Answer
${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_3}\left( t \right)$ have the same projection onto the $xy$-plane.
Work Step by Step
The projection of a space curve onto the $xy$-plane is obtained by setting the $z$-component equal to zero. Thus, the projections of these space curves onto the $xy$-plane are listed in the following table:
$\begin{array}{*{20}{c}}
{Curve}&{Projection {\ } onto {\ } xy - plane}\\
{{{\bf{r}}_1}\left( t \right) = \left( {t,{t^2},{{\rm{e}}^t}} \right)}&{\left( {t,{t^2},0} \right)}\\
{{{\bf{r}}_2}\left( t \right) = \left( {{{\rm{e}}^t},{t^2},t} \right)}&{\left( {{{\rm{e}}^t},{t^2},0} \right)}\\
{{{\bf{r}}_3}\left( t \right) = \left( {t,{t^2},\cos t} \right)}&{\left( {t,{t^2},0} \right)}
\end{array}$
From the table above, we see that ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_3}\left( t \right)$ have the same projection onto the $xy$-plane.
