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A very simple question in vector

  1. Sep 17, 2010 #1
    Let [itex] \vec{A} [/itex] be a vector with length [itex] |\vec{ A}| [/itex]

    [tex] \hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|} [/tex]

    1) What is [tex] \frac{\vec{A}}{|\vec{ A}|}[/tex] If [itex]|\vec{A}| [/itex] = 0?

    2) What is [tex] |\frac{\vec{A}}{|\vec{ A}|}| [/tex] If [itex]|\vec{A}| [/itex] = 0?

    My answer for 1) is [itex] \hat{A}[/itex] and 2) equal to 1

    Please tell me why?
    Last edited: Sep 17, 2010
  2. jcsd
  3. Sep 18, 2010 #2


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    Well, the only vector whose length is zero, is the zero vector ([tex]\vec 0[/tex]).
    And the one vector that you can't stretch or shrink to unit length, is the unit vector.
    So IMO, [tex]\hat 0[/tex] 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?
  4. Sep 18, 2010 #3
    This is from PDE book by Strauss p194. On the Green's function on a sphere.

    The equation is:

    [tex] 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\|}[/tex]

    The book gave

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

    when [itex]\vec{x_0}=0[/itex]

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

    Please help.
  5. Sep 18, 2010 #4


    Staff: Mentor

    I don't see what the problem is. G(x, 0) doesn't involve [itex]x_0[/itex], so it doesn't matter that [itex]x_0[/itex] happens to be zero.
  6. Sep 18, 2010 #5
    But you can clearly see the first term of the denominator equal zero and the second term is a. This means:

    [tex] |\frac{\vec{x_0}}{r_0}|=1[/tex]

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


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    Science Advisor

    But that is NOT in what you quoted orginally. And, surely, where it the text says
    [tex] |\frac{\vec{x_0}}{r_0}|=1[/tex]
    it also has a provision for the case that [itex]|\vec{x_0}|\ne 0[/itex].
  8. Sep 19, 2010 #7
    The text book did not say anything, but if you look at the formula given by the book, [itex] |\frac{\vec{x_0}}{r_0}|=1[/itex] is the only way for the forumlas to be true. That is where I am confused. Here is what the book gave:

    If [itex]x_0\neq 0[/itex]:

    [tex] 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\|}[/tex]

    The book gave when [itex]\vec{x_0}=0[/itex]

    [tex]G(\vec{x},0) = -\frac{1}{4\pi \left\|\vec{x} \right\|} + \frac{1}{4\pi a} [/tex]
    Last edited: Sep 19, 2010
  9. Sep 19, 2010 #8
    Anyone please? This is on p194 of the PDE book by Strauss equation (11).

  10. Sep 20, 2010 #9
    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 [tex] \hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|} [/tex] 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 [tex]\hat{A} \;=\; \frac{\vec{A}}{|\vec{ A}|}[/tex] 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.
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