Cross product simplification If you have two vectors, $A$ and $B$, then we can write the cross product as $A \times B$. Simplify the following expressions:


*

*$A \times (A \times (A \times B))$

*$A \times (A \times (A \times (A \times B)))$


I'm not sure how this can be simplified.
 A: *

*$\|A\|^2 A \times B$. 


Follow Benjamin's hint for the second (with "the vector" replaced by "a vector"). 
To explain the answer to 1: $u = A \times B$ is a vector perpendicular to the plane of $A$ and $B$. (If either $A$, $B$, or $A \times B$ is zero, my answer's trivially correct, so I'm assuming they're not). Let $v$ be another vector in that same plane, perpendicular to $u$. 
Then $w = A \times u$ is some multiple of $v$, because 
$w = A \times u$ is perpendicular to $A$ and $u$. 
By the same logic, $A \times w$ is perpendicular to $v$, but lies in the $uv$ plane, so it's a multiple of $u$. So now we know that 
$$
A \times (A \times (A \times B))
$$
is some multiple of $A \times B$. The only question is "What multiple?" 
If $A$ is a unit vector, then it's clear that
$$
A \times (A \times B) = |A| |A \times B| q
$$
where $q$ is a unit vector perpendicular to $A$ and $A \times B$, has the same length as $A\times B$. (Because $|A| = 1$). So for a unit vector, my answer's correct. 
What about if $A = cD$, where $D$ is a unit vector? Then each cross product multiplies by $c = |A|$, so we get a factor of $c^2$. 
More sophisticated answer for this second part: the function $A \mapsto A \times h$ is evidently linear in $A$. 
