If $\overline{a}=\overline{i}+\overline{j}+\overline{k}$  ,  $\overline{c}=\overline{j}-\overline{k}$ then  $\overline{b}$   such that $\overline{a}\times \overline{b}=\overline{c}$   and $\overline{a}.\overline{b}=3$   is

Let   $\stackrel{-}{b}=x\stackrel{-}{i}+y\stackrel{-}{J}+z\stackrel{-}{k}$   $\stackrel{-}{a}\times \stackrel{-}{b}=\stackrel{-}{c}$   $\left|\begin{array}{ccc}\stackrel{-}{i}& \stackrel{-}{j}& \stackrel{-}{k}\\ 1& 1& 1\\ x& y& z\end{array}\right|=\stackrel{-}{J}-\stackrel{-}{k}$   $\stackrel{-}{i}(z-y)-\stackrel{-}{J}(z-x)+\stackrel{-}{k}(y-x)=\stackrel{-}{J}-\stackrel{-}{k}$$z-y=0$  $x-z=1,y-x=-1$   $z=yx-y=1$   $x=1+y$   $\stackrel{-}{a}.\stackrel{-}{b}=3$  $x+y+z=3$

$1+y+y+y=3$  $y=\frac{2}{3}z=\frac{2}{3}$   $x=1+\frac{2}{3}=\frac{5}{3}$  $\stackrel{-}{b}=\frac{1}{3}[5\stackrel{-}{i}+2\stackrel{-}{J}+2\stackrel{-}{k}]$