Answer
1. For the vector field: ${{\bf{e}}_r}$
${{\bf{e}}_r}\left( P \right) = \frac{1}{{\sqrt 2 }}\left( {1,1,0} \right)$
${{\bf{e}}_r}\left( Q \right) = \frac{1}{3}\left( {2,1,2} \right)$
2. For the vector field: $\frac{{{{\bf{e}}_r}}}{r}$
$\frac{{{{\bf{e}}_r}}}{r}\left( P \right) = \frac{1}{2}\left( {1,1,0} \right)$
$\frac{{{{\bf{e}}_r}}}{r}\left( Q \right) = \frac{1}{9}\left( {2,1,2} \right)$
3. For the vector field: $\frac{{{{\bf{e}}_r}}}{{{r^2}}}$
$\frac{{{{\bf{e}}_r}}}{{{r^2}}}\left( P \right) = \frac{1}{{2\sqrt 2 }}\left( {1,1,0} \right)$
$\frac{{{{\bf{e}}_r}}}{{{r^2}}}\left( Q \right) = \frac{1}{{27}}\left( {2,1,2} \right)$
Work Step by Step
We have $P = \left( {1,1,0} \right)$ and $Q = \left( {2,1,2} \right)$.
1. For the vector field: ${{\bf{e}}_r}$
We have ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right)$. Using $r = \sqrt {{x^2} + {y^2} + {z^2}} $, we get
${{\bf{e}}_r} = \frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\left( {x,y,z} \right)$
So,
${{\bf{e}}_r}\left( P \right) = \frac{1}{{\sqrt {{1^2} + {1^2} + 0} }}\left( {1,1,0} \right) = \frac{1}{{\sqrt 2 }}\left( {1,1,0} \right)$
${{\bf{e}}_r}\left( Q \right) = \frac{1}{{\sqrt {{2^2} + {1^2} + {2^2}} }}\left( {2,1,2} \right) = \frac{1}{3}\left( {2,1,2} \right)$
2. For the vector field: $\frac{{{{\bf{e}}_r}}}{r}$
We have ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right)$. So, $\frac{{{{\bf{e}}_r}}}{r} = \left( {\frac{x}{{{r^2}}},\frac{y}{{{r^2}}},\frac{z}{{{r^2}}}} \right)$. Using ${r^2} = {x^2} + {y^2} + {z^2}$, we get
$\frac{{{{\bf{e}}_r}}}{r} = \frac{1}{{{x^2} + {y^2} + {z^2}}}\left( {x,y,z} \right)$
So,
$\frac{{{{\bf{e}}_r}}}{r}\left( P \right) = \frac{1}{{{1^2} + {1^2} + 0}}\left( {1,1,0} \right) = \frac{1}{2}\left( {1,1,0} \right)$
$\frac{{{{\bf{e}}_r}}}{r}\left( Q \right) = \frac{1}{{{2^2} + {1^2} + {2^2}}}\left( {2,1,2} \right) = \frac{1}{9}\left( {2,1,2} \right)$
3. For the vector field: $\frac{{{{\bf{e}}_r}}}{{{r^2}}}$
We have ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right)$. So, $\frac{{{{\bf{e}}_r}}}{{{r^2}}} = \left( {\frac{x}{{{r^3}}},\frac{y}{{{r^3}}},\frac{z}{{{r^3}}}} \right)$. Using ${r^2} = {x^2} + {y^2} + {z^2}$, we get
$\frac{{{{\bf{e}}_r}}}{{{r^2}}} = \frac{1}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}\left( {x,y,z} \right)$
So,
$\frac{{{{\bf{e}}_r}}}{{{r^2}}}\left( P \right) = \frac{1}{{{{\left( {{1^2} + {1^2} + 0} \right)}^{3/2}}}}\left( {1,1,0} \right) = \frac{1}{{2\sqrt 2 }}\left( {1,1,0} \right)$
$\frac{{{{\bf{e}}_r}}}{{{r^2}}}\left( Q \right) = \frac{1}{{{{\left( {{2^2} + {1^2} + {2^2}} \right)}^{3/2}}}}\left( {2,1,2} \right) = \frac{1}{{27}}\left( {2,1,2} \right)$
