## Calculus (3rd Edition)

$v$, $v_1$ and $v_3$ are equivalent.
We have $$v=\overrightarrow{P_0Q_0}=Q_0-P_0=(0,1,-4)-(1,-2,5)=\langle -1,3,-9\rangle,$$ $$v_1=\overrightarrow{P_0Q_0}=Q_0-P_0=(0,5,-5)-(1,2,4)=\langle -1,3,-9\rangle,$$ $$v_2=\overrightarrow{P_0Q_0}=Q_0-P_0=(0,-8,13)-(1,5,4)=\langle -1,-13,-9\rangle,$$ $$v_3=\overrightarrow{P_0Q_0}=Q_0-P_0=(-1,3,-9)-(0,0,0)=\langle -1,3,-9 \rangle,$$ $$v_4=\overrightarrow{P_0Q_0}=Q_0-P_0=(1,7,4)-(2,4,5)=\langle -1,3,-1 \rangle.$$ Hence, $v$, $v_1$ and $v_3$ are equivalent.