1. ## Simple question about vector.

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

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

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

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

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

Is this correct?

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

I think:

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

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.

4. Hello, yungman!

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

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

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

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

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

The unit vector does not exist.

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

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: .$\displaystyle \vec v \:=\:\langle a,b\rangle$

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

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

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

5. 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.

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

7. Thanks.

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

The equation is:

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

The book gave

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

when $\displaystyle \vec{x_0}=0$

This imply my first question is not zero.

10. Originally Posted by Soroban
Hello, yungman!

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

If $\displaystyle |\vec A| = 0$, the vector has length 0; it is a point, $\displaystyle \bullet$
. . 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: .$\displaystyle \vec v \:=\:\langle a,b\rangle$

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

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

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

The equation is:

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

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

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

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

The book gave

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

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