Show that $a\times (a\times b)=-(a\bullet a)b$ Let $a$ and $b$ be orthogonal vectors in $\mathbb R^3$
I am trying to show that $a\times (a\times b)=-(a\cdot a)b$ where $\cdot$ is dot product.
We have $a\cdot b=0$. Also I think that $a\times (a\times b)$ is parallel to $b$ then it must be a scalar product of $b$.
Also we know that $a\times b=|a||b|\sin\theta \ n$ where $\theta$ is the angle between $a$ and $b$ and $n$ is a unit normal. So $a\times b=|a||b| n$. Then $a\times (a\times b)=a\cdot a |b|n_2$ where $n_2$ is a unit normal to $a$ and $a\times b$.
I cannot continue to proof. Maybe what I have wrote above are nonsense. I am sorry in advance
Thanks for any help
 A: Your intuition "$a\times(a\times b)$ is parallel to $b$" is good ; for this kind of problems, it helps to :

*

*Draw what's happening ;

*Give a name to the objects we're manipulating ;

*If we see some vectors that are parallel but have a problem of scalar multiplicator in our equation, begin with vectors with norm 1 first.

Let's first prove the result when $a$ and $b$ have norm $1$. Let's note $c= a \times b$ : we then know that $(a,b,c)$ is a direct orthonormal basis and therefore : $a \times b = c$, $b \times c = a$ and $ c \times a = b$. Thus
$$a \times (a\times b) = a \times c= -(c\times a) = -b.$$
Now, if $a$ and $b$ dont have norm $1$ (and are not zero), we can write $a = \|a\|.a'$ and $b=\|b\|.b'$ where $a'$ and $b'$ are vectors with norm 1. Then :
$$\begin{array}{llll}
a \times (a\times b) &=& \|a\|.a' \times (\|a\|.a' \times \|b\|.b') \\
&=& \|a\|^2\|b\| .a' \times (a' \times b') &\text{as $\|a\|$ and $\|b\|$ are numbers}\\ 
&=& \|a\|^2\|b\| .(-b') &\text{from before as $a'$ and $b'$ have norm 1}\\
&=& \|a\|^2 .(-\|b\|.b') \\
&=& \|a\|^2 .(-b) \\
&=& -\|a\|^2 .b \\
&=& -(a \bullet a).b \\
\end{array}$$
