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Problem :

Let $\vec{v}=\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{j} \times \vec{k}))))))))$ Then find the value of $||\vec{v}||$

I am not getting any idea how to proceed in this, please suggest , will be of great help. Thanks.

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  • $\begingroup$ What are $\vec i$, $\vec j$, and $\vec k$? $\endgroup$
    – Lee Mosher
    Dec 28, 2016 at 16:23
  • $\begingroup$ If those are indeed the unit basis vectors, have you tried starting with the case of just taking the cross product of two of them at a time? $\endgroup$
    – J. M. ain't a mathematician
    Dec 28, 2016 at 16:23

4 Answers 4

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Hint:

$$\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{j} \times \vec{k}))))))))$$

$$\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times \vec{i})))))))$$

$$-\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times\vec{k}))))))$$

$$\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times (\vec{k}\times \vec{j})))))$$

$$-\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times (\vec{j}\times \vec{i}))))$$

$$\vec{i}\times (\vec{j}\times (\vec{k}\times (\vec{i}\times \vec{k})))$$

$$-\vec{i}\times (\vec{j}\times (\vec{k}\times \vec{j}))$$

$$\vec{i}\times (\vec{j}\times \vec i)$$

$$-\vec{i}\times \vec k$$

$$\vec{j}$$

You must check all intermediate results.

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  • $\begingroup$ Shameful downvote, this is the only correct answer here. $\endgroup$
    – user65203
    Dec 28, 2016 at 16:44
  • $\begingroup$ OP is asking for $\|v\|$. You might want to take the norm in the very last step? $\endgroup$
    – user9464
    Dec 28, 2016 at 16:59
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    $\begingroup$ @Jack I should flag as a hint as I indeed leave this trivial step to the reader. But those who blindly answered $1$ are right by chance and missed the possibility of a zero. $\endgroup$
    – user65203
    Dec 28, 2016 at 17:01
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Work from inside to out, starting with $j \times k=i,$ at first glance it seems you'll end up at $\pm$ one of $i,j,k$ so norm is $1.$

Added: The quaternions are a division ring, in particular are associative. Since $ijk=ii=-1,$ the first two of $ijkijkijjk$ make $+1,$ and then $ijjk=-ik=j.$

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    $\begingroup$ Could be $0$ as well. $\endgroup$
    – user65203
    Dec 28, 2016 at 16:31
  • $\begingroup$ @YvesDaoust Yes might be zero. But seems not as the things repeat in blocks of threes and the innermost isn't zero. I get $-j$ for this. $\endgroup$
    – coffeemath
    Dec 29, 2016 at 0:33
  • $\begingroup$ A complete answer should warn the OP about this possibility and prove that it doesn't occur. The final vector is $+\vec j$. $\endgroup$
    – user65203
    Dec 29, 2016 at 9:21
  • $\begingroup$ @YvesDaost Note that I worded my answer as "it seems you'll end up at..." making it clear OP should actually do the steps. [and believe it or not I did get $+j$ but miscopied from my paper] $\endgroup$
    – coffeemath
    Dec 29, 2016 at 23:45
  • $\begingroup$ @YvesDaoust Sorry, misspelled your username in above response to your comment, just now doing another so you get notiied. $\endgroup$
    – coffeemath
    Dec 30, 2016 at 0:03
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You just have to be methodical about it.

$$ \begin{align} \vec{v}&=\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times(\vec{j}\times\vec{k}))))))))&&\vec{j}\times\vec{k}=\vec{i}\\&=\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times\vec{i})))))))&&\vec{j}\times\vec{i}=-\vec{k}\\&=\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(-\vec{k})))))))&&\vec{i}\times\left(-\vec{k}\right)=\vec{j}\\&=\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times(\vec{k}\times\vec{j})))))&&\vec{k}\times\vec{j}=-\vec{i}\\&=\vec{i}\times(\vec{j}\times(\vec{k}\times(\vec{i}\times(\vec{j}\times(-\vec{i})))))&&\vec{j}\times\left(-\vec{i}\right)=\vec{k}\\&\vdots&& \end{align} $$

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  • $\begingroup$ From my findings, $-\vec i\times\vec k$. $\endgroup$
    – user65203
    Dec 28, 2016 at 16:43
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As always, the cross product supplies you with a vector that is orthogonal to both vectors, with a magnitude equal to the parallelogram made by them.

Crossing any of the three unit vectors will give a new positive or negative $i, j, k$. As such, no matter how many you cross, you will always have a unit vector. This has a magnitude of $1$.

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    $\begingroup$ "no matter how many you cross": what about $\vec{i}\times (\vec{j} \times \vec{k})$ ? $\endgroup$
    – user65203
    Dec 28, 2016 at 16:30
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    $\begingroup$ Ah, that's an excellent point, should have considered it. It doesn't happen in this question though. $\endgroup$
    – Kaynex
    Dec 28, 2016 at 16:56

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