# A very simple question in vector

1. Sep 17, 2010

### yungman

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

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

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

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

My answer for 1) is $\hat{A}$ and 2) equal to 1

Last edited: Sep 17, 2010
2. Sep 18, 2010

### CompuChip

Well, the only vector whose length is zero, is the zero vector ($$\vec 0$$).
And the one vector that you can't stretch or shrink to unit length, is the unit vector.
So IMO, $$\hat 0$$ is not defined and the answers to both your questions are basically the same as the answer to "what is x / 0?".

By the way, how do you mean "please tell me why." Should we tell you why you answered A and 1?

3. Sep 18, 2010

### yungman

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 $\vec{x_0}=0$

If what you said is true, this will not be correct.

4. Sep 18, 2010

### Staff: Mentor

I don't see what the problem is. G(x, 0) doesn't involve $x_0$, so it doesn't matter that $x_0$ happens to be zero.

5. Sep 18, 2010

### yungman

But you can clearly see the first term of the denominator equal zero and the second term is a. This means:

$$|\frac{\vec{x_0}}{r_0}|=1$$

$$r_0=|\vec{x_0}|$$

Last edited: Sep 18, 2010
6. Sep 19, 2010

### HallsofIvy

Staff Emeritus
But that is NOT in what you quoted orginally. And, surely, where it the text says
$$|\frac{\vec{x_0}}{r_0}|=1$$
it also has a provision for the case that $|\vec{x_0}|\ne 0$.

7. Sep 19, 2010

### yungman

The text book did not say anything, but if you look at the formula given by the book, $|\frac{\vec{x_0}}{r_0}|=1$ is the only way for the forumlas to be true. That is where I am confused. Here is what the book gave:

If $x_0\neq 0$:

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

The book gave when $\vec{x_0}=0$

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

Last edited: Sep 19, 2010
8. Sep 19, 2010

### yungman

Anyone please? This is on p194 of the PDE book by Strauss equation (11).
Thanks

Alan

9. Sep 20, 2010

### Tac-Tics

I don't know the book you're talking about, and it really doesn't matter for your question.

Division by zero, even when we're talking about vectors, is undefined.

If you arrive at a statement $$\hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|}$$ and |A| = 0 in your calculations, you have made a mistake earlier.

It could very well be that the mistake is in your textbook.

Resolving this invalid division is simply a matter of restricting the domain appropriately. Many authors will take this for granted and perform operations which are invalid at a handful of points.

For example, if I have two functions f and g, and f(x) g(x) = 1, I CANNOT say that f(x) = 1/g(x), because g(x) might evaluate to zero at some places. More correctly, I must say f(x) = 1/g(x) for all x where g(x) /= 0.... but that's a lot to write out, and often clutters the more important point being made by the author.

Note that $$\hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|}$$ is how you "normalize" a vector (find a vector in the same direction, but whose length is 1). The zero vector is the only vector that cannot be normalized, because it doesn't have a direction.