A proof to a vector identity I do not know how to prove this, can anybody help me out with that?
Consider five vectors:
$\vec{a},\vec{b},\vec{c}, \vec{p}, \vec{q} \in \mathbb{R}^3$
then:
$$(\vec{p}\cdot\vec{q})(\vec{a}\cdot(\vec{b}\times\vec{c}))
= (\vec{p}\cdot\vec{a})(\vec{q}\cdot(\vec{b}\times\vec{c}))
+ (\vec{p}\cdot\vec{b})(\vec{q}\cdot(\vec{c}\times\vec{a}))
+ (\vec{p}\cdot\vec{c})(\vec{q}\cdot(\vec{a}\times\vec{b}))$$
 A: Assume $\vec{a},\vec{b},\vec{c}$ are linearly independent. Therefore they form a basis $\mathscr{G}$ for three-dimensional vector space $\mathbb{V}$. Then there is a dual basis $\mathscr{G}'={\vec{a}',\vec{b}',\vec{c}'}$ and 
$$\vec{a}'=\frac{1}{g}\vec{b}\times \vec{c}$$
$$\vec{b}'=\frac{1}{g}\vec{c}\times \vec{a}$$
$$\vec{c}'=\frac{1}{g}\vec{a}\times \vec{b}$$
where
$$g=(\vec{a}\times \vec{b}).\vec{c} = (\vec{b}\times \vec{c}).\vec{a} = (\vec{c}\times \vec{a}).\vec{b}$$
Now find the components of $\vec{q}$ on the basis $\mathscr{G}'$,
$$q_1=\frac{1}{g}\vec{q}.(\vec{b}\times \vec{c})$$
$$q_2=\frac{1}{g}\vec{q}.(\vec{c}\times \vec{a})$$
$$q_3=\frac{1}{g}\vec{q}.(\vec{a}\times \vec{b})$$
And components of $\vec{p}$ on $\mathscr{G}$,
$$p_1=\vec{p}.\vec{a}$$
$$p_2=\vec{p}.\vec{b}$$
$$p_3=\vec{p}.\vec{c}$$
On the other hand, when $\mathscr{G}'$ and $\mathscr{G}$ are dual bases, the Euclidian inner product of $\vec{p}$ and $\vec{q}$ can be written
$$\vec{p}.\vec{q} = \sum_{i=1}^3p_iq_i = \frac{1}{g}(\vec{p}.\vec{a})(\vec{q}.(\vec{b}\times \vec{c})) +\frac{1}{g}(\vec{p}.\vec{b})(\vec{q}.(\vec{c}\times \vec{a}))+\frac{1}{g}(\vec{p}.\vec{c})(\vec{q}.(\vec{a}\times \vec{b}))$$
and finaly by the definition of $g$
$$(\vec{p}.\vec{q})((\vec{a}\times \vec{b}).\vec{c})  = (\vec{p}.\vec{a})(\vec{q}.(\vec{b}\times \vec{c})) + (\vec{p}.\vec{b})(\vec{q}.(\vec{c}\times \vec{a}))+(\vec{p}.\vec{c})(\vec{q}.(\vec{a}\times \vec{b}))$$
A: Given a set of linearly independent base vectors $\{a,b,c\}$ denote their triple product as $\gamma=\det(a,\,b,\,c)$.
Construct a second basis which is orthogonal (but not orthonormal) to the first 
$$\eqalign{
x &= b\times c \cr
y &= c\times a \cr
z &= a\times b \cr
}$$
Now we have the very useful results 
$$\eqalign{
 \gamma &= a^Tx = b^Ty = c^Tz \cr
 \gamma I &= ax^T+by^T+cz^T \cr
}$$
Using the two bases allow us to write the individual terms on the RHS of the problem very compactly.
For example 
$$(p\cdot a)(q\cdot (b\times c))=(p\cdot a)(q\cdot x)=(pq^T):(ax^T)$$
where the colon represents the double-dot product, aka the matrix inner product.
The current problem can be handled easily
$$\eqalign{
\phi
 &= pq^T:(ax^T+by^T+cz^T) \cr
 &= pq^T:\gamma I \cr
 &= \gamma p^Tq \cr
 &= (a^Tx) \,\, (p^Tq) \cr
}$$
A: The calculation is lengthy, so I list steps instead. The identity is linear in $\vec p, \vec q$ respectively,  so it only needs to verify the identity holds for $\vec p, \vec q$ take on the unit coordinate vectors $\vec i, \vec j, \vec k$.
The rest is a length writing, but manageable this way. With the symmetry, it only needs to verify two case: $\vec p = \vec q = \vec i$, and $\vec p = \vec i \ne \vec q = \vec j$.
