Vector Equation involving cross product I am trying to solve a problem from Vector Analysis, which should be fairly easy, but somehow I can't solve it.
Solve for $X_i$
$$kX_i+\epsilon_{ijk}X_jP_k=Q_i$$
Also I am trying to solve  the following coupled equation:
$$aX_i+\epsilon_{ijk}Y_jP_k=A_i$$
$$bY_i+\epsilon_{ijk}X_jP_k=B_i$$
I have absolutely no clue how to solve these, but I am thinking it can be done by contracting with a Levi-Civita Tensor.
There are more problems, but I think if I can get clues to solve these then I should be able to solve the others as well.
thanks
 A: First part
$$
kX_i+\epsilon_{ijk}X_jP_k=Q_i\tag A
$$
Multiply (A) by $\epsilon_{mil}P_l$ so that
$$
\begin{align}
\epsilon_{mil}P_lQ_i&=k\epsilon_{mil}X_iP_l+\epsilon_{mil}P_l\epsilon_{ijk}X_jP_k\\
&=k\epsilon_{mil}X_iP_l+\epsilon_{mil}\epsilon_{ijk}P_lX_jP_k\\
&=k\epsilon_{mil}X_iP_l+(\delta_{lj}\delta_{mk}-\delta_{lk}\delta_{mj})P_lX_jP_k\\
&=k\underbrace{\epsilon_{mil}X_iP_l}_{Q_m-kX_m\;(\text{from (A)})}+P_jX_jP_m-P_kX_mP_k\\
&=k(Q_m-kX_m)+P_jX_jP_m-P^2X_m\\
\Longrightarrow\qquad -kQ_m-\epsilon_{mli}P_lQ_i&=-(k^2+P^2)X_m+P_jX_jP_m\tag a
\end{align}
$$
Multiplying (A) by $P_i$, we obtain
$$
kX_iP_i+\underbrace{\epsilon_{ijk}X_jP_kP_i }_0=Q_iP_i
$$
that is 
$$
kX_iP_i=Q_iP_i\tag b
$$
and substituting (b) in (a) we find
$$\boxed{
X_m=\frac{1}{k^2+P^2}\left(kQ_m+\frac{P_jQ_j}{k}P_m+\epsilon_{mli}P_lQ_i\right)}\tag B
$$
Second part
$$
\begin{align}
aX_i+\epsilon_{ijk}Y_jP_k=&A_i\tag 1\\
bY_i+\epsilon_{ijk}X_jP_k=&B_i\tag 2
\end{align}
$$
Multiplying $(1)$ by $\epsilon_{lim}P_m$ and following the same method to obtain eq. (a) as in First Part, we find
$$
a\epsilon_{lim}X_iP_m+Y_mP_l-P^2Y_l=\epsilon_{lim}A_iP_m\tag 3
$$
and substituting $\epsilon_{lim}P_m$ from the eq. $(2)$ we obtain
$$
a(B_l-bY_l)+Y_mP_mP_l-P^2Y_l=\epsilon_{lim}A_iP_m\tag 4
$$
Multiplying $(2)$ by $P_i$ we have
$$
bY_iP_i+\underbrace{\epsilon_{ijk}X_jP_kP_i}_0=B_iP_i
$$
that is
$$
bY_iP_i=B_iP_i\tag 5
$$
Substituting $(5)$ in $(4)$
$$
a(B_l-bY_l)+\frac{1}{b}B_mP_mP_l-P^2Y_l=\epsilon_{lim}A_iP_m
$$
and then
$$\boxed{
Y_l=\frac{1}{ab+P^2}\left(aB_l+\frac{1}{b}B_mP_mP_l-\epsilon_{lim}A_iP_m\right)}\tag{$\alpha$}
$$
In the same way we find $X_l$ or by simmetry we can change $a\leftrightarrow b,\,X_i\leftrightarrow Y_i,\,A_i\leftrightarrow B_i,\,$ and find
$$\boxed{
X_l=\frac{1}{ab+P^2}\left(bA_l+\frac{1}{a}A_mP_mP_l-\epsilon_{lim}B_iP_m\right)}\tag{$\beta$}
$$
A: The first expression can be written
$$k\vec X+\vec X\times \vec P=\vec Q \tag 1$$
Expressing $\vec X$ as $\vec X=a\vec P+b\vec Q+c(\vec Q\times \vec P)$ and substituting into $(1)$ yields
$$ak\vec P+bk\vec Q+ck(\vec Q\times \vec P)+b(\vec Q\times \vec P)+c\left((\vec P\cdot \vec Q)\vec P-P^2 \vec Q\right)=\vec Q \tag 2$$
From $(2)$, we find 
$$\begin{align}
k\,a+(\vec P\cdot \vec Q)c&=0\\\\kb-cP^2&=1\\\\ b+k\,c&=0
\end{align}$$
whereupon solving for $a$, $b$ and $c$ reveals
$$\begin{align}
a&=\frac{(1/k)(\vec P\cdot \vec Q)}{P^2+k^2}\\\\b&=\frac{k}{P^2+k^2}\\\\c&=-\frac{1}{P^2+k^2}
\end{align}$$
Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{\vec X=\frac{1}{P^2+k^2}\left(\frac{(\vec P\cdot \vec Q)}{k}\vec P+k\vec Q-(\vec Q\times \vec P) )\right)}$$
A: Well, for the first, if you rewrite in vector notation, you get 
$$
\newcommand{\xb} {\mathbf{x} }
\newcommand{\bb} {\mathbf{b} }
\newcommand{\qb} {\mathbf{q} }
k\xb + \xb \times \bb = \mathbf{q}
$$
If write $x = \alpha \bb + \bb^\perp$, where $\bb^\perp$ is some vector orthogonal to $\bb$, and then take the dot product of both sides with $\bb$, you get
$$
\alpha k \|\bb \|^2 = \qb \cdot \bb
$$
from which you can solve for $\alpha$. 
Then, projecting $\qb$ onto the subspace orthogonal to $\bb$ to get $\qb'$, and simialrly projectin the left hand side onto the subspace, you get
$$
k \bb^\perp + \bb^\perp \times \bb = \qb'.
$$
This is a linear system in the unknown $\bb'$, which you can solve. 
