Solving vector equation 3

Question

Solve for $\bar{x}$ and $\bar{y}$
$$\bar{x}+\bar{y}=\bar{a},~~ \bar{x}\times \bar{y}=\bar{b},~~ \bar{x}.\bar{a}=1$$

Attempt:

$\bar{x}+\bar{y}=\bar{a}$ dot by $\bar{a}$, we get $1+\bar{a}.\bar{y}=|\bar{a}|^2$ i.e $\bar{a}.\bar{y}=|\bar{a}|^2-1$

$\bar{x}\times \bar{y}=\bar{b}$ pre-cross by $\bar{a}$, we get $$(\bar{a}.\bar{y})\bar{x}-(\bar{a}.\bar{x})\bar{y}=\bar{a}\times \bar{b}\implies (|\bar{a}|^2-1)\bar{x}-\bar{y}=\bar{a}\times \bar{b}$$

I need another such equation to solve $\bar{x}$ and $\bar{y}$. Please help.

Answer 1

Eliminate $y$ by substituting the first equation into the second $$\eqalign{ y &= a-x \cr b &= x\times y &= x\times(a-x) &= x\times a \cr\cr }$$ Now we need to isolate $x$, so let's cross the last equation by $a$ $$\eqalign{ a\times b &= a\times(x\times a) \cr &= a^2x-(a\cdot x)a \cr &= a^2x-a \cr }$$ Now you can solve for $x$ entirely in terms of $a$ and $b$.

Then you can substitute that result to solve for $y$.

Answer 2

You found one equation and use second equation which is given $i.e.$

$x+y=a$ . Use these two equations to find $x$ and $y$

Hope this will be helpful !