Let U=... V=... Find ||...||

**Jishent** · December 2nd 2010, 11:25 PM

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

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

Find:

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

Starters?

Thanks in advance

**Ackbeet** · December 3rd 2010, 02:47 AM

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?

**HallsofIvy** · December 3rd 2010, 05:02 AM

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

**Jishent** · December 3rd 2010, 12:50 PM

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

Thanks

**Ackbeet** · December 3rd 2010, 12:53 PM

You would agree that

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

So

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

**Plato** · December 3rd 2010, 12:56 PM

If $v$ is a nonzero vector then $\left\| {\dfrac{v}<br /> {{\left\| v \right\|}}} \right\| = 1$
That is all there is to it.

**Jishent** · December 3rd 2010, 12:57 PM

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

**Ackbeet** · December 3rd 2010, 01:00 PM

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.

**Plato** · December 3rd 2010, 01:04 PM

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

Therefore $\left\| {\dfrac{v}{{\left\| v \right\|}}} \right\|=\left| {\frac{1}{{\left\| v \right\|}}} \right|\left\| v \right\| = \frac{1}<br /> {{\left\| v \right\|}}\left\| v \right\| = 1$

**Jishent** · December 3rd 2010, 01:04 PM

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

**Ackbeet** · December 3rd 2010, 01:06 PM

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

Be careful with your parentheses!

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