Graph the curve $C$ traced out by the vector function $$\begin{equation*} \mathbf{r}(t)=(2+3t)\mathbf{i}+(3-t)\mathbf{j}+2t\mathbf{k} \end{equation*}$$

Figure 5 $\mathbf{r}(t)=(2+3t)\mathbf{i}+(3-t)\mathbf{j}+2t\mathbf{k}$

Solution The components of $\mathbf{r=r}(t)$ are the parametric equations $$\begin{equation*} x(t)=2+3t\qquad y(t)=3-t\qquad z(t)=2t \end{equation*}$$

These are the parametric equations of a line in space containing the point $(2,3,0)$ (corresponding to $t=0$) and in the direction of the vector $3\mathbf{i}-\mathbf{j}+2\mathbf{k}$. Solving for $t$, we obtain the symmetric equations of this line $$\begin{equation*} \frac{x-2}{3}=\frac{y-3}{-1}=\frac{z}{2} \end{equation*}$$

Since $\mathbf{r}(0)=2\mathbf{i}\,{+}\,3\mathbf{j}$ and $\mathbf{r}(1)=5\mathbf{i}\,{+}\,2\mathbf{j}\,{+}\,2\mathbf{k}$, the positive direction of $C$ is given by $\mathbf{r}(1)-\mathbf{r}(0)=3\mathbf{i}-\mathbf{j}\,{+}\,2\mathbf{k}$. See Figure 5.