Decomposition of a vector on basis $\vec{q}\times \vec{r}, \vec{p}\times \vec{r},\vec{p}\times \vec{q}$

Question

How to prove that $\begin{bmatrix} \vec{p} &\vec{q} &\vec{r} \end{bmatrix}\vec{a}=\left ( \vec{q}\times \vec{r} \right )\left ( \vec{p}.\vec{a} \right )+\left ( \vec{p}\times \vec{r} \right )\left ( \vec{q}.\vec{a} \right )+\left ( \vec{p}\times \vec{q} \right )\left ( \vec{r}.\vec{a} \right )$?

Answer 1

Here is a derivation. See conventions in a remark below.

The result to be established can be expressed under the form :

$$\tau a = (pu^T+qv^T+rw^T)^T a\tag{*}$$

with notations :

$$\tau:=\det(p,q,r)\tag{1}$$

$$\left\{\eqalign{ u &:= q\times r \cr v &:= r\times p \cr w &:= p\times q \cr }\right.\tag{2}$$

It is easy to see that :

$$\begin{pmatrix} - \ u^T \ - \\ \hline - \ v^T \ - \\ \hline - \ w^T \ - \end{pmatrix}\left( \begin{array}{c|c|c}&&\\p&q&r\\&&\end{array}\right)=\tau I \tag{3}$$

As the inverse is unique, we have therefore as well :

$$\tau I =\left( \begin{array}{c|c|c}&&\\p&q&r\\&&\end{array}\right)\begin{pmatrix} - \ u^T \ - \\ \hline - \ v^T \ - \\ \hline - \ w^T \ - \end{pmatrix}= pu^T+qv^T+rw^T\tag{4}$$

(3) and (4) with words : the matrix whose columns are the components of cross products $u,v,w$ is the adjugate of the matrix with column vectors $p,q,r$.

It suffices now to apply the two extreme expressions of (4) to (column) vector $a$, to obtain (*).

Remark : Classically, a vector is assimilated to a column vector ; besides, notation $v^T$ means the (T)ranspose of a column vector $v$, i.e., a row vector. With these notations, for example $v^Tu$ represents a number, dot product of vectors $u$ and $v$, whereas $uv^T$ represents a $3 \times 3$ matrix.