Answer
(a) Since the length of ${{\bf{e}}_P}$ is $1$, so ${{\bf{e}}_P}$ is a unit vector field.
(b) ${{\bf{e}}_P}\left( {1,1} \right)$ for $P = \left( {a,b} \right) = \left( {3,2} \right)$
${{\bf{e}}_P}\left( {1,1} \right) = \frac{1}{{\sqrt 5 }}\left( { - 2, - 1} \right)$
(c) A potential function for ${{\bf{e}}_P}$ is
$f = \sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} $
Work Step by Step
We have the unit radial vector field based at $P = \left( {a,b} \right)$ given by
${{\bf{e}}_P} = \frac{{\left( {x - a,y - b} \right)}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}$
(a) Evaluate the dot product:
${{\bf{e}}_P}\cdot{{\bf{e}}_P} = ||{{\bf{e}}_P}|{|^2} = \left( {\frac{{\left( {x - a,y - b} \right)}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}} \right)\cdot\left( {\frac{{\left( {x - a,y - b} \right)}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}} \right)$
$||{{\bf{e}}_P}|{|^2} = \frac{{{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}}}{{{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}}} = 1$
$||{{\bf{e}}_P}|| = 1$
Since the length of ${{\bf{e}}_P}$ is $1$, so ${{\bf{e}}_P}$ is a unit vector field.
(b) Evaluate ${{\bf{e}}_P}\left( {1,1} \right)$ for $P = \left( {a,b} \right) = \left( {3,2} \right)$.
${{\bf{e}}_P}\left( {1,1} \right) = \frac{{\left( {1 - 3,1 - 2} \right)}}{{\sqrt {{{\left( {1 - 3} \right)}^2} + {{\left( {1 - 2} \right)}^2}} }} = \frac{1}{{\sqrt 5 }}\left( { - 2, - 1} \right)$
(c) Let $f$ be the potential function for ${{\bf{e}}_P}$. Then
$\nabla f = {{\bf{e}}_P} = \frac{{\left( {x - a,y - b} \right)}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}$
Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right)$, so
$\frac{{\partial f}}{{\partial x}} = \frac{{x - a}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}$, ${\ \ \ \ }$ $\frac{{\partial f}}{{\partial y}} = \frac{{y - b}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}$
Examining these partial derivatives, they suggest that $f = \sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} $.
Since $\nabla f = \frac{{\left( {x - a,y - b} \right)}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}$, we conclude that $f$ is a potential function for ${{\bf{e}}_P}$.
