Answer
The head of vector ${\bf{v}}'$ is $\left( {3,5} \right)$.
The head of vector ${{\bf{v}}_0}$ is $\left( {1,1} \right)$.
Work Step by Step
We have ${\bf{v}} = \overrightarrow {PQ} = \left( {2,2} \right) - \left( {1,1} \right) = \left( {1,1} \right)$.
${\bf{v}}'$ is based at $\left( {2,4} \right)$ and let its head be the point $\left( {a,b} \right)$. It is equivalent to ${\bf{v}}$ if $\left( {a,b} \right) - \left( {2,4} \right) = \left( {1,1} \right)$. So, the head of vector ${\bf{v}}'$ is $\left( {a,b} \right) = \left( {3,5} \right)$.
Since ${{\bf{v}}_0}$ is based at $\left( {0,0} \right)$. Let its head be the point $\left( {c,d} \right)$. It is equivalent to ${\bf{v}}$ if $\left( {c,d} \right) - \left( {0,0} \right) = \left( {1,1} \right)$. So, the head of vector ${{\bf{v}}_0}$ is $\left( {c,d} \right) = \left( {1,1} \right)$.
