Vector Equation involving cross product

Question

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

Answer 1

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 $$\boxed{ X_l=\frac{1}{ab+P^2}\left(bA_l+\frac{1}{a}A_mP_mP_l-\epsilon_{lim}B_iP_m\right)}\tag{$\beta$} $$

Answer 2

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

$$\bbox[5px,border:2px solid #C0A000]{\vec X=\frac{1}{P^2+k^2}\left(\frac{(\vec P\cdot \vec Q)}{k}\vec P+k\vec Q-(\vec Q\times \vec P) )\right)}$$

Answer 3

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.