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Kleppner/Kolenkow Vector Problem

  1. Dec 17, 2015 #1
    Let ##\vec {A}## be an arbitrary vector and let ##\hat {n}## be a unit vector in some fixed direction. Show that [tex] \vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}. [/tex]

    Here is my attempt, which I do believe is fine; I just want to know if there is a better way to approach the problem:

    The first term, ##(\vec{A} \cdot \hat {n})\hat{n}## is a new vector ##\vec {C}## with magnitude ##A \cos \theta## in the ##\hat {n}## direction. For the second term, I simply did the operations from left-to-right, first getting ##\vec {C} = ( \hat {n} \times \vec {A})##, which if looking down on the plane made by ##\vec {A}## and ##\hat {n}##, would point upwards out of the page (I drew ##\vec {A} to be above \hat {n}##). Then, I took ##\vec {C} \times \hat {n}## and called this new vector ##\vec {D}## which has a magnitude ##A \sin \theta## and points in a direction ##\hat d##, which is perpendicular to ##\hat {n}##. So, I end up with [tex] \vec {A} = A(\cos \theta \hat {n} + \sin \theta \hat{d}). [/tex]

    So, I've shown that the statement ##\vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}## just breaks ##\vec {A}## up into its horizontal and vertical components. At least, that's what I make of my work. What do you guys think?

    Thanks in advance!
     
  2. jcsd
  3. Dec 17, 2015 #2
    I should say that, while I've heard of the triple product, I intentionally didn't use it because (1) it hasn't been taught yet in my book and (2) I don't truly know what it means. I figured there had to be another way to do the problem, so I sought it out.
     
  4. Dec 18, 2015 #3

    Fredrik

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    Looks good.

    The main alternative is to use the definitions of the cross product and the dot product to show that ##(\hat n\times\vec A)\times\hat n =\vec A-(\vec A\cdot\hat n)\hat n##.
     
  5. Dec 19, 2015 #4
    Sorry, I'm not sure I see the difference between what we are saying.
     
  6. Dec 19, 2015 #5

    Fredrik

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    The definitions I had in mind are these formulas:
    \begin{align*}
    &\vec x\times\vec y =(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1)\\
    &\vec x\cdot\vec y=\sum_{i=1}^3 x_i y_i.
    \end{align*} You didn't use either of these directly. Instead you used two results that can be considered equivalent definitions: If ##\theta## denotes the angle between the vectors, then ##\vec x\cdot\vec y## is equal to ##|\vec x||\vec y|\cos\theta##, and ##\vec x\times\vec y## is the vector of magnitude ##|\vec x||\vec y|\sin\theta## in the direction perpendicular to both ##\vec x## and ##\vec y## identified by the right-hand rule.
     
  7. Dec 19, 2015 #6
    Oh, I see! Sorry, I was thinking of the ##AB \cos \theta## and ##AB \sin \theta## formule.
     
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