Q:

If $\vec{x}+\vec{y}=\vec{a},\vec{x}\times\vec{y}=\vec{b}$ and $\vec{x}.\vec{a}=1$ then find $\vec{x},\vec{y}$

If $\vec{x}+\vec{y}=\vec{a},\vec{x}\times\vec{y}=\vec{b}$ and $\vec{x}.\vec{a}=1$ then prove that $\vec{x}=\dfrac{\vec{a}+(\vec{a}\times\vec{b})}{a},\vec{y}=\dfrac{(a^2-1)\vec{a}-(\vec{a}\times\vec{b})}{a^2}$

My Attempt
$\vec{a},\vec{x},\vec{y},\vec{a}\times\vec{b}$ are coplanar,
$$
\vec{x}=t\vec{a}+n(\vec{a}\times\vec{b})\\
\vec{x}.\vec{a}=ta^2=1\implies t=\frac{1}{a^2}\\
\vec{x}=\frac{\vec{a}}{a^2}+n(\vec{a}\times\vec{b})\\
$$
I think I am stuck with the last step ?

A:

We have $\vec y=\vec a-\vec x$ put this in the next eq. to get $\vec x \times (\vec a- \vec x)=\vec b$, then we get $\vec x \times \vec a=\vec b.$ Next take the cross prodi\uct of both sides with $\vec a$ from left: $$\vec a \times (\vec x \times \vec a) = \vec a \times \vec b  \Rightarrow (\vec a. \vec a)\vec x-(\vec a .\vec x)\vec a 
\Rightarrow a^2\vec x- \vec a=\vec a \times \vec b \Rightarrow \vec x =\frac{\vec a \times \vec b+\vec a}{a^2}.$$ Then you get $\vec y=\vec a -\vec x$. Here we have used the vector triple product as: $$\vec A \times (\vec B \times \vec C)=(\vec A. \vec C) \vec B- (\vec A. \vec B) \vec C$$

