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Thread: Simple question about vector.

  1. #1
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    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?
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  2. #2
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    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?
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  3. #3
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    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.
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  4. #4
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    Hello, yungman!

    I think your questions are asking for undefined quantities.


    $\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.


    I must admit that your claim has merit . . .

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  5. #5
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    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.
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  6. #6
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    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.
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  7. #7
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    Thanks.
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  8. #8
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    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.
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  9. #9
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    Anyone please?
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  10. #10
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    Quote Originally Posted by Soroban View Post
    Hello, yungman!

    I think your questions are asking for undefined quantities.



    $\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.


    I must admit that your claim has merit . . .

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