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I'm trying to prove whether the followings statements are true or not. I would appreciate your help, as I'm not sure how to begin.

Given: $ u,x_n \in \mathbb{R}^3$ and for every $n$, let $x_{n+1}=u \times x_n$.

Are the following statements true or not?

  1. if $u$ is a unit vector, then for every $n > 1$, $x_{n+2} = -x_n$

  2. if $u$ is a unit vector then $x_5$ is the projection of $x_1$ on a plane perpendicular to $u$.

Thanks.

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    Have you tried a few examples? What can you say about the angle between $x_n$ and $u$ for $n\geq 2$? What can you say about the length of $x_{n+1}$ versus that of $x_n$ ($n\geq 2$)?2012-04-26
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    We know that the length of $x_{n+1}$ equals $|x_{n}|*|u|sin(a)$. Can you please explain more?2012-04-26

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Assign $u$ to be the unit normal vector to some plane $S$ in $\mathbb{R}^3$, then $u\times x_n \in T(S)$ for any $n$, which means $u\times x_n $ is a vector field on the plane $S$. After the first cross product, all $x_n$'s ($n > 1$) lie on the plane $S$, because $x_{n}\cdot u = (u\times x_{n-1})\cdot u = 0$ for all $n> 1$.

Now $x_{n+2} = u\times(u\times x_n)$, for $n>1$ where $x_n$ already is a vector field on this plane, geometrically speaking, doing cross product twice is the same as rotating $x_n$ twice counter-clockwisely with respect $u$ by $\pi/2$, and you get $-x_n$.

Proof-wise, use the cross product definition $u\times x_n = |u||x_n|\sin(\theta) \boldsymbol{\nu}_n$, where $\theta$ is the angle between $u$ and $x_n$, for $n>1$, this is $\pi/2$, because $x_{n}\cdot u =0$ for all $n> 1$ which is proved above, and dot product of two vectors is the cosine of the angle between them if normalized by their norms.

$\boldsymbol{\nu}_n$ is the unit vector perpendicular to the plane containing $u$ and $x_n$ directioning by right-hand rule, recalling that $x_n$ is already a vector field on the plane $S$ that has unit normal $u$, $\boldsymbol{\nu}_n$ is in the direction of the counter-clockwise $\pi/2$-rotation of $x_n$ on this plane. Hence $u\times x_n = |x_n| \boldsymbol{\nu}_n$.

Next we have $u\times(u\times x_n) = |x_n| u\times\boldsymbol{\nu}_n = |x_n|\,|u|\, |\boldsymbol{\nu}_n|\,\sin(\theta)\boldsymbol{\nu}_{n+1}$, $\theta$ is again $\pi/2$ due to the same reason that the dot product is zero, and $\boldsymbol{\nu}_{n+1}$ is now the unit vector perpendicular to the plane containing $u$ and $u\times x_n$ directioning by right-hand rule, repeat the right hand rule argument above, we know that $\boldsymbol{\nu}_{n+1}$ is in the direction of $x_n$'s counter-clockwise $\pi$-rotation with respect to $u$, therefore $u\times(u\times x_n) = |x_n|\boldsymbol{\nu}_{n+1} = -x_n$.

Remark: In vector calculus, the projection of any pointwisely well-defined vector field $\boldsymbol{v}$ onto a smooth surface $S$ is defined as $\boldsymbol{n}\times(\boldsymbol{v}\times\boldsymbol{n})$, where $\boldsymbol{n}$ is the outer unit normal of this surface, now if $x_n$ is already a vector field on $S$, its projection is itself.

For the second statement, apply the first $x_5 = -x_3 = u\times(x_1\times u)$, either repeat the argument above, or directly use above remark.

Hence both statements are true, given that $u,x_1 \in \mathbb{R}^3$.

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    Very funny a correct answer got downvoted.2012-04-26
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    Thanks, looks very convincing :) Can i ask you two more things regarding this question? 1) if $||u|| < 1$ then does $x_n$ converges? 2) if $||u|| <= 1$ then does $||x_n||$ converges?2012-04-27
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    @Guy 1) Yes, since $\|x_n\|\to 0$ (need to be proved) . 2)If you mean "$\|x_n\|$ converges to zero", then no, as you can see the case above when $\|u\|=1$, for $n\geq 3$, $x_n$ are all rotations of $x_2$ on the plane $S$, and have the same non-zero length. But $\|x_n\|$ does converge to $\|x_2\|$.2012-04-27
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    Thanks again. Last thing: if ||u||>1 and $||x_1||>1$, would $||x_n||->\infty$?2012-04-28
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    @Guy If $u$ is not parallel to $x_1$ then yes, but if $x_1$ is parallel to $u$, ie $x_1\perp S$, then $x_2$ is the zero vector and henceforth.2012-04-28
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I've edited this answer because I've given an incorrect asnwer. Thanks to the precise observations of @Jon. I would like to keep this here because it shows why we are taking the hypotesis $n>1$. The statement is false for $n\geqslant 1$. Take for example $u=(1,1,0)$ and $x_1=(0,1,0)$. The cross product $x_2=u\times x_1$ will belong to the orthogonal complement of the plane $xy$. But when you take the cross product $x_3=u\times x_2$, it will belong to the plane ortoghonal to the subspace generated by $u$ and $x_2$, to which $x_1$ is not orthogonal, so there is no chance of being true $x_3=-x_1$

I think you could try this: the cross product of $u=(a,b,c)$ by a vector $v=(x,y,z)$ can be viewed as the following matricial product:

$$u\times v = \left(\begin{array}{ccc} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{array}\right)\cdot \left( \begin{array}{c} x \\ y \\ z \end{array} \right)=\left( \begin{array}{c} ay-bz \\ -ax+cz \\ bx-cy \end{array} \right)$$

and with this it could be easier to prove properties about the cross product.

I'm thinking about this to give my first example: if we take $u$ an unitary vector and $x_1$, the cross product $x_2$ will lie in the orthogonal complement of the plane of the screen, in my drawing below. So, when we take the cross product of $u$ and $x_2$, it will lie in the black line, so it could not be parallel to $x_1$:

enter image description here

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    Perhaps it could be some incorrect signs of $+$ and $-$ in this expression...2012-04-26
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    Read the problem, $x_{n+2} = -x_n$ only holds for $n>1$, $x_3$ is the exact opposite direction of the projection vector of $x_1$ onto the plane which has $u$ as a normal vector, and $u$ should be a unit vector otherwise the scale wouldn't work as projections.2012-04-26
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    and based on your "counter-example", if you normalize $u$ to be the unit vector: $u = (1/\sqrt{2},1/\sqrt{2},0)$, then you will find that $x_4 = -x_2$, and $x_5$ is the projection of $x_1$ onto the plane $\{(x,y,z): (x,y,z)\cdot u = 0 \}$2012-04-26
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    Hmmm it seems that someone is discounting their frustrations :)2012-04-26
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    @Jon, my example is not a counterexample. I only would like to demand comments in a little more polite way. When someone say something between quotes, it only means you are somehow making joke with other's thoughts. It would be sufficient to say what was incorrect with my example and everyone would be fine.2012-04-26
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    And before you think I voted down in your answer, I voted up, because it is correct. I don't know why or who voted down and I think it is wrong.2012-04-26
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    @matgaio Please pardon my rude comments just now, I was kinda infuriated by the downvote of my answer and got triggered :'( .2012-04-26
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    That's ok. We are here for a common purpose, wich is knowing a little bit more of mathematics and that's what we need to try to do. That's all ok.2012-04-26
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    @matgaio An now I am totally calmed down, hope my rudeness doesn't make you unhappy, have a nice day!2012-04-26
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    That's fine. We all do some strange kind of things when strange thing s happens to us. It happens with all of us all the time.2012-04-26
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    @matgaio Thanks for the understanding my friend! :)2012-04-27