Find $\vec{x},\vec{y},\vec{z}$ in terms of $\vec{a},\vec{b}$ and $\vec{x}.\vec{b}$ If $\vec{x}\times\vec{y}=\vec{a},\quad \vec{y}\times\vec{z}=\vec{b},\quad \vec{x}\cdot \vec{b}=\gamma,\quad \vec{x}\cdot \vec{y}=1$ and $\vec{y}\cdot\vec{z}=1$ then find $\vec{x},\vec{y},\vec{z}$ in terms of $\vec{a},\vec{b}$ and $\gamma$

I tried many things,but could not crack this problem.Please help me in solving this problem.Thanks.
 A: The first two statements mean that $y$ is perpendicular to both $a$ and $b$. Therefore 


*

*$y=\lambda(a\times b)$


We also have:


*$x \cdot a=y \cdot a= y\cdot b= z\cdot b= 0$


Now using the vector triple product formula, $$x\times y=x\times \lambda(a\times b)=\lambda[a(x\cdot b)-b(x \cdot a)]$$
$$\Rightarrow a=\lambda a \gamma\Rightarrow \lambda=\frac {1}{\gamma}$$
Hence, so far, we have $$y=\frac{1}{\gamma}(a\times b)$$
Now consider $$y \times a=y\times(x\times y)=x(y\cdot y)-y(x\cdot y)=x|y|^2-y$$
Similarly, $$y\times b=y-z|y|^2$$
Adding these gives $$y\times(a+b)=(x-z)|y|^2$$
Hence $$x-z=\gamma \frac{(a\times b)\times(a+b)}{|a\times b|^2}$$
If instead we subtract, we get $$y\times (a-b)=(x+z)|y|^2-2y$$
This leads to$$x+z=\gamma \frac{(a\times b)\times(a-b)+2(a\times b)}{|a\times b|^2}$$
Finally you can get expressions for $x$ and $z$ by solving these, but I will leave that to you.
A: The following is an incomplete (geometric) solution to the problem.
$(E)$ If $\vec{A}\times\vec{X}=\vec{B}$ and $\vec{A}\cdot\vec{B}=0$ then $\vec{X}=\frac{\vec{B}\times\vec{A}}{||\vec{A}||^2}+\lambda\vec{A}$ for any scalar $\lambda$.$(F)$ If $\vec{A}\cdot\vec{X}=b$ then $\vec{X}=\frac{b\vec{A}}{||\vec{A}||^2}+\overrightarrow{W}$ where $\overrightarrow{W}$ is an arbitrary vector orthogonal to $\vec{A}$.
$\vec{x}\cdot\vec{y}=\vec{z}\cdot\vec{y}=1$ implies that $\vec{y}\cdot\left(\vec{x}-\vec{z}\right)=0$.$\vec{x}\times\vec{y}=\vec{a}$ and $\vec{y}\times\vec{z}=\vec{b}$ implies that $\left(\vec{x}-\vec{z}\right)\times\vec{y}=\vec{a}+\vec{b}$. From $(E)$, we deduce that $$\vec{y}=\frac{\left(\vec{a}+\vec{b}\right)\times\left(\vec{x}-\vec{z}\right)}{||\vec{x}-\vec{z}||^2}+\lambda\left(\vec{x}-\vec{z}\right)$$ From $\vec{y}\cdot\left(\vec{x}-\vec{z}\right)=0$ we deduce that $\lambda=0$ so we have $$\frac{\vec{y}}{||\vec{y}||^2}=\frac{\vec{a}+\vec{b}}{||\vec{a}+\vec{b}||^2}\times\left(\vec{x}-\vec{z}\right)$$
