Simple question about vector.

**yungman** · September 17th 2010, 03:58 PM

Let $\vec{A}$ be a vector with length $|\vec{ A}|$

$\hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|}$

What is $\frac{\vec{A}}{|\vec{ A}|}$ If $|\vec{A}|$ = 0?

What is $| \frac{\vec{A}}{|\vec{ A}|}|$ If $|\vec{A}|$ = 0?

**Plato** · September 17th 2010, 04:51 PM

I think what this post means is the following.
$\hat A = \left\{ {\begin{array}{rl}<br /> {\frac{1}{{\left\| A \right\|}}A,} & {\left\| A \right\| \ne 0} \\<br /> {0,} & {else} \\ \end{array} } \right.$

Is this correct?

**yungman** · September 17th 2010, 05:03 PM

Thanks for the reply. That was not what I am driving at.

I think:

1) should still be $\hat{A}$

2) should be equal to 1!!!!

This is because a unit vector is a unit vector whether the magnitude of the original vector is zero or some function.

I just want to see what other people think.

**Soroban** · September 17th 2010, 06:31 PM

Hello, yungman!

I think your questions are asking for undefined quantities.

$\text{Let }\vec{A}\text{ be a vector with length }|\vec{ A}|$

$\hat{A} \;=\; \dfrac{\vec{A}}{|\vec{ A}|}$

$\text{What is }\dfrac{\vec{A}}{|\vec{ A}|}\text{ if }|\vec{A}| \,=\, 0\;?$

$\hat A \:=\:\dfrac{\vec A}{|\vec A|}\,\text{ is a unit vector in the direction of }\vec A.$

If $|\vec A| = 0$ , the vector has length 0; it is a point, $\bullet$
. . It has no direction.

The unit vector does not exist.

$\text{What is }\left|\dfrac{\vec{A}}{|\vec{ A}|}\right| \text{ if }|\vec{A}| \,=\, 0\:?$

Now you are asking for the magnitude of a unit vector that does not exist.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Having said that, I did a "mind experiment."

We have a vector: . $\vec v \:=\:\langle a,b\rangle$

Its unit vector is:. $\vec u \:=\:\dfrac{\vec v}{|\vec v|} \;=\;\dfrac{\langle a,b\rangle}{\sqrt{a^2+b^2}}$

And you claim that: . $\displaystyle \lim_{a,b\to0} |\vec u| \;=\;\lim_{a,b\to0}\left|\frac{\langle a,b\rangle}{\sqrt{a^2+b^2}}\right| \;=\;1$

That is, that all vectors (however small) have a unit vector of length 1.

I must admit that your claim has merit . . .

**yungman** · September 17th 2010, 07:03 PM

Thanks for you reply. I thought the first part by definition is a unit vector, regardless where the magnitude is zero!! This should be as valid as the second part that the magnitude is 1.

**HallsofIvy** · September 18th 2010, 05:25 AM

No, it is not. If $|\vec{A}|= 0$ then $\frac{\vec{A}}{|\vec{A}|}$ does not exist because you cannot divide by 0. As long as $|\vec{A}|> 0$ , you can think of $\frac{\vec{A}}{|\vec{A}|}$ and "the unit vector in the direction of vector A" but, as Soroban says, the 0 vector has no direction.

**yungman** · September 18th 2010, 02:26 PM

Thanks.

**yungman** · September 18th 2010, 03:06 PM

This is from PDE book by Strauss p194. On the Green's function on a sphere.

The equation is:

$G(\vec{x},\vec{x_0}) = -\frac{1}{4\pi \left\|\vec{x} -\vec{x_0}\right\|} + \frac{1}{4\pi \left\|\frac{r_0}{a} \vec{x} -\frac{a}{r_0}\vec{x_0}\right\|}$

The book gave

$G(\vec{x},0) = -\frac{1}{4\pi \left\|\vec{x} \right\|} + \frac{1}{4\pi a}$

when $\vec{x_0}=0$

This imply my first question is not zero.

**yungman** · September 19th 2010, 09:01 AM

Anyone please?

**yungman** · September 21st 2010, 07:38 AM

Originally Posted by Soroban

Hello, yungman!

I think your questions are asking for undefined quantities.

$\hat A \:=\:\dfrac{\vec A}{|\vec A|}\,\text{ is a unit vector in the direction of }\vec A.$

If $|\vec A| = 0$ , the vector has length 0; it is a point, $\bullet$
. . It has no direction.

The unit vector does not exist.

Now you are asking for the magnitude of a unit vector that does not exist.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Having said that, I did a "mind experiment."

We have a vector: . $\vec v \:=\:\langle a,b\rangle$

Its unit vector is:. $\vec u \:=\:\dfrac{\vec v}{|\vec v|} \;=\;\dfrac{\langle a,b\rangle}{\sqrt{a^2+b^2}}$

And you claim that: . $\displaystyle \lim_{a,b\to0} |\vec u| \;=\;\lim_{a,b\to0}\left|\frac{\langle a,b\rangle}{\sqrt{a^2+b^2}}\right| \;=\;1$

That is, that all vectors (however small) have a unit vector of length 1.

I must admit that your claim has merit . . .

The equation is:

$G(\vec{x},\vec{x_0}) = -\frac{1}{4\pi \left\|\vec{x} -\vec{x_0}\right\|} + \frac{1}{4\pi \left\|\frac{r_0}{a} \vec{x} -\frac{a}{r_0}\vec{x_0}\right\| }$

Where $r_0 = \left\|\vec{x_0} \right\|$

$\frac{1}{4\pi \left\|\frac{r_0}{a} \vec{x} -\frac{a}{r_0}\vec{x_0}\right\|} = \frac{1}{4\pi \left\| \frac{r_0}{a} \vec{x}\right\| - 4\pi \left\|\frac{a}{r_0}\vec{x_0}\right\| }$

$\left\|\frac{a}{r_0}\vec{x_0}\right\| = \left\|a\frac{\vec{x_0}}{r_0}\right\| = a$ as $\left\|\frac{\vec{x_0}}{r_0}\right\| = 1$ even $r_0 = 0$

The book gave

$G(\vec{x},0) = -\frac{1}{4\pi \left\|\vec{x} \right\|} + \frac{1}{4\pi a}$

when $\left\|\vec{x_0}\right\|=0$

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