How to solve the equation and directly determine the value of $\vec{v}$ in the equation? $$\vec{v} - a[\vec{v}{\times}\hat{y}] = b\vec{E}$$ where, $\vec{v}$ and $\vec{E}$ are in the $\hat{x}$ direction, and $a$ and $b$ are scalars.
edit: edited the vector
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Given that $\vec v$ is in the $x$-direction, $\pmatrix{v_x\\0\\0}\times\pmatrix{0\\1\\0}=\pmatrix{0\\0\\v_x}=\vec c$. To find $\vec v$, solve for the components algebraically and separately. In this case, that’s just $v_x$; if $\vec v$ is in the $x$-direction, then $v_y=v_z=0$.
$$\begin{align} v_x-ac_x &= bE_x \\ v_x &= bE_x +ac_x \\ &= bE_x +0 \\ \end{align}$$
You can convert $\vec{c} = \vec{a}\times \vec{b}$ into a matrix-vector product with the following trick (it is called the cross product operator matrix).
$$ \begin{aligned} \vec{c} & = [\vec{a}\times] \vec{b} \\ [\vec{a}\times] & = \pmatrix{0 & -a_z & a_y\\ a_z & 0 & -a_x \\ -a_y & a_x & 0} \end{aligned} $$
So the LHS of the equation above is
$$ \vec{v} -a (\vec{v} \times \hat{y}) = \vec{v} + a ( \hat{y} \times \vec{v} ) = \left( \mathbf{1}+ a [ \hat{y}\times] \right) \vec{v} $$
where $\mathbf{1}$ is 3×3 the identity matrix, and $[\hat{y}\times]$ the 3×3 cross product operator.
This makes an equation like $\vec{v} -a (\vec{v} \times \hat{y}) = b \vec{E}$ solvable
$$ \boxed{ \vec{v} = \left( \mathbf{1} + a [ \hat{y}\times] \right)^{-1} b \vec{E} }$$