \alpha \right|\left\| v \right\|$.

Therefore $\displaystyle \left\| {\dfrac{v}{{\left\| v \right\|}}} \right\|=\left| {\frac{1}{{\left\| v \right\|}}} \right|\left\| v \right\| = \frac{1} {{\left\| v \right\|}}\left\| v \right\| = 1$

10. Because I am a high school student trying to learn University alg by using a text book that doesn't teach anything.

11. Technically, it's $\displaystyle \|u-v\|/\|u-v\|=1.$