Answer
The required vector field is
${\bf{F}}\left( {x,y} \right) = \left( {\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}, - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$
or
${\bf{F}}\left( {x,y} \right) = \left( { - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }},\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$.
Work Step by Step
Write ${\bf{F}}\left( {x,y} \right) = \left( {{F_1},{F_2}} \right)$.
Since $||{\bf{F}}\left( {x,y} \right)|| = 1$, so
$||{\bf{F}}\left( {x,y} \right)|| = \sqrt {\left( {{F_1},{F_2}} \right)\cdot\left( {{F_1},{F_2}} \right)} = {F_1}^2 + {F_2}^2 = 1$
Since ${\bf{F}}\left( {x,y} \right)$ is orthogonal to $G\left( {x,y} \right) = \left( {x,y} \right)$ for all $x$, $y$, so
${\bf{F}}\left( {x,y} \right)\cdot G\left( {x,y} \right) = \left( {{F_1},{F_2}} \right)\cdot\left( {x,y} \right) = x{F_1} + y{F_2} = 0$
From this equation with the assumptions that $x \ne 0$ and $y \ne 0$, we obtain ${F_2} = - \dfrac{x}{y}{F_1}$. Substituting it into ${F_1}^2 + {F_2}^2 = 1$ gives
${F_1}^2 + {\left( { - \dfrac{x}{y}{F_1}} \right)^2} = 1$
${F_1}^2\left( {1 + \dfrac{{{x^2}}}{{{y^2}}}} \right) = 1$
${F_1}^2\left( {\dfrac{{{x^2} + {y^2}}}{{{y^2}}}} \right) = 1$
Therefore, ${F_1} = \pm \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}$.
Substituting ${F_1} = \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}$ into ${F_1}^2 + {F_2}^2 = 1$ gives
$\dfrac{{{y^2}}}{{{x^2} + {y^2}}} + {F_2}^2 = 1$
${F_2}^2 = 1 - \dfrac{{{y^2}}}{{{x^2} + {y^2}}} = \dfrac{{{x^2}}}{{{x^2} + {y^2}}}$
Therefore, ${F_2} = \pm \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}$.
Since ${F_2} = - \dfrac{x}{y}{F_1}$ for all $x$, $y$; we conclude that ${F_1}$ and ${F_2}$ have different signs.
The required vector field is ${\bf{F}}\left( {x,y} \right) = \left( {\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}, - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$ or ${\bf{F}}\left( {x,y} \right) = \left( { - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }},\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$.
