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We know that cross product gives a vector that is orthogonal to other two vectors. Let this vector denoted by $$|\vec{v} \times \vec{u}| = \vec{n}$$ Then $$\vec{n}\cdot \vec{u} = 0 $$ Everything okay up to here. Then how we choose a vector from two possible orthogonal vectors, $$\vec{n}$$ or $$\vec{-n}$$ Why following right hand rule?

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  • $\begingroup$ It's just a convention to have things well-defined. There may as well have been a "left hand rule". $\endgroup$ – David Mitra Jul 3 '15 at 12:41
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There is no mathematical way to distinguish between the two directions, they are perfectly equivalent. But in the physical applications, there is a difference. Therefore by convention we use the right hand rule.

Why right and not left could be for any of the following reasons:

  • Historically the right hand was considered "better" than the left
  • In electromagnetism, the verse of the current and the direction of the compass follow the right hand rule (at least considering electrons negative, which is again a convention)
  • It groups the counterclockwise direction (='positive', again by convention) with the "out of the sheet, towards the observer" direction, both appear more comfortable when reading or writing on a table.
  • (added) As Emilio Novati pointed out, it could also be because the rotation of the Earth and the North direction form again a right-hand rule.
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This is just a convention. We could use the opposite convention, and everything would work fine. However, the usual convention has the following nice property. For any two linearly independent vectors $u,v$, if we let $A$ denote the matrix whose columns are $u,v,u\times v$, then $\det A>0$.

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  • $\begingroup$ Actually this is circular, because the $Z$ direction is chosen exactly so that $\vec z = \vec x \times \vec y$ (it uses a cross product!), which implies what you say. $\endgroup$ – geodude Jul 3 '15 at 12:48
  • $\begingroup$ Hate negative things. :) $\endgroup$ – Salihcyilmaz Jul 3 '15 at 12:52
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It is a matter of convention deriving from the convention on orientation of $x,y,z$ axis ( $ \vec i, \vec j, \vec k$) , so that

$\vec k= \vec i \times \vec j= \left |\begin{matrix} \vec i&\vec j&\vec k\\ 1&0&0\\0&1&0 \end{matrix} \right|$

I don't know why the right hand was chosen as positive, but i love to think that this has something to do with the spin of our Earth ( oriented counterclockwise).

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