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 ?
On
In this solution, vectors will be written without arrows. But I will write $\alpha$ for $|a|$ instead to prevent confusion.
First, recall the identity $p\times(q\times r)=q(p\cdot r)-r(p\cdot q)$. When $p=x+y$, $q=x$, and $r=y$, we have $$a\times b=(x+y)\times(x\times y)=x\big((x+y)\cdot y\big)-y\big((x+y)\cdot x\big).$$ So $$a\times b=x(a\cdot y)-y(a\cdot x).\ \ \ \ \ (1)$$ Because $a\cdot x=1$ we have $$\alpha^2=a\cdot a=a\cdot (x+y)=a\cdot x+a\cdot y=1+a\cdot y.$$ Thus $a\cdot y=\alpha^2-1$. Plugging this into (1) we get $$a\times b=x(\alpha^2-1)-y(1)=(\alpha^2-1)x-y.$$ But $x+y=a$, so $$a\times b=(\alpha^2-1)x-(a-x)=\alpha^2x-a.$$ Hence $x=\frac{a+a\times b}{\alpha^2}$, so $y=a-x=\frac{(\alpha^2-1)a-a\times b}{\alpha^2}$.
In general, if $x+y=a$, $x\times y=b$, and $x\cdot c=\gamma$, then $y\cdot c=a\cdot c-\gamma$, so from (1) we have $$a\times b=(a\cdot c-\gamma)x-\gamma y=(a\cdot c)x-\gamma a.$$ Hence $x=\frac{\gamma a+a\times b}{a\cdot c}$ and $y=\frac{(a\cdot c-\gamma )a-a\times b}{a\cdot c}$, provided that $a\cdot c\ne 0$.
On
As $\overrightarrow b = \overrightarrow x \times \overrightarrow y$ is perpendicular to $\overrightarrow a = \overrightarrow x + \overrightarrow y$ consider an orthonormal basis $(\hat a, \hat b, \widehat {a\times b})$. We must express $x, y$ in this basis.
This may be done by solving the system using dot products. For example:
$$a\hat a\cdot (\overrightarrow x+\overrightarrow y) = 1+\overrightarrow y\cdot a\hat a= a^2$$
Can you take it from here?
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Another solution in OP's way
Since $\vec y= \vec a-\vec x$, puu\tting this in the next Eq. we get $ \vec a \times (\vec a- \vec x)=0,$ and hence we have $$\vec x \times \vec a= \vec b ~~~~(1) ~~~~\Rightarrow \vec a . \vec b=0. $$ $\vec a, \vec b, (\vec a \times \vec b)$ are linearly independent vectors which can form a basis so we can say $$\vec x= u \vec a+ v \vec b +w (\vec a \times \vec b)~~~(2)$$ Putting this in (1), we get $$ [ u \vec a+ v \vec b +w (\vec a \times \vec b)] \times a= \vec b \Rightarrow v(\vec b \times \vec a)+ w (\vec a \times \vec b) \times \vec a=\vec b$$ Or $$ -v(\vec a \times b) -w \vec a \times (\vec a \times \vec b)= \vec b.~~~(3)$$ Let us use $\vec a \times (\vec a \times \vec b) =(\vec a.\vec b)\vec a-(\vec a. \vec a) \vec b =-a^2 \vec b$ in (3). Then $$-v (\vec a \times \vec b)+w a^2 \vec b =\vec b ~~~~(4)$$ let us compare the co-efficients of the bases $\vec a, \vec b, (\vec a \times \vec b)$ in both sides of (4). We get $v=0$, $w=\frac{1}{a^2}.$ Soi from Eq. (2), we get $$\vec x = u \vec a +\frac{(\vec a \times \vec b)}{a^2}.$$ But given that $\vec a .\vec x=1$, we get $u=\frac{1}{a^2}$ as $a.(\vec a \times \vec b)=0.$ So finally $$\vec x= \frac{\vec a+(\vec a \times \vec b)}{a^2}.$$
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: $$\vec A \times (\vec B \times \vec C)=(\vec A. \vec C) \vec B- (\vec A. \vec B) \vec C$$