Vectors question 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?


*

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

*if $u$ is a unit vector then $x_5$ is the projection of $x_1$ on a plane perpendicular to $u$.
Thanks.
 A: 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 $u,x_1 \in \mathbb{R}^3$.
A: 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 $x_1$:

