$$\vec{a} + \vec{b} \times \vec{x} = \lambda\vec{x}$$
$\vec{a}, \vec{b}$ are known vectors and $\lambda$ is a known scalar. How can i solve that equation ($\times$ between $\vec{b}$ and $\vec{x}$ is a vector product)?

  • 2
    $\begingroup$ If you mean by vector product; cross product so you can set it in the formula by \times. Moreover your vectors have which dimensions? $\endgroup$
    – Mikasa
    Mar 24, 2013 at 12:25
  • $\begingroup$ They are not given. $\endgroup$
    – A6SE
    Mar 24, 2013 at 12:34
  • 4
    $\begingroup$ Cross product is defined only in $3$ dimensions. $\endgroup$
    – Berci
    Mar 24, 2013 at 12:35
  • 1
    $\begingroup$ This is how I tried to do this: $$(\vec{a}+\vec{b}\times\vec{x})*\vec{b}=\lambda\vec{x}*\vec{b}$$ $$\vec{a}*\vec{b}+(\vec{b}\times\vec{x})*\vec{b}=\lambda\vec{x}*\vec{b}$$ $$\vec{a}*\vec{b}+\vec{x}(-\vec{b}\times\vec{b})=\lambda\vec{x}*\vec{b}$$ $$\vec{a}*\vec{b}=\lambda\vec{x}*\vec{b}$$ Now I have no idea what to do next. $\endgroup$
    – A6SE
    Mar 24, 2013 at 12:40
  • $\begingroup$ What happens if we cross product the equation with b? $\endgroup$
    – A6SE
    Mar 24, 2013 at 12:46

2 Answers 2


$\vec{a} + \vec{b} \times \vec{x} = \lambda\vec{x}$

If you write it using the definition of cross product you get:

$a_1\vec{i} + a_2\vec{j} + a_3\vec{k} + (b_2x_3-b_3x_2)\vec{i} - (b_1x_3-b_3x_1)\vec{j}+(b_1x_2-b_2x_1)\vec{k}=\lambda(x_1\vec{i}+x_2\vec{j}+x_3\vec{k})$

Now use the definition of equality of vectors and you`re left with three equations in three unknowns and sure you know how to handle this linear system of equations.


First, assume that $a$ and $b$ are linearly independent (not parallel). Then, $a,\ b,\ a\times b$ is a basis of the space, and write $$x=\alpha a+\beta b+\gamma(a\times b)$$ Then we have to solve $$a-\alpha(a\times b)+\gamma\cdot b\times (a\times b) = \lambda \alpha a+\lambda\beta b+ \lambda\gamma(a\times b) \\ a+\gamma\cdot((b\cdot a)b-(b\cdot b)a)-\alpha(a\times b)= \lambda \alpha a+\lambda\beta b+ \lambda\gamma(a\times b) \\ (1-(b\cdot b))a+(\gamma(b\cdot a))b-\alpha(a\times b)= \lambda \alpha a+\lambda\beta b+ \lambda\gamma(a\times b) $$ By the uniqueness of coordinates, we thus have $1-(b\cdot b)=\lambda\alpha$, $\ \gamma(b\cdot a)=\lambda\beta$ and $\ -\alpha=\lambda\gamma$.

So, providing $\lambda\ne 0$, we have $\alpha=\frac{1-(b\cdot b)}{\lambda}$, $\ \gamma=-\frac{\alpha}{\lambda}\ $ and $\ \beta=\frac{\gamma(b\cdot a)}{\lambda}$.


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