Solve the vector cross product equation Consider the vector equation $a \times x = b$ in $\Bbb R^3$ , where $a\ne0$. 
Show that $$x = \frac{b\times a}{|a|^2} + ka$$ 
is the solution to the equation for any scalar $k$
i really dont know where to start , i really appreciate any help
 A: Suppose that $a,b$ are linear independent. Let $R$ be a rotation such that $Ra=k_1e_1,Rb=k_2e_2+k_3e_3$, where $e_1,e_2,e_3$ are standard base of $R^3$ and $k_1,k_2$ are scalars and $k_1\neq0,k_2\neq0$. By the property of cross product , one has
$$ R(a\times b)=Ra\times Rb=k_1e_1\times (k_2e_2+k_3e_3)=k_1k_2e_3-k_1k_3e_1$$
and hence
$$ R^{-1}(k_2e_3-k_3e_1)=\frac{1}{k_1}a\times b. $$
From $a\times x=b$, one has
$$ R(a\times x)=Rb.$$
 one obtains
$$ (Ra)\times (Rx)=Rb$$
or
$$ k_1e_1\times (Rx)=k_2e_2. $$
Writing $Rx=c_1e_1+c_2e_2+c_3e_3$, one has
$$ k_1e_1\times (c_1e_1+c_2e_2+c_3e_3)=k_2e_2+k_3e_3$$
or
$$ k_1c_2e_3-k_1c_3e_2=k_2e_2+k_3e_3 $$
So 
$$ c_2=\frac{k_3}{k_1},c_3=-\frac{k_2}{k_1}. $$
and henceforth
$$ x=R^{-1}(c_1e_1+c_2e_2+c_3e_3)=c_1R^{-1}e_1+R^{-1}(c_2e_2+c_3e_3)=c_1a+\frac{1}{k_1^2}a\times b=c_1a+\frac{1}{\|a\|^2}a\times b. $$
A: For any constant $\lambda$, $a \times (x - \lambda a) = a\times x = b$  and hence $x-\lambda a$ also satisfies the given condition. Now,
\begin{align*}
a \times (a \times x) = (a \cdot x ) a - (a \cdot a) x = a \times b
\end{align*}
and hence
\begin{align*}
x = \frac{b \times a}{|a|^2} + \frac{a \cdot x}{|a|^2} a
\end{align*}
For an arbitrary constant $k$ apply the above argument to $x - \left(\frac{a\cdot x - k }{|a|^2} \right)a$
to get $x = \frac{b \times a}{|a|^2} + ka$
A: If you set $$a = \begin{pmatrix} a_1 \\a_2 \\ a_3 \end{pmatrix} \quad b = \begin{pmatrix} b_1 \\b_2 \\ b_3 \end{pmatrix}$$ you can write all the operations on the 3 components, and verify the given relation.
Another way of starting  is to substitute the given $x$ in $a \times x$, and then use the properties of the cross product (linearity etc) to simplify the equation, and see if you get what you want.
Let $x$ be a solution of the equation.
$$a \times x = b \Rightarrow a\cdot(a\times x)= x\cdot (a \times a)=0 = (a\cdot b)$$
In this case, if there is a solution that verifies the relation, necessarly, $a \cdot b = 0$. 
Now, let us assume that $a \cdot b = 0$.
$$a \times (b \times a) = b(a\cdot a) - a (a \cdot b) = b ||a||^2$$
Now, you can continue from here
