I have a vector equation, given by:
$$ \vec{y}=\vec{y}\times\vec{a}+\vec{b} $$
and I'm trying to solve it for $\vec{y}$. I've tried to take $\times\vec{a}$ of both sides and use vector identities to get me somewhere, but it's really not looking promising and it's got me wondering whether there actually is a way to solve this problem without prior information about $\vec{y}$, $\vec{a}$ and $\vec{b}$


2 Answers 2


Crossing both sides with $\mathbf a$ gives \begin{align*} \mathbf y \times \mathbf a &= (\mathbf y \times \mathbf a) \times \mathbf a + \mathbf b \times \mathbf a \\ \mathbf y - \mathbf b &= (\mathbf y \cdot \mathbf a)\mathbf a - (\mathbf a \cdot \mathbf a)\mathbf y + \mathbf b \times \mathbf a. \end{align*} Dotting both sides with $\mathbf a$ gives $$\mathbf y \cdot \mathbf a = \mathbf b \cdot \mathbf a.$$ Thus, \begin{align*} \mathbf y - \mathbf b &= (\mathbf b \cdot \mathbf a)\mathbf a - (\mathbf a \cdot \mathbf a)\mathbf y + \mathbf b \times \mathbf a \\ (1 + (\mathbf a \cdot \mathbf a))\mathbf y &= (\mathbf b \cdot \mathbf a)\mathbf a + \mathbf b +\mathbf b \times \mathbf a \\ \mathbf y &= \frac{(\mathbf b \cdot \mathbf a)\mathbf a + \mathbf b + \mathbf b \times \mathbf a}{1 + |\mathbf a|^2}. \end{align*}

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    $\begingroup$ You could also dot with $\vec{a}$ to start. Doing the dot and cross products then can be viewed as looking at the parallel and perpendicular components of $\vec{y}$ with respect to $\vec{a}$ which will identify the solution. $\endgroup$ Commented Mar 11 at 15:32

I will drop the vector arrow. We can rewrite the cross product as $$ y \times a=Ay $$ for some $3 \times3$ matrix $A$. Then, we can re-write the equation to $$ y=y \times a + b \iff Iy=Ay+b $$ with $I$ being the $3 \times 3$ identity matrix. Then you can re-write this as a linear system $$ (I-A)y=b $$ and check for its space of solutions.


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