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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 $u$.


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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 $x_n$ ($n\geq 2$)? – Daan Michiels Apr 26 '12 at 16:38
We know that the length of $x_{n+1}$ equals $|x_{n}|*|u|sin(a)$. Can you please explain more? – Guy Apr 26 '12 at 17:15
up vote 3 down vote accepted

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. – Shuhao Cao Apr 26 '12 at 20:10
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? – Guy Apr 27 '12 at 14:29
@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 $\|x_2\|$. – Shuhao Cao Apr 27 '12 at 17:44
Thanks so much! – Guy Apr 28 '12 at 11:22
Thanks again. Last thing: if ||u||>1 and $||x_1||>1$, would $||x_n||->\infty$? – Guy Apr 28 '12 at 11:27

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... – matgaio Apr 26 '12 at 17:43
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. – Shuhao Cao Apr 26 '12 at 20:13
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 $\{(x,y,z): (x,y,z)\cdot u = 0 \}$ – Shuhao Cao Apr 26 '12 at 20:23
Hmmm it seems that someone is discounting their frustrations :) – user27456 Apr 26 '12 at 21:36
@matgaio Thanks for the understanding my friend! :) – Shuhao Cao Apr 27 '12 at 0:31

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