The following problem was the last one, and the hardest, on a high school test I had recently. I wasn’t be able to solve it on time and now I am trying to figure out its solution.
Could anyone help me? maybe with a hint or even a full solution?
Here is the problem:
Let $\vec{a},\vec{b}$ be two space vectors, with $\vec{a},\vec{b} \neq \vec{\mathbb{0}}$ solve the equation: $$2\vec{x}+\langle \vec{a},\vec{x} \rangle \vec{a}\times \vec{b}=\vec{b} $$
Thank you in advance.
On
Following the general philosophy of "dot with everything in sight", $$ 2x + \langle a,x \rangle (a \times b) = b \\ 2\langle a,x \rangle = \langle a, b \rangle \\ 2 \langle b,x \rangle = \langle b,b \rangle $$ If we set $x=\frac{1}{2}b+y$, clearly the latter two equations imply that $\langle a,y\rangle = \langle b,y\rangle=0$. Then the original equation becomes $$ b+2y + (\tfrac{1}{2}\langle a,b \rangle + \langle a,y \rangle) (a \times b) = b, $$ or $$ y = -\tfrac{1}{4}\langle a,b \rangle (a \times b). $$ Therefore $$ x=\tfrac{1}{2}b-\tfrac{1}{4}\langle a,b \rangle (a \times b) $$ is the unique solution.
Hint: Multiply both sides with the vector $\vec a$. You get $$2\langle \vec a,\vec x\rangle +\langle \vec a,\vec x\rangle \langle\vec a\times \vec b,\vec a\rangle =\langle\vec a,\vec b\rangle$$ $$\langle \vec a,\vec x\rangle=\frac{1}{2}\langle\vec a,\vec b\rangle$$ Then multiply by $\vec b$
$$\langle \vec b,\vec x\rangle =\frac{1}{2}\vec b^2$$