Ιf a vector $w$ satisfies an equation : $$ a\times(w\times a)+w=b$$ and $a,b$ are known vectors, then, show that : $$ a\cdot w=a\cdot b$$ and $$ w\times a =\frac{1}{1+|a|^ 2}\cdot(b\times a) $$
Finally, find the solution of the equation.
So, I tried to use the given equation and with numerical operations,reach the form of the equation in the first question but i didn't have any result that helps me find these equations in the first question.I think that i have to use a transformation but i am not sure. I would appreciate any help, hint or thorough solution/discussion to help me get how I work on this problem.
Note that $a \cdot b - a \cdot w = a \cdot (a \times (w \times a))$. This is a scalar triple product between $a$, $a$, and $w \times a$, which is $0$ (the parallelopiped generated by these vectors is degenerate, due to the repeated vector). We also have, \begin{align*} b \times a &= (a \times (w \times a)) \times a + w \times a \\ &= (|a|^2 w - (a \cdot w)a) \times a + w \times a \\ &= (|a|^2 + 1)(w \times a) - (a \cdot w)(a \times a) \\ &= (|a|^2 + 1)(w \times a), \end{align*} as required.