A proof to a vector identity

Question

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:

$$(\vec{p}\cdot\vec{q})(\vec{a}\cdot(\vec{b}\times\vec{c})) = (\vec{p}\cdot\vec{a})(\vec{q}\cdot(\vec{b}\times\vec{c})) + (\vec{p}\cdot\vec{b})(\vec{q}\cdot(\vec{c}\times\vec{a})) + (\vec{p}\cdot\vec{c})(\vec{q}\cdot(\vec{a}\times\vec{b}))$$

Answer 1

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 $g$

$$(\vec{p}.\vec{q})((\vec{a}\times \vec{b}).\vec{c}) = (\vec{p}.\vec{a})(\vec{q}.(\vec{b}\times \vec{c})) + (\vec{p}.\vec{b})(\vec{q}.(\vec{c}\times \vec{a}))+(\vec{p}.\vec{c})(\vec{q}.(\vec{a}\times \vec{b}))$$

Answer 2

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 $\vec p = \vec i \ne \vec q = \vec j$.

Answer 3

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
$$\eqalign{ \phi &= pq^T:(ax^T+by^T+cz^T) \cr &= pq^T:\gamma I \cr &= \gamma p^Tq \cr &= (a^Tx) \,\, (p^Tq) \cr }$$