# How to solve the equation and directly determine the value of vec(v) in the equation? vec(v)-a[vec(v) xx vec(y)]=b vec(E) where, vec(v) and vec(E) are in the hat(x) direction, and a and b are scalars.

How to solve the equation and directly determine the value of $\stackrel{\to }{v}$ in the equation?
$\stackrel{\to }{v}-a\left[\stackrel{\to }{v}×\stackrel{^}{y}\right]=b\stackrel{\to }{E}$
where, $\stackrel{\to }{v}$ and $\stackrel{\to }{E}$ are in the $\stackrel{^}{x}$ direction, and a and b are scalars.
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Miguel Shah
You can convert $\stackrel{\to }{c}=\stackrel{\to }{a}×\stackrel{\to }{b}$ into a matrix-vector product with the following trick (it is called the cross product operator matrix).
$\begin{array}{rl}\stackrel{\to }{c}& =\left[\stackrel{\to }{a}×\right]\stackrel{\to }{b}\\ \left[\stackrel{\to }{a}×\right]& =\left(\begin{array}{ccc}0& -{a}_{z}& {a}_{y}\\ {a}_{z}& 0& -{a}_{x}\\ -{a}_{y}& {a}_{x}& 0\end{array}\right)\end{array}$
So the LHS of the equation above is
$\stackrel{\to }{v}-a\left(\stackrel{\to }{v}×\stackrel{^}{y}\right)=\stackrel{\to }{v}+a\left(\stackrel{^}{y}×\stackrel{\to }{v}\right)=\left(\mathbf{1}+a\left[\stackrel{^}{y}×\right]\right)\stackrel{\to }{v}$
where 1 is 3×3 the identity matrix, and $\left[\stackrel{^}{y}×\right]$ the 3×3 cross product operator.
This makes an equation like v⃗ −a(v⃗ ×y^)=bE⃗ $\stackrel{\to }{v}-a\left(\stackrel{\to }{v}×\stackrel{^}{y}\right)=b\stackrel{\to }{E}$ solvable
$\overline{)\stackrel{\to }{v}={\left(\mathbf{1}+a\left[\stackrel{^}{y}×\right]\right)}^{-1}b\stackrel{\to }{E}}$
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Hana Buck
Given that $\stackrel{\to }{v}$ is in the x-direction, $\left(\begin{array}{c}{v}_{x}\\ 0\\ 0\end{array}\right)×\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ {v}_{x}\end{array}\right)=\stackrel{\to }{c}$ . To find $\stackrel{\to }{v}$, solve for the components algebraically and separately. In this case, that’s just ${v}_{x}$; if $\stackrel{\to }{v}$ is in the x-direction, then${v}_{y}={v}_{z}=0$
$\begin{array}{rl}{v}_{x}-a{c}_{x}& =b{E}_{x}\\ {v}_{x}& =b{E}_{x}+a{c}_{x}\\ & =b{E}_{x}+0\end{array}$